RS Aggarwal Solutions for Class 12 Chapter 05 Matrices

Question 1. If \( A = \begin{bmatrix} 5 & -2 & 6 & 1 \\ 7 & 0 & 8 & -3 \\ \sqrt{2} & \frac{3}{5} & 4 & 3 \end{bmatrix} \) then write
(i) the number of rows in A,
(ii) the number of columns in A,
(iii) the order of the matrix A,
(iv) the number of all entries in A,
(v) the elements \( a_{23}, a_{31}, a_{14}, a_{33}, a_{22} \) of A.
Answer:
(i) The matrix contains 3 rows.
(ii) The matrix contains 4 columns.
(iii) The order of the matrix is the product of the number of rows and the number of columns, which gives \( 3 \times 4 \).
(iv) The total number of entries equals the product of the number of rows and the number of columns: \( 3 \times 4 = 12 \).
(v) \( a_{ij} \) represents the element in the \( i^{th} \) row and \( j^{th} \) column.
\( a_{23} = 8 \)
\( a_{31} = \sqrt{2} \)
\( a_{14} = 1 \)
\( a_{33} = 4 \)
\( a_{22} = 0 \)

Question 2. Write the order of each of the following matrices:
(i) \( A = \begin{bmatrix} 3 & 5 & 4 & -2 \\ 0 & \sqrt{3} & -1 & \frac{4}{9} \end{bmatrix} \)
(ii) \( B = \begin{bmatrix} 6 & -5 \\ \frac{1}{2} & \frac{3}{4} \\ -2 & -1 \end{bmatrix} \)
(iii) \( C = \begin{bmatrix} 7 - \sqrt{2} & 5 & 0 \end{bmatrix} \)
(iv) \( D = \begin{bmatrix} 8 & -3 \end{bmatrix} \)
(v) \( E = \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} \)
(vi) \( F = \begin{bmatrix} 6 \end{bmatrix} \)
Answer:
(i) The order equals the number of rows multiplied by the number of columns: \( 2 \times 4 \).
(ii) The order is found by multiplying the number of rows by the number of columns: \( 4 \times 2 \).
(iii) The order is the product of rows and columns: \( 1 \times 4 \).
(iv) For matrix D, the order is: \( 1 \times 2 \).
(v) The order equals the number of rows times the number of columns: \( 3 \times 1 \).
(vi) For matrix F, the order is: \( 1 \times 1 \).

Question 3. If a matrix has 18 elements, what are the possible orders it can have?
Answer: The total number of entries equals the product of rows and columns, which is 18. If the order is \( a \times b \), then \( a \times b = 18 \). The factor pairs of 18 are found by identifying all divisors. The possible orders are \( 1 \times 18, 2 \times 9, 3 \times 6, 6 \times 3, 9 \times 2, 18 \times 1 \).

Question 4. Find all possible orders of matrices having 7 elements.
Answer: The total number of entries equals the product of rows and columns, which gives 7. If the order is \( a \times b \), then \( a \times b = 7 \). Since 7 is a prime number, its only factor pairs are found by dividing 7 by its divisors. The possible orders are \( 1 \times 7 \) and \( 7 \times 1 \).

Question 5. Construct a \( 3 \times 2 \) matrix whose elements are given by \( a_{ij} = (2i - j) \).
Answer: We are given the formula \( a_{ij} = 2i - j \). We calculate each element by substituting the appropriate row and column indices.
\( a_{11} = 2(1) - 1 = 1 \)
\( a_{12} = 2(1) - 2 = 0 \)
\( a_{21} = 2(2) - 1 = 3 \)
\( a_{22} = 2(2) - 2 = 2 \)
\( a_{31} = 2(3) - 1 = 5 \)
\( a_{32} = 2(3) - 2 = 4 \)
Therefore, the required matrix is: \( A = \begin{bmatrix} 1 & 0 \\ 3 & 2 \\ 5 & 4 \end{bmatrix} \)

Question 6. Construct a \( 4 \times 3 \) matrix whose elements are given by \( a_{ij} = \frac{i}{j} \).
Answer: We use the formula \( a_{ij} = \frac{i}{j} \) to find each element. For a \( 4 \times 3 \) matrix, we have 4 rows and 3 columns.
\( a_{11} = 1, \quad a_{12} = \frac{1}{2}, \quad a_{13} = \frac{1}{3} \)
\( a_{21} = 2, \quad a_{22} = 1, \quad a_{23} = \frac{2}{3} \)
\( a_{31} = 3, \quad a_{32} = \frac{3}{2}, \quad a_{33} = 1 \)
\( a_{41} = 4, \quad a_{42} = 2, \quad a_{43} = \frac{4}{3} \)
Therefore, the required matrix is: \( A = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ 2 & 1 & \frac{2}{3} \\ 3 & \frac{3}{2} & 1 \\ 4 & 2 & \frac{4}{3} \end{bmatrix} \)

Question 7. Construct a \( 2 \times 2 \) matrix whose elements are \( a_{ij} = \frac{(i + 2j)^2}{2} \).
Answer: We apply the formula \( a_{ij} = \frac{(i + 2j)^2}{2} \) to compute each element. For a \( 2 \times 2 \) matrix:
\( a_{11} = \frac{(1 + 2)^2}{2} = \frac{9}{2} \)
\( a_{12} = \frac{(1 + 4)^2}{2} = \frac{25}{2} \)
\( a_{21} = \frac{(2 + 2)^2}{2} = 8 \)
\( a_{22} = \frac{(2 + 4)^2}{2} = 18 \)
Therefore, the required matrix is: \( A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix} \)

Question 8. Construct a \( 2 \times 3 \) matrix whose elements are \( a_{ij} = \frac{(i - 2j)^2}{2} \).
Answer: We use the formula \( a_{ij} = \frac{(i - 2j)^2}{2} \). For a \( 2 \times 3 \) matrix:
\( a_{11} = \frac{(1 - 2)^2}{2} = \frac{1}{2} \)
\( a_{12} = \frac{(1 - 4)^2}{2} = \frac{9}{2} \)
\( a_{13} = \frac{(1 - 6)^2}{2} = \frac{25}{2} \)
\( a_{21} = \frac{(2 - 2)^2}{2} = 0 \)
\( a_{22} = \frac{(2 - 4)^2}{2} = 2 \)
\( a_{23} = \frac{(2 - 6)^2}{2} = 8 \)
Therefore, the required matrix is: \( A = \begin{bmatrix} \frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\ 0 & 2 & 8 \end{bmatrix} \)

Question 9. Construct a \( 3 \times 4 \) matrix whose elements are given by \( a_{ij} = \frac{1}{2}|-3i + j| \).
Answer: We apply the formula \( a_{ij} = \frac{1}{2}|-3i + j| \). For a \( 3 \times 4 \) matrix, we compute each element:
\( a_{11} = \frac{1}{2}|-3 + 1| = 1 \)
\( a_{12} = \frac{1}{2}|-3 + 2| = \frac{1}{2} \)
\( a_{13} = \frac{1}{2}|-3 + 3| = 0 \)
\( a_{14} = \frac{1}{2}|-3 + 4| = \frac{1}{2} \)
\( a_{21} = \frac{1}{2}|-6 + 1| = \frac{5}{2} \)
\( a_{22} = \frac{1}{2}|-6 + 2| = 2 \)
\( a_{23} = \frac{1}{2}|-6 + 3| = \frac{3}{2} \)
\( a_{24} = \frac{1}{2}|-6 + 4| = 1 \)
\( a_{31} = \frac{1}{2}|-9 + 1| = 4 \)
\( a_{32} = \frac{1}{2}|-9 + 2| = \frac{7}{2} \)
\( a_{33} = \frac{1}{2}|-9 + 3| = 3 \)
\( a_{34} = \frac{1}{2}|-9 + 4| = \frac{5}{2} \)
Therefore, the required matrix is: \( A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix} \)

Question 1. If \( A = \begin{bmatrix} 2 & -3 & 5 \\ -1 & 0 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 2 & -2 \\ 4 & -3 & 1 \end{bmatrix} \), verify that \( (A + B) = (B + A) \).
Answer: We first compute \( A + B \) by adding corresponding elements of the two matrices. Each element in the sum is obtained by adding the elements from the same position in A and B:
\( A + B = \begin{bmatrix} 2 + 3 & -3 + 2 & 5 - 2 \\ -1 + 4 & 0 - 3 & 3 + 1 \end{bmatrix} = \begin{bmatrix} 5 & -1 & 3 \\ 3 & -3 & 4 \end{bmatrix} \)
Next, we compute \( B + A \) by adding the elements in reverse order:
\( B + A = \begin{bmatrix} 3 + 2 & 2 - 3 & -2 + 5 \\ 4 - 1 & -3 + 0 & 1 + 3 \end{bmatrix} = \begin{bmatrix} 5 & -1 & 3 \\ 3 & -3 & 4 \end{bmatrix} \)
Since both results are identical, we have demonstrated that \( A + B = B + A \). This property, known as the commutative property of matrix addition, holds for all matrices of compatible dimensions.

Question 2. If \( A = \begin{bmatrix} 3 & 5 \\ -2 & 0 \\ 6 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & -3 \\ 4 & 2 \\ -2 & 3 \end{bmatrix} \) and \( C = \begin{bmatrix} 0 & 2 \\ 3 & -4 \\ 1 & 6 \end{bmatrix} \), verify that \( (A + B) + C = A + (B + C) \).
Answer: We compute the left side \( (A + B) + C \) first. Adding matrices A and B element-wise:
\( A + B = \begin{bmatrix} 2 & 2 \\ 2 & 2 \\ 4 & 2 \end{bmatrix} \)
Then adding matrix C to this result:
\( (A + B) + C = \begin{bmatrix} 2 & 4 \\ 5 & -2 \\ 5 & 8 \end{bmatrix} \)
Now we compute the right side \( A + (B + C) \). First, we add matrices B and C:
\( B + C = \begin{bmatrix} -1 & -1 \\ 7 & -2 \\ -1 & 9 \end{bmatrix} \)
Then adding this to matrix A:
\( A + (B + C) = \begin{bmatrix} 2 & 4 \\ 5 & -2 \\ 5 & 8 \end{bmatrix} \)
Both sides yield the same result, confirming that \( (A + B) + C = A + (B + C) \). This demonstrates the associative property of matrix addition.

Question 3. If \( A = \begin{bmatrix} 3 & 1 & 2 \\ 1 & 2 & -3 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 0 & 4 \\ 5 & -3 & 2 \end{bmatrix} \), find \( (2A - B) \).
Answer: We first multiply matrix A by the scalar 2, which means multiplying each element by 2:
\( 2A = \begin{bmatrix} 6 & 2 & 4 \\ 2 & 4 & -6 \end{bmatrix} \)
Now we subtract matrix B from \( 2A \) by subtracting corresponding elements:
\( 2A - B = \begin{bmatrix} 6 - (-2) & 2 - 0 & 4 - 4 \\ 2 - 5 & 4 - (-3) & -6 - 2 \end{bmatrix} = \begin{bmatrix} 8 & 2 & 0 \\ -3 & 7 & -8 \end{bmatrix} \)

Question 4. Let \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \). Find:
(i) \( A + 2B \)
(ii) \( B - 4C \)
(iii) \( A - 2B + 3C \)
Answer:
(i) We multiply matrix B by 2, then add it to A:
\( 2B = \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix} \)
\( A + 2B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix} = \begin{bmatrix} 4 & 10 \\ -1 & 12 \end{bmatrix} \)
(ii) We multiply matrix C by 4, then subtract from B:
\( 4C = \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \)
\( B - 4C = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} - \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} = \begin{bmatrix} 9 & -17 \\ -14 & -11 \end{bmatrix} \)
(iii) We compute each term: \( A, 2B, \) and \( 3C \), then combine them:
\( A - 2B + 3C = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix} + \begin{bmatrix} -6 & 15 \\ 9 & 12 \end{bmatrix} = \begin{bmatrix} -6 & 13 \\ 16 & 4 \end{bmatrix} \)

Question 5. Let \( A = \begin{bmatrix} 0 & 1 & -2 \\ 5 & -1 & -4 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -3 & -1 \\ 0 & -2 & 5 \end{bmatrix} \) and \( C = \begin{bmatrix} 2 & -5 & 1 \\ -4 & 0 & 6 \end{bmatrix} \). Compute \( 5A - 3B + 4C \).
Answer: We compute each scalar multiple separately, then combine them. First, multiply A by 5:
\( 5A = \begin{bmatrix} 0 & 5 & -10 \\ 25 & -5 & -20 \end{bmatrix} \)
Next, multiply B by 3:
\( 3B = \begin{bmatrix} 3 & -9 & -3 \\ 0 & -6 & 15 \end{bmatrix} \)
Then, multiply C by 4:
\( 4C = \begin{bmatrix} 8 & -20 & 4 \\ -16 & 0 & 24 \end{bmatrix} \)
Now combine: \( 5A - 3B + 4C = \begin{bmatrix} 0 & 5 & -10 \\ 25 & -5 & -20 \end{bmatrix} - \begin{bmatrix} 3 & -9 & -3 \\ 0 & -6 & 15 \end{bmatrix} + \begin{bmatrix} 8 & -20 & 4 \\ -16 & 0 & 24 \end{bmatrix} = \begin{bmatrix} 5 & -6 & -3 \\ 9 & 1 & -11 \end{bmatrix} \)

Question 6. If \( 5A = \begin{bmatrix} 5 & 10 & -15 \\ 2 & 3 & 4 \\ 1 & 0 & -5 \end{bmatrix} \), find A.
Answer: To find A, we divide each element of the given matrix by 5. This is equivalent to multiplying the matrix by \( \frac{1}{5} \):
\( A = \frac{1}{5} \begin{bmatrix} 5 & 10 & -15 \\ 2 & 3 & 4 \\ 1 & 0 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 2 & -3 \\ \frac{2}{5} & \frac{3}{5} & \frac{4}{5} \\ \frac{1}{5} & 0 & -1 \end{bmatrix} \)

Question 7. Find matrices A and B, if \( A + B = \begin{bmatrix} 1 & 0 & 2 \\ 5 & 4 & -6 \\ 7 & 3 & 8 \end{bmatrix} \) and \( A - B = \begin{bmatrix} -5 & -4 & 8 \\ 11 & 2 & 0 \\ -1 & 7 & 4 \end{bmatrix} \).
Answer: To find both matrices, we add the two equations together to isolate A. Adding the corresponding elements of the two given equations:
\( (A + B) + (A - B) = 2A = \begin{bmatrix} -4 & -4 & 10 \\ 16 & 6 & -6 \\ 6 & 10 & 12 \end{bmatrix} \)
Dividing by 2:
\( A = \begin{bmatrix} -2 & -2 & 5 \\ 8 & 3 & -3 \\ 3 & 5 & 6 \end{bmatrix} \)
Now we find B by subtracting the second equation from the first:
\( (A + B) - (A - B) = 2B = \begin{bmatrix} 6 & 4 & -6 \\ -6 & 2 & -6 \\ 8 & -4 & 4 \end{bmatrix} \)
Dividing by 2:
\( B = \begin{bmatrix} 3 & 2 & -3 \\ -3 & 1 & -3 \\ 4 & -2 & 2 \end{bmatrix} \)

Question 8. Find matrices A and B, if \( 2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} \) and \( 2B + A = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} \).
Answer: To solve for the two unknowns, we manipulate both equations. We multiply the first equation by 2 and then add it to the second equation:
\( 2(2A - B) + (2B + A) = 2 \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} \)
\( 5A = \begin{bmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} = \begin{bmatrix} 15 & -10 & 5 \\ -10 & 5 & -5 \end{bmatrix} \)
Dividing by 5:
\( A = \begin{bmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{bmatrix} \)
To find B, we substitute A into the second equation:
\( 2B = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} - A = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{bmatrix} - \begin{bmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 4 & 4 \\ 0 & 0 & -6 \end{bmatrix} \)
Dividing by 2:
\( B = \begin{bmatrix} 0 & 2 & 2 \\ 0 & 0 & -3 \end{bmatrix} \)

Question 9. Find matrix X, if \( \begin{bmatrix} 3 & 5 & -9 \\ -1 & 4 & -7 \end{bmatrix} + X = \begin{bmatrix} 6 & 2 & 3 \\ 4 & 8 & 6 \end{bmatrix} \).
Answer: To isolate X, we subtract the first matrix from both sides of the equation:
\( X = \begin{bmatrix} 6 & 2 & 3 \\ 4 & 8 & 6 \end{bmatrix} - \begin{bmatrix} 3 & 5 & -9 \\ -1 & 4 & -7 \end{bmatrix} = \begin{bmatrix} 3 & -3 & 12 \\ 5 & 4 & 13 \end{bmatrix} \)

Question 10. If \( A = \begin{bmatrix} -2 & 3 \\ 4 & 5 \\ 1 & -6 \end{bmatrix} \) and \( B = \begin{bmatrix} 5 & 2 \\ -7 & 3 \\ 6 & 4 \end{bmatrix} \), find a matrix C such that \( A + B - C = O \).
Answer: From the equation \( A + B - C = O \), we can rearrange to get \( C = A + B \). Adding matrices A and B element-wise:
\( C = \begin{bmatrix} -2 & 3 \\ 4 & 5 \\ 1 & -6 \end{bmatrix} + \begin{bmatrix} 5 & 2 \\ -7 & 3 \\ 6 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ -3 & 8 \\ 7 & -2 \end{bmatrix} \)

Question 11. Find the matrix X such that \( 2A - B + X = O \), where \( A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 1 \\ 0 & 3 \end{bmatrix} \).
Answer: From the equation \( 2A - B + X = O \), we rearrange to get \( X = B - 2A \). First, we compute \( 2A \):
\( 2A = \begin{bmatrix} 6 & 2 \\ 0 & 4 \end{bmatrix} \)
Then we find X by subtracting \( 2A \) from B:
\( X = \begin{bmatrix} -2 & 1 \\ 0 & 3 \end{bmatrix} - \begin{bmatrix} 6 & 2 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -8 & -1 \\ 0 & -1 \end{bmatrix} \)

Question 12. If \( A = \begin{bmatrix} 1 & -3 & 2 \\ 2 & 0 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \end{bmatrix} \), find a matrix C such that \( (A + B + C) \) is a zero matrix.
Answer: For the sum to be a zero matrix, we need \( A + B + C = O \), which means \( C = -(A + B) \). First, we add matrices A and B:
\( A + B = \begin{bmatrix} 1 & -3 & 2 \\ 2 & 0 & 2 \end{bmatrix} + \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 1 \\ 3 & 0 & 1 \end{bmatrix} \)
Now we negate this result to find C:
\( C = -\begin{bmatrix} 3 & -4 & 1 \\ 3 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 4 & -1 \\ -3 & 0 & -1 \end{bmatrix} \)

Question 13. If \( A = \text{diag}[2, -5, 9] \), \( B = \text{diag}[-3, 7, 14] \) and \( C = \text{diag}[4, -6, 3] \), find:
(i) \( A + 2B \)
(ii) \( B + C - A \)
(iii) \( 2A + B - 5C \)
Answer:
(i) For diagonal matrices, scalar multiples and sums can be computed by operating on the diagonal entries directly. We multiply the diagonal entries of B by 2, then add them to those of A:
\( A + 2B = \text{diag}[2 + 2(-3), -5 + 2(7), 9 + 2(14)] = \text{diag}[2 - 6, -5 + 14, 9 + 28] = \text{diag}[-4, 9, 37] \)
(ii) We perform the operations on the diagonal entries:
\( B + C - A = \text{diag}[-3 + 4 - 2, 7 - 6 - (-5), 14 + 3 - 9] = \text{diag}[-1, 6, 8] \)
(iii) We compute each term on the diagonal:
\( 2A + B - 5C = \text{diag}[2(2) - 3 - 5(4), 2(-5) + 7 - 5(-6), 2(9) + 14 - 5(3)] = \text{diag}[4 - 3 - 20, -10 + 7 + 30, 18 + 14 - 15] = \text{diag}[-19, 27, 17] \)

Question 14. Find the value of x and y, when:
(i) \( \begin{bmatrix} x + y \\ x - y \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 2x + 5 & 7 \\ 0 & 3y - 7 \end{bmatrix} = \begin{bmatrix} x - 3 & 7 \\ 0 & -5 \end{bmatrix} \)
(iii) \( 2 \begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
Answer:
(i) Two matrices are equal when their corresponding elements are identical. This gives us the system: \( x + y = 8 \) and \( x - y = 4 \). Adding these equations gives \( 2x = 12 \), so \( x = 6 \). Substituting back: \( y = 2 \).
(ii) Comparing corresponding elements: \( 2x + 5 = x - 3 \) gives \( x = -8 \), and \( 3y - 7 = -5 \) gives \( 3y = 2 \), so \( y = \frac{2}{3} \).
(iii) We first perform the scalar multiplication and addition on the left side. Multiplying the first matrix by 2 and adding the second matrix gives \( \begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} \). Equating to the right side matrix: \( 2x + 3 = 7 \) gives \( x = 2 \), and \( 2y - 4 = 14 \) gives \( y = 9 \).

Question 15. Find the value of \( (x + y) \) from the following equation:
\( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
Answer: We first perform the scalar multiplication and then add the matrices. Multiplying the first matrix by 2 gives \( \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} \). Adding the second matrix gives \( \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} \). Comparing with the right side: the element in position (1,1) gives \( 2 + y = 5 \), so \( y = 3 \). The element in position (2,2) gives \( 2x + 2 = 8 \), so \( x = 3 \). Therefore, \( x + y = 6 \).

Question 16. If \( \begin{bmatrix} x - y & 2y \\ 2y + z & x + y \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix} \), then write the value of \( (x + y) \).
Answer: We equate corresponding elements of the two matrices. From the matrix equality, we obtain four equations: \( x - y = 1 \), \( 2y = 4 \), \( 2y + z = 9 \), and \( x + y = 5 \). From the second equation, \( y = 2 \). From the fourth equation, \( x + y = 5 \). We can verify consistency: from the first equation, \( x = 1 + y = 3 \), and from \( x + y = 5 \), we get \( 3 + 2 = 5 \), which is correct. Therefore, \( x + y = 5 \).

Question 1A. Compute AB and BA, whichever exists when \( A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \\ 0 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 1 \end{bmatrix} \).
Answer: Matrix A is of order \( 3 \times 2 \) and matrix B is of order \( 2 \times 3 \). For the product AB to exist, the number of columns in A must equal the number of rows in B. Since A has 2 columns and B has 2 rows, AB exists and will have order \( 3 \times 3 \). For the product BA, the number of columns in B must equal the number of rows in A. Since B has 3 columns and A has 3 rows, BA also exists and will have order \( 2 \times 2 \).

For matrix AB, we compute each element as the dot product of a row from A with a column from B:
\( AB = \begin{bmatrix} 1(0) + 2(2) & 1(1) + 2(-1) & 1(-1) + 2(1) \\ -1(0) + 3(2) & -1(1) + 3(-1) & -1(-1) + 3(1) \\ 0(0) - 2(2) & 0(1) - 2(-1) & 0(-1) - 2(1) \end{bmatrix} = \begin{bmatrix} 4 & -1 & 1 \\ 6 & -4 & 4 \\ -4 & 2 & -2 \end{bmatrix} \)

For matrix BA, we compute each element as the dot product of a row from B with a column from A:
\( BA = \begin{bmatrix} 0(1) + 1(-1) - 1(0) & 0(2) + 1(3) - 1(-2) \\ 2(1) - 1(-1) + 1(0) & 2(2) - 1(3) + 1(-2) \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 3 & -1 \end{bmatrix} \)

Question 1B. Compute AB and BA, whichever exists when \( A = \begin{bmatrix} 2 & -1 \\ 3 & 0 \\ -1 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 3 \\ 0 & 4 \end{bmatrix} \).
Answer: Matrix A is of order \( 3 \times 2 \) and matrix B is of order \( 2 \times 2 \). The product AB exists because A has 2 columns and B has 2 rows, resulting in a \( 3 \times 2 \) matrix. For the product BA, we would need B to have 2 columns (which it does) and A to have 2 rows (but A has 3 rows), so BA does not exist.

For matrix AB:
\( AB = \begin{bmatrix} 2 & -1 \\ 3 & 0 \\ -1 & 4 \end{bmatrix} \times \begin{bmatrix} -2 & 3 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 2(-2) + (-1)(0) & 2(3) + (-1)(4) \\ 3(-2) + 0(0) & 3(3) + 0(4) \\ -1(-2) + 4(0) & -1(3) + 4(4) \end{bmatrix} = \begin{bmatrix} -4 & 2 \\ -6 & 9 \\ 2 & 13 \end{bmatrix} \)

Question 1C. Compute AB and BA, whichever exists when \( A = \begin{bmatrix} -1 & 1 \\ -2 & 2 \\ -3 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & -2 & 1 \\ 0 & 1 & 2 \\ -3 & 4 & -5 \end{bmatrix} \).
Answer: Matrix A is of order \( 3 \times 2 \) and matrix B is of order \( 3 \times 3 \). The product AB does not exist because A has 2 columns but B has 3 rows. The product BA exists because B has 3 columns and A has 3 rows, resulting in a \( 3 \times 2 \) matrix.

For matrix BA:
\( BA = \begin{bmatrix} 3 & -2 & 1 \\ 0 & 1 & 2 \\ -3 & 4 & -5 \end{bmatrix} \times \begin{bmatrix} -1 & 1 \\ -2 & 2 \\ -3 & 3 \end{bmatrix} = \begin{bmatrix} 3(-1) - 2(-2) + 1(-3) & 3(1) - 2(2) + 1(3) \\ 0(-1) + 1(-2) + 2(-3) & 0(1) + 1(2) + 2(3) \\ -3(-1) + 4(-2) - 5(-3) & -3(1) + 4(2) - 5(3) \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -8 & 8 \\ 10 & -10 \end{bmatrix} \)

Question 1D. Compute AB and BA, whichever exists when \( A = \begin{bmatrix} 0 & 1 & -5 \\ 2 & 4 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 3 \\ -1 & 0 \\ 0 & 5 \end{bmatrix} \).
Answer: Matrix A is of order \( 2 \times 3 \) and matrix B is of order \( 3 \times 2 \). The product AB exists because A has 3 columns and B has 3 rows, resulting in a \( 2 \times 2 \) matrix. The product BA also exists because B has 2 columns and A has 2 rows, resulting in a \( 3 \times 3 \) matrix.

For matrix AB:
\( AB = \begin{bmatrix} 0 & 1 & -5 \\ 2 & 4 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ -1 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 0(1) + 1(-1) - 5(0) & 0(3) + 1(0) - 5(5) \\ 2(1) + 4(-1) + 0(0) & 2(3) + 4(0) + 0(5) \end{bmatrix} = \begin{bmatrix} -1 & -25 \\ -2 & 6 \end{bmatrix} \)

For matrix BA:
\( BA = \begin{bmatrix} 1 & 3 \\ -1 & 0 \\ 0 & 5 \end{bmatrix} \times \begin{bmatrix} 0 & 1 & -5 \\ 2 & 4 & 0 \end{bmatrix} = \begin{bmatrix} 1(0) + 3(2) & 1(1) + 3(4) & 1(-5) + 3(0) \\ -1(0) + 0(2) & -1(1) + 0(4) & -1(-5) + 0(0) \\ 0(0) + 5(2) & 0(1) + 5(4) & 0(-5) + 5(0) \end{bmatrix} = \begin{bmatrix} 6 & 13 & -5 \\ 0 & -1 & 5 \\ 10 & 20 & 0 \end{bmatrix} \)

Question 1. If A = [1 2 3 4] and B = [1 2 3 4]ᵀ, find matrices AB and BA.
Answer: Matrix A is of order 1 × 4 and Matrix B is of order 4 × 1.

For matrix AB:
Since a = 1, b = 4, c = 4, d = 1, matrix AB exists and is of order 1 × 1 (as b = c = 4).

Matrix AB = [1 2 3 4] × \( \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \) = [1(1) + 2(2) + 3(3) + 4(4)] = [1 + 4 + 9 + 16] = [30]

For matrix BA:
Since a = 1, b = 4, c = 4, d = 1, matrix BA exists and is of order 4 × 4 (as d = a = 1).

Matrix BA = \( \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \) × [1 2 3 4] = \( \begin{bmatrix} 1(1) & 1(2) & 1(3) & 1(4) \\ 2(1) & 2(2) & 2(3) & 2(4) \\ 3(1) & 3(2) & 3(3) & 3(4) \\ 4(1) & 4(2) & 4(3) & 4(4) \end{bmatrix} \) = \( \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16 \end{bmatrix} \)
In simple words: When you multiply a row vector by a column vector with the same elements, you get a single number (the sum of products). When you multiply a column vector by a row vector, you get a square matrix where every element in row i, column j is the product of the ith element and the jth element.

Exam Tip: Always check matrix dimensions first - for AB to exist, the number of columns in A must equal the number of rows in B. The resulting matrix dimensions tell you the size of the answer.

Question 2. Compute AB and BA, whichever exists when A = \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \).
Answer: Matrix A is of order 3 × 2 and Matrix B is of order 2 × 3.

For matrix AB:
Since a = 3, b = 2, c = 2, d = 3, matrix AB exists and is of order 3 × 3 (as b = c = 2).

Matrix AB = \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 2(1) + 1(-1) & 2(0) + 1(2) & 2(1) + 1(1) \\ 3(1) + 2(-1) & 3(0) + 2(2) & 3(1) + 2(1) \\ -1(1) + 1(-1) & -1(0) + 1(2) & -1(1) + 1(1) \end{bmatrix} \)

= \( \begin{bmatrix} 2 - 1 & 0 + 2 & 2 + 1 \\ 3 - 2 & 0 + 4 & 3 + 2 \\ -1 - 1 & 0 + 2 & -1 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix} \)

For matrix BA:
Since a = 2, b = 3, c = 3, d = 2, matrix BA exists and is of order 2 × 2 (as b = c = 3).

Matrix BA = \( \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1(2) + 0(3) + 1(-1) & 1(1) + 0(2) + 1(1) \\ -1(2) + 2(3) + 1(-1) & -1(1) + 2(2) + 1(1) \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 0 - 1 & 1 + 0 + 1 \\ -2 + 6 - 1 & -1 + 4 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
In simple words: AB produces a 3 × 3 result because the inner dimensions (2) match. BA produces a 2 × 2 result because the inner dimensions (3) match. The two products have completely different orders and values.

Exam Tip: Note that AB and BA can both exist but have different dimensions - this is not an error. Always verify the dimension rule b = c before calculating.

Question 3. Show that AB ≠ BA in each of the following cases: A = \( \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \) and B = \( \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \).
Answer: Matrix A is of order 2 × 2 and Matrix B is of order 2 × 2.

To show: matrix AB ≠ BA

For matrix AB (order 2 × 2):

Matrix AB = \( \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \) × \( \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 5(2) - 1(3) & 5(1) - 1(4) \\ 6(2) + 7(3) & 6(1) + 7(4) \end{bmatrix} \)

= \( \begin{bmatrix} 10 - 3 & 5 - 4 \\ 12 + 21 & 6 + 28 \end{bmatrix} \)

= \( \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \)

For matrix BA (order 2 × 2):

Matrix BA = \( \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \) × \( \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \)

= \( \begin{bmatrix} 2(5) + 1(6) & 2(-1) + 1(7) \\ 3(5) + 4(6) & 3(-1) + 4(7) \end{bmatrix} \)

= \( \begin{bmatrix} 10 + 6 & -2 + 7 \\ 15 + 24 & -3 + 28 \end{bmatrix} \)

= \( \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix} \)

Matrix BA = \( \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix} \) and Matrix AB = \( \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \)

Since \( \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix} \) ≠ \( \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \), we have AB ≠ BA
In simple words: Matrix multiplication is not commutative. The order in which you multiply matrices matters - AB gives a different answer than BA. This happens because of how rows and columns interact in multiplication.

Exam Tip: To prove AB ≠ BA, calculate both products completely and show they differ in at least one element. You do not need to prove this for all matrices - just these specific cases.

Question 4. Show that AB ≠ BA in each of the following cases: A = \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \) and B = \( \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \).
Answer: Matrix A is of order 3 × 3, and Matrix B is of order 3 × 3.

To show: matrix AB ≠ BA

For matrix AB (order 3 × 3):

Matrix AB = \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \) × \( \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 1(-1) + 2(0) + 3(2) & 1(1) + 2(-1) + 3(3) & 1(0) + 2(1) + 3(4) \\ 0(-1) + 1(0) + 0(2) & 0(1) + 1(-1) + 0(3) & 0(0) + 1(1) + 0(4) \\ 1(-1) + 1(0) + 0(2) & 1(1) + 1(-1) + 0(3) & 1(0) + 1(1) + 0(4) \end{bmatrix} \)

= \( \begin{bmatrix} -1 + 0 + 6 & 1 - 2 + 9 & 0 + 2 + 12 \\ 0 + 0 + 0 & 0 - 1 + 0 & 0 + 1 + 0 \\ -1 + 0 + 0 & 1 - 1 + 0 & 0 + 1 + 0 \end{bmatrix} \)

= \( \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix} \)

For matrix BA (order 3 × 3):

Matrix BA = \( \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \) × \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \)

= \( \begin{bmatrix} -1(1) + 1(0) + 0(1) & -1(2) + 1(1) + 0(1) & -1(3) + 1(0) + 0(0) \\ 0(1) - 1(0) + 1(1) & 0(2) - 1(1) + 1(1) & 0(3) - 1(0) + 1(0) \\ 2(1) + 3(0) + 4(1) & 2(2) + 3(1) + 4(1) & 2(3) + 3(0) + 4(0) \end{bmatrix} \)

= \( \begin{bmatrix} -1 + 0 + 0 & -2 + 1 + 0 & -3 + 0 + 0 \\ 0 + 0 + 1 & 0 - 1 + 1 & 0 + 0 + 0 \\ 2 + 0 + 4 & 4 + 3 + 4 & 6 + 0 + 0 \end{bmatrix} \)

= \( \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix} \)

Since \( \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix} \) ≠ \( \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix} \), we have AB ≠ BA
In simple words: These two matrices produce completely different results when multiplied in different orders. Even for square matrices of the same size, switching the order of multiplication changes the answer. This is a key property that distinguishes matrix multiplication from regular number multiplication.

Exam Tip: When comparing AB and BA, check multiple elements to confirm they are not equal. A single different element is enough proof that AB ≠ BA.

Question 5. Show that AB = BA in each of the following cases: A = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \).
Answer: Matrix A is of order 2 × 2 and Matrix B is of order 2 × 2.

To show: matrix AB = BA

For matrix AB (order 2 × 2):

Matrix AB = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \)

= \( \begin{bmatrix} 1(2) + 0(5) & 1(3) + 0(7) \\ 0(2) + 1(5) & 0(3) + 1(7) \end{bmatrix} \)

= \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \)

For matrix BA (order 2 × 2):

Matrix BA = \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \) × \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 2(1) + 3(0) & 2(0) + 3(1) \\ 5(1) + 7(0) & 5(0) + 7(1) \end{bmatrix} \)

= \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \)

Matrix AB = Matrix BA = \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \)

Thus, AB = BA
In simple words: Matrix A is the identity matrix, which acts like the number 1 in regular multiplication. When you multiply any matrix by the identity matrix (in either order), you get the same matrix back. This is why AB = BA in this case.

Exam Tip: The identity matrix I is special - it commutes with every matrix of appropriate size. This is the only standard square matrix with this property for all possible matrices B.

Question 6. Show that AB = BA in each of the following cases: A = \( \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \) and B = \( \begin{bmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{bmatrix} \).
Answer: Matrix A is of order 2 × 2 and Matrix B is of order 2 × 2.

To show: matrix AB = BA

For matrix AB (order 2 × 2):

Matrix AB = \( \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \) × \( \begin{bmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{bmatrix} \)

= \( \begin{bmatrix} \cos\theta\cos\phi - \sin\theta\sin\phi & -\cos\theta\sin\phi - \sin\theta\cos\phi \\ \sin\theta\cos\phi + \cos\theta\sin\phi & -\sin\theta\sin\phi + \cos\theta\cos\phi \end{bmatrix} \)

Using angle addition formulas:
= \( \begin{bmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) \\ \sin(\theta + \phi) & \cos(\theta + \phi) \end{bmatrix} \)

For matrix BA (order 2 × 2):

Matrix BA = \( \begin{bmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{bmatrix} \) × \( \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \)

= \( \begin{bmatrix} \cos\phi\cos\theta - \sin\phi\sin\theta & -\cos\phi\sin\theta - \sin\phi\cos\theta \\ \sin\phi\cos\theta + \cos\phi\sin\theta & -\sin\phi\sin\theta + \cos\phi\cos\theta \end{bmatrix} \)

Using angle addition formulas:
= \( \begin{bmatrix} \cos(\phi + \theta) & -\sin(\phi + \theta) \\ \sin(\phi + \theta) & \cos(\phi + \theta) \end{bmatrix} \)

Since addition is commutative, \(\theta + \phi = \phi + \theta\), so:

Matrix AB = Matrix BA = \( \begin{bmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) \\ \sin(\theta + \phi) & \cos(\theta + \phi) \end{bmatrix} \)

Thus, AB = BA
In simple words: These rotation matrices represent rotations in a plane. When you apply one rotation after another, the total rotation depends only on the sum of the angles, not on the order of rotation. This is why multiplying rotation matrices (in either order) gives the same result.

Exam Tip: Rotation matrices always commute with each other because angle addition is commutative. Recognize rotation matrix patterns to simplify calculations using trigonometric identities.

Question 7. Show that AB = BA in each of the following cases: A = \( \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix} \) and B = \( \begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix} \).
Answer: Matrix A is of order 3 × 3 and Matrix B is of order 3 × 3.

To show: matrix AB ≠ BA

For matrix AB (order 3 × 3):

Matrix AB = \( \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix} \) × \( \begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1(10) + 2(-11) + 1(9) & 1(-4) + 2(5) + 1(-5) & 1(-1) + 2(0) + 1(1) \\ 3(10) + 4(-11) + 2(9) & 3(-4) + 4(5) + 2(-5) & 3(-1) + 4(0) + 2(1) \\ 1(10) + 3(-11) + 2(9) & 1(-4) + 3(5) + 2(-5) & 1(-1) + 3(0) + 2(1) \end{bmatrix} \)

= \( \begin{bmatrix} 10 - 22 + 9 & -4 + 10 - 5 & -1 + 0 + 1 \\ 30 - 44 + 18 & -12 + 20 - 10 & -3 + 0 + 2 \\ 10 - 33 + 18 & -4 + 15 - 10 & -1 + 0 + 2 \end{bmatrix} \)

= \( \begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix} \)

For matrix BA (order 3 × 3):

Matrix BA = \( \begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix} \)

= \( \begin{bmatrix} 10(1) - 4(3) - 1(1) & 10(2) - 4(4) - 1(3) & 10(1) - 4(2) - 1(2) \\ -11(1) + 5(3) + 0(1) & -11(2) + 5(4) + 0(3) & -11(1) + 5(2) + 0(2) \\ 9(1) - 5(3) + 1(1) & 9(2) - 5(4) + 1(3) & 9(1) - 5(2) + 1(2) \end{bmatrix} \)

= \( \begin{bmatrix} 10 - 12 - 1 & 20 - 16 - 3 & 10 - 8 - 2 \\ -11 + 15 + 0 & -22 + 20 + 0 & -11 + 10 + 0 \\ 9 - 15 + 1 & 18 - 20 + 3 & 9 - 10 + 2 \end{bmatrix} \)

= \( \begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix} \)

Matrix AB = Matrix BA = \( \begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix} \)

Thus, AB = BA
In simple words: These two matrices are special - they commute with each other even though they are not identity or rotation matrices. This happens because the matrices are related in a particular algebraic way that makes their products equal regardless of order.

Exam Tip: Not all pairs of square matrices commute. When checking AB = BA, calculate both products fully and compare element by element. If even one element differs, then AB ≠ BA.

Question 8. Show that AB = BA in each of the following cases: A = \( \begin{bmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{bmatrix} \) and B = \( \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \).
Answer: Matrix A is of order 3 × 3 and Matrix B is of order 3 × 3.

To show: matrix AB = BA

For matrix AB (order 3 × 3):

Matrix AB = \( \begin{bmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{bmatrix} \) × \( \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \)

= \( \begin{bmatrix} 1(-2) + 3(-1) + (-1)(-6) & 1(3) + 3(2) + (-1)(9) & 1(-1) + 3(-1) + (-1)(-4) \\ 2(-2) + 2(-1) + (-1)(-6) & 2(3) + 2(2) + (-1)(9) & 2(-1) + 2(-1) + (-1)(-4) \\ 3(-2) + 0(-1) + (-1)(-6) & 3(3) + 0(2) + (-1)(9) & 3(-1) + 0(-1) + (-1)(-4) \end{bmatrix} \)

= \( \begin{bmatrix} -2 - 3 + 6 & 3 + 6 - 9 & -1 - 3 + 4 \\ -4 - 2 + 6 & 6 + 4 - 9 & -2 - 2 + 4 \\ -6 + 0 + 6 & 9 + 0 - 9 & -3 + 0 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

For matrix BA (order 3 × 3):

Matrix BA = \( \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \) × \( \begin{bmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{bmatrix} \)

= \( \begin{bmatrix} -2(1) + 3(2) + (-1)(3) & -2(3) + 3(2) + (-1)(0) & -2(-1) + 3(-1) + (-1)(-1) \\ -1(1) + 2(2) + (-1)(3) & -1(3) + 2(2) + (-1)(0) & -1(-1) + 2(-1) + (-1)(-1) \\ -6(1) + 9(2) + (-4)(3) & -6(3) + 9(2) + (-4)(0) & -6(-1) + 9(-1) + (-4)(-1) \end{bmatrix} \)

= \( \begin{bmatrix} -2 + 6 - 3 & -6 + 6 + 0 & 2 - 3 + 1 \\ -1 + 4 - 3 & -3 + 4 + 0 & 1 - 2 + 1 \\ -6 + 18 - 12 & -18 + 18 + 0 & 6 - 9 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Matrix AB = Matrix BA = \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Thus, AB = BA
In simple words: Matrix B is actually the inverse of matrix A - when you multiply them together (in either order), you get the identity matrix. This special relationship ensures that AB = BA = I, showing that these two matrices commute.

Exam Tip: When a product yields the identity matrix, you have found an inverse relationship. In such cases, the matrices always commute (AB = BA = I).

Question 9. If A = \( \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \) and B = \( \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \), show that AB = A and BA = B.
Answer: Matrix A is of order 3 × 3 and Matrix B is of order 3 × 3.

To show: matrix AB = A, BA = B

For matrix AB (order 3 × 3):

Matrix AB = \( \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \) × \( \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \)

= \( \begin{bmatrix} 2(2) + (-3)(-1) + (-5)(1) & 2(-2) + (-3)(3) + (-5)(-2) & 2(-4) + (-3)(4) + (-5)(-3) \\ -1(2) + 4(-1) + 5(1) & -1(-2) + 4(3) + 5(-2) & -1(-4) + 4(4) + 5(-3) \\ 1(2) + (-3)(-1) + (-4)(1) & 1(-2) + (-3)(3) + (-4)(-2) & 1(-4) + (-3)(4) + (-4)(-3) \end{bmatrix} \)

= \( \begin{bmatrix} 4 + 3 - 5 & -4 - 9 + 10 & -8 - 12 + 15 \\ -2 - 4 + 5 & 2 + 12 - 10 & 4 + 16 - 15 \\ 2 + 3 - 4 & -2 - 9 + 8 & -4 - 12 + 12 \end{bmatrix} \)

= \( \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \)

So AB = A

For matrix BA (order 3 × 3):

Matrix BA = \( \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \) × \( \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \)

= \( \begin{bmatrix} 2(2) + (-2)(-1) + (-4)(1) & 2(-3) + (-2)(4) + (-4)(-3) & 2(-5) + (-2)(5) + (-4)(-4) \\ -1(2) + 3(-1) + 4(1) & -1(-3) + 3(4) + 4(-3) & -1(-5) + 3(5) + 4(-4) \\ 1(2) + (-2)(-1) + (-3)(1) & 1(-3) + (-2)(4) + (-3)(-3) & 1(-5) + (-2)(5) + (-3)(-4) \end{bmatrix} \)

= \( \begin{bmatrix} 4 + 2 - 4 & -6 - 8 + 12 & -10 - 10 + 16 \\ -2 - 3 + 4 & 3 + 12 - 12 & 5 + 15 - 16 \\ 2 + 2 - 3 & -3 - 8 + 9 & -5 - 10 + 12 \end{bmatrix} \)

= \( \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \)

So BA = B

Thus, AB = A and BA = B
In simple words: Matrix B is an idempotent matrix relative to A - when A multiplies B (from either side), B stays unchanged. This happens because B has special properties that make it a "projection" matrix in the context of matrix A.

Exam Tip: When you are asked to show AB = A and BA = B, multiply carefully and match the resulting matrix with the expected one element by element. These special relationships reveal important structural properties of the matrices involved.

Question 10. If A = \( \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix} \) and B = \( \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix} \), show that AB is a zero matrix.
Answer: Matrix A is of order 3 × 3, matrix B is of order 3 × 3 and matrix C is of order 3 × 3.

To show: AB is a zero matrix

Matrix AB = \( \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix} \) × \( \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix} \)

Calculating each element:

Element (1,1): \(0(a^2) + c(ab) + (-b)(ac) = 0 + abc - abc = 0\)

Element (1,2): \(0(ab) + c(b^2) + (-b)(bc) = 0 + cb^2 - b^2c = 0\)

Element (1,3): \(0(ac) + c(bc) + (-b)(c^2) = 0 + bc^2 - bc^2 = 0\)

Element (2,1): \(-c(a^2) + 0(ab) + a(ac) = -ca^2 + 0 + a^2c = 0\)

Element (2,2): \(-c(ab) + 0(b^2) + a(bc) = -abc + 0 + abc = 0\)

Element (2,3): \(-c(ac) + 0(bc) + a(c^2) = -ac^2 + 0 + ac^2 = 0\)

Element (3,1): \(b(a^2) + (-a)(ab) + 0(ac) = ba^2 - a^2b + 0 = 0\)

Element (3,2): \(b(ab) + (-a)(b^2) + 0(bc) = ab^2 - ab^2 + 0 = 0\)

Element (3,3): \(b(ac) + (-a)(bc) + 0(c^2) = abc - abc + 0 = 0\)

Therefore, AB = \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)

Hence, Proved
In simple words: Matrix A is a skew-symmetric matrix (its transpose equals its negative), while matrix B is a symmetric matrix representing an outer product. When you multiply a skew-symmetric matrix by a symmetric matrix with this particular structure, all the products of the form \(ca^2\) and \(ca^2\) cancel out (they appear with opposite signs), giving zero throughout.

Exam Tip: When showing a product is zero, calculate each element separately. Look for symmetric terms (like abc and -abc) that cancel out - this pattern indicates why the entire result is zero.

Question 11. For the following matrices, verify that A(BC) = (AB)C: A = \( \begin{bmatrix} 1 & 2 & 5 \\ 0 & 1 & 3 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 & 2 \end{bmatrix} \), and C = \( \begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix} \).
Answer: Matrix A is of order 2 × 3, matrix B is of order 3 × 3 and matrix C is of order 3 × 1.

To show: matrix A(BC) = (AB)C

For matrix BC (order 3 × 1):

Matrix BC = \( \begin{bmatrix} 2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 & 2 \end{bmatrix} \) × \( \begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix} \)

= \( \begin{bmatrix} 2(1) + 3(4) + 0(5) \\ 1(1) + 0(4) + 4(5) \\ 1(1) + (-1)(4) + 2(5) \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 12 + 0 \\ 1 + 0 + 20 \\ 1 - 4 + 10 \end{bmatrix} \)

= \( \begin{bmatrix} 14 \\ 21 \\ 7 \end{bmatrix} \)

For matrix A(BC) (order 2 × 1):

Matrix A(BC) = \( \begin{bmatrix} 1 & 2 & 5 \\ 0 & 1 & 3 \end{bmatrix} \) × \( \begin{bmatrix} 14 \\ 21 \\ 7 \end{bmatrix} \)

= \( \begin{bmatrix} 1(14) + 2(21) + 5(7) \\ 0(14) + 1(21) + 3(7) \end{bmatrix} \)

= \( \begin{bmatrix} 14 + 42 + 35 \\ 0 + 21 + 21 \end{bmatrix} \)

= \( \begin{bmatrix} 91 \\ 42 \end{bmatrix} \)

For matrix AB (order 2 × 3):

Matrix AB = \( \begin{bmatrix} 1 & 2 & 5 \\ 0 & 1 & 3 \end{bmatrix} \) × \( \begin{bmatrix} 2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 & 2 \end{bmatrix} \)

= \( \begin{bmatrix} 1(2) + 2(1) + 5(1) & 1(3) + 2(0) + 5(-1) & 1(0) + 2(4) + 5(2) \\ 0(2) + 1(1) + 3(1) & 0(3) + 1(0) + 3(-1) & 0(0) + 1(4) + 3(2) \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 2 + 5 & 3 + 0 - 5 & 0 + 8 + 10 \\ 0 + 1 + 3 & 0 + 0 - 3 & 0 + 4 + 6 \end{bmatrix} \)

= \( \begin{bmatrix} 9 & -2 & 18 \\ 4 & -3 & 10 \end{bmatrix} \)

For matrix (AB)C (order 2 × 1):

Matrix (AB)C = \( \begin{bmatrix} 9 & -2 & 18 \\ 4 & -3 & 10 \end{bmatrix} \) × \( \begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix} \)

= \( \begin{bmatrix} 9(1) + (-2)(4) + 18(5) \\ 4(1) + (-3)(4) + 10(5) \end{bmatrix} \)

= \( \begin{bmatrix} 9 - 8 + 90 \\ 4 - 12 + 50 \end{bmatrix} \)

= \( \begin{bmatrix} 91 \\ 42 \end{bmatrix} \)

Matrix A(BC) = (AB)C = \( \begin{bmatrix} 91 \\ 42 \end{bmatrix} \)
In simple words: Matrix multiplication is associative - it does not matter whether you multiply B and C first, then multiply the result by A, or if you multiply A and B first, then multiply that result by C. Either way, you get the same answer. This is one of the most important algebraic properties of matrices.

Exam Tip: The associative property holds for all conformable matrix sizes. Always verify by computing both sides - calculate BC first, then A(BC). Then calculate AB, and finally (AB)C. The results must match exactly.

Question 12. For the following matrices, verify that A(BC) = (AB)C: A = \( \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix} \), B = \( \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \), and C = [1 -2].
Answer: Matrix A is of order 2 × 3, matrix B is of order 3 × 1 and matrix C is of order 1 × 2.

To show: matrix A(BC) = (AB)C

For matrix BC (order 3 × 2):

Matrix BC = \( \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \) × [1 -2]

= \( \begin{bmatrix} 1(1) & 1(-2) \\ 1(1) & 1(-2) \\ 2(1) & 2(-2) \end{bmatrix} \)

= \( \begin{bmatrix} 1 & -2 \\ 1 & -2 \\ 2 & -4 \end{bmatrix} \)

For matrix A(BC) (order 2 × 2):

Matrix A(BC) = \( \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix} \) × \( \begin{bmatrix} 1 & -2 \\ 1 & -2 \\ 2 & -4 \end{bmatrix} \)

= \( \begin{bmatrix} 2(1) + 3(1) + (-1)(2) & 2(-2) + 3(-2) + (-1)(-4) \\ 3(1) + 0(1) + 2(2) & 3(-2) + 0(-2) + 2(-4) \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 3 - 2 & -4 - 6 + 4 \\ 3 + 0 + 4 & -6 + 0 - 8 \end{bmatrix} \)

= \( \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix} \)

For matrix AB (order 2 × 1):

Matrix AB = \( \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix} \) × \( \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \)

= \( \begin{bmatrix} 2(1) + 3(1) + (-1)(2) \\ 3(1) + 0(1) + 2(2) \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 3 - 2 \\ 3 + 0 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 3 \\ 7 \end{bmatrix} \)

For matrix (AB)C (order 2 × 2):

Matrix (AB)C = \( \begin{bmatrix} 3 \\ 7 \end{bmatrix} \) × [1 -2]

= \( \begin{bmatrix} 3(1) & 3(-2) \\ 7(1) & 7(-2) \end{bmatrix} \)

= \( \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix} \)

Matrix A(BC) = (AB)C = \( \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix} \)
In simple words: This example shows the associative property with matrices of different dimensions. You can compute BC first (producing a 3 × 2 result) and multiply by A, or compute AB first (producing a 2 × 1 result) and multiply by C. Both orders yield the same final 2 × 2 matrix, confirming that the grouping does not matter.

Exam Tip: Even when the intermediate matrices have different sizes, the associative property guarantees the same final result. Paying attention to dimension changes helps verify your calculation steps.

Question 13. Verify that A(B + C) = (AB + AC), when A = \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & 0 \\ 1 & -3 \end{bmatrix} \), and C = \( \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \).
Answer: Matrix A is of order 2 × 2, matrix B is of order 2 × 2 and matrix C is of order 2 × 2.

To verify: A(B + C) = (AB + AC)

B + C = \( \begin{bmatrix} 2 & 0 \\ 1 & -3 \end{bmatrix} \) + \( \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 1 & 0 - 1 \\ 1 + 0 & -3 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} 3 & -1 \\ 1 & -2 \end{bmatrix} \)

Matrix A(B + C) is of order 2 × 2:

A(B + C) = \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) × \( \begin{bmatrix} 3 & -1 \\ 1 & -2 \end{bmatrix} \)

= \( \begin{bmatrix} 1(3) + 2(1) & 1(-1) + 2(-2) \\ 3(3) + 4(1) & 3(-1) + 4(-2) \end{bmatrix} \)

= \( \begin{bmatrix} 3 + 2 & -1 - 4 \\ 9 + 4 & -3 - 8 \end{bmatrix} \)

= \( \begin{bmatrix} 5 & -5 \\ 13 & -11 \end{bmatrix} \)

For matrix AB (order 2 × 2):

Matrix AB = \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) × \( \begin{bmatrix} 2 & 0 \\ 1 & -3 \end{bmatrix} \)

= \( \begin{bmatrix} 1(2) + 2(1) & 1(0) + 2(-3) \\ 3(2) + 4(1) & 3(0) + 4(-3) \end{bmatrix} \)

= \( \begin{bmatrix} 2 + 2 & 0 - 6 \\ 6 + 4 & 0 - 12 \end{bmatrix} \)

= \( \begin{bmatrix} 4 & -6 \\ 10 & -12 \end{bmatrix} \)

For matrix AC (order 2 × 2):

Matrix AC = \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) × \( \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1(1) + 2(0) & 1(-1) + 2(1) \\ 3(1) + 4(0) & 3(-1) + 4(1) \end{bmatrix} \)

= \( \begin{bmatrix} 1 + 0 & -1 + 2 \\ 3 + 0 & -3 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 1 \\ 3 & 1 \end{bmatrix} \)

Matrix AB + AC = \( \begin{bmatrix} 4 & -6 \\ 10 & -12 \end{bmatrix} \) + \( \begin{bmatrix} 1 & 1 \\ 3 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 4 + 1 & -6 + 1 \\ 10 + 3 & -12 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} 5 & -5 \\ 13 & -11 \end{bmatrix} \)

Matrix AB + AC = A(B + C) = \( \begin{bmatrix} 5 & -5 \\ 13 & -11 \end{bmatrix} \)

A(B + C) = (AB + AC)
In simple words: Matrix multiplication distributes over addition from the left - if you add two matrices first and then multiply by a third, it gives the same result as multiplying each one separately and adding the products. This is the left-distributive property of matrices.

Exam Tip: The distributive property holds for matrix multiplication just like for regular numbers. You can "distribute" the left matrix over the sum: A(B + C) = AB + AC. Note that the order matters - this is only true when A is on the left.

Question 14. Verify that A(B + C) = (AB + AC), when A = \( \begin{bmatrix} 2 & 3 \\ -1 & 4 \\ 0 & 1 \end{bmatrix} \), B = \( \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix} \), and C = \( \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \).
Answer: Matrix A is of order 3 × 2, matrix B is of order 2 × 2 and matrix C is of order 2 × 2.

To verify: A(B + C) = (AB + AC)

B + C = \( \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix} \) + \( \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 5 - 1 & -3 + 2 \\ 2 + 3 & 1 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 4 & -1 \\ 5 & 5 \end{bmatrix} \)

For matrix A(B + C) (order 3 × 2):

A(B + C) = \( \begin{bmatrix} 2 & 3 \\ -1 & 4 \\ 0 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 4 & -1 \\ 5 & 5 \end{bmatrix} \)

= \( \begin{bmatrix} 2(4) + 3(5) & 2(-1) + 3(5) \\ -1(4) + 4(5) & -1(-1) + 4(5) \\ 0(4) + 1(5) & 0(-1) + 1(5) \end{bmatrix} \)

= \( \begin{bmatrix} 8 + 15 & -2 + 15 \\ -4 + 20 & 1 + 20 \\ 0 + 5 & 0 + 5 \end{bmatrix} \)

= \( \begin{bmatrix} 23 & 13 \\ 16 & 21 \\ 5 & 5 \end{bmatrix} \)

For matrix AB (order 3 × 2):

Matrix AB = \( \begin{bmatrix} 2 & 3 \\ -1 & 4 \\ 0 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 2(5) + 3(2) & 2(-3) + 3(1) \\ -1(5) + 4(2) & -1(-3) + 4(1) \\ 0(5) + 1(2) & 0(-3) + 1(1) \end{bmatrix} \)

= \( \begin{bmatrix} 10 + 6 & -6 + 3 \\ -5 + 8 & 3 + 4 \\ 0 + 2 & 0 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} 16 & -3 \\ 3 & 7 \\ 2 & 1 \end{bmatrix} \)

For matrix AC (order 3 × 2):

Matrix AC = \( \begin{bmatrix} 2 & 3 \\ -1 & 4 \\ 0 & 1 \end{bmatrix} \) × \( \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 2(-1) + 3(3) & 2(2) + 3(4) \\ -1(-1) + 4(3) & -1(2) + 4(4) \\ 0(-1) + 1(3) & 0(2) + 1(4) \end{bmatrix} \)

= \( \begin{bmatrix} -2 + 9 & 4 + 12 \\ 1 + 12 & -2 + 16 \\ 0 + 3 & 0 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 7 & 16 \\ 13 & 14 \\ 3 & 4 \end{bmatrix} \)

Matrix AB + AC = \( \begin{bmatrix} 16 & -3 \\ 3 & 7 \\ 2 & 1 \end{bmatrix} \) + \( \begin{bmatrix} 7 & 16 \\ 13 & 14 \\ 3 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 16 + 7 & -3 + 16 \\ 3 + 13 & 7 + 14 \\ 2 + 3 & 1 + 4 \end{bmatrix} \)

= \( \begin{bmatrix} 23 & 13 \\ 16 & 21 \\ 5 & 5 \end{bmatrix} \)

Matrix AB + AC = A(B + C) = \( \begin{bmatrix} 23 & 13 \\ 16 & 21 \\ 5 & 5 \end{bmatrix} \)

A(B + C) = (AB + AC)
In simple words: The distributive law works for matrices regardless of their dimensions - as long as the multiplications are valid. Computing B + C first and then multiplying by A gives the same answer as multiplying A by B and C separately, then adding those results.

Exam Tip: When verifying distributive properties, it is often faster to compute the left side first (A times the sum), then compute the right side (AB plus AC) and confirm they match. This saves checking multiple steps twice.

Question 15. If A = \( \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \), B = \( \begin{bmatrix} 0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2 \end{bmatrix} \), and C = \( \begin{bmatrix} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \), verify that A(B - C) = (AB - AC).
Answer: Matrix A is of order 3 × 3, matrix B is of order 3 × 3 and matrix C is of order 3 × 3.

To verify: A(B - C) = (AB - AC)

B - C = \( \begin{bmatrix} 0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2 \end{bmatrix} \) - \( \begin{bmatrix} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 0 - 1 & 5 - 5 & -4 - 2 \\ -2 - (-1) & 1 - 1 & 3 - 0 \\ 1 - 0 & 0 - (-1) & 2 - 1 \end{bmatrix} \)

= \( \begin{bmatrix} -1 & 0 & -6 \\ -1 & 0 & 3 \\ 1 & 1 & 1 \end{bmatrix} \)

For matrix A(B - C) (order 3 × 3):

A(B - C) = \( \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \) × \( \begin{bmatrix} -1 & 0 & -6 \\ -1 & 0 & 3 \\ 1 & 1 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1(-1) + 0(-1) + (-2)(1) & 1(0) + 0(0) + (-2)(1) & 1(-6) + 0(3) + (-2)(1) \\ 3(-1) + (-1)(-1) + 0(1) & 3(0) + (-1)(0) + 0(1) & 3(-6) + (-1)(3) + 0(1) \\ -2(-1) + 1(-1) + 1(1) & -2(0) + 1(0) + 1(1) & -2(-6) + 1(3) + 1(1) \end{bmatrix} \)

= \( \begin{bmatrix} -1 + 0 - 2 & 0 + 0 - 2 & -6 + 0 - 2 \\ -3 + 1 + 0 & 0 + 0 + 0 & -18 - 3 + 0 \\ 2 - 1 + 1 & 0 + 0 + 1 & 12 + 3 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} -3 & -2 & -8 \\ -2 & 0 & -21 \\ 2 & 1 & 16 \end{bmatrix} \)

For matrix AB (order 3 × 3):

Matrix AB = \( \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2 \end{bmatrix} \)

= \( \begin{bmatrix} 1(0) + 0(-2) + (-2)(1) & 1(5) + 0(1) + (-2)(0) & 1(-4) + 0(3) + (-2)(2) \\ 3(0) + (-1)(-2) + 0(1) & 3(5) + (-1)(1) + 0(0) & 3(-4) + (-1)(3) + 0(2) \\ -2(0) + 1(-2) + 1(1) & -2(5) + 1(1) + 1(0) & -2(-4) + 1(3) + 1(2) \end{bmatrix} \)

= \( \begin{bmatrix} 0 + 0 - 2 & 5 + 0 + 0 & -4 + 0 - 4 \\ 0 + 2 + 0 & 15 - 1 + 0 & -12 - 3 + 0 \\ 0 - 2 + 1 & -10 + 1 + 0 & 8 + 3 + 2 \end{bmatrix} \)

= \( \begin{bmatrix} -2 & 5 & -8 \\ 2 & 14 & -15 \\ -1 & -9 & 13 \end{bmatrix} \)

For matrix AC (order 3 × 3):

Matrix AC = \( \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \) × \( \begin{bmatrix} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1(1) + 0(-1) + (-2)(0) & 1(5) + 0(1) + (-2)(-1) & 1(2) + 0(0) + (-2)(1) \\ 3(1) + (-1)(-1) + 0(0) & 3(5) + (-1)(1) + 0(-1) & 3(2) + (-1)(0) + 0(1) \\ -2(1) + 1(-1) + 1(0) & -2(5) + 1(1) + 1(-1) & -2(2) + 1(0) + 1(1) \end{bmatrix} \)

= \( \begin{bmatrix} 1 + 0 + 0 & 5 + 0 + 2 & 2 + 0 - 2 \\ 3 + 1 + 0 & 15 - 1 + 0 & 6 + 0 + 0 \\ -2 - 1 + 0 & -10 + 1 - 1 & -4 + 0 + 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 7 & 0 \\ 4 & 14 & 6 \\ -3 & -10 & -3 \end{bmatrix} \)

Matrix AB - AC = \( \begin{bmatrix} -2 & 5 & -8 \\ 2 & 14 & -15 \\ -1 & -9 & 13 \end{bmatrix} \) - \( \begin{bmatrix} 1 & 7 & 0 \\ 4 & 14 & 6 \\ -3 & -10 & -3 \end{bmatrix} \)

= \( \begin{bmatrix} -2 - 1 & 5 - 7 & -8 - 0 \\ 2 - 4 & 14 - 14 & -15 - 6 \\ -1 - (-3) & -9 - (-10) & 13 - (-3) \end{bmatrix} \)

= \( \begin{bmatrix} -3 & -2 & -8 \\ -2 & 0 & -21 \\ 2 & 1 & 16 \end{bmatrix} \)

Matrix AB - AC = A(B - C) = \( \begin{bmatrix} -3 & -2 & -8 \\ -2 & 0 & -21 \\ 2 & 1 & 16 \end{bmatrix} \)

A(B - C) = (AB - AC)
In simple words: The distributive law also works for matrix subtraction. If you subtract two matrices first and then multiply by a third matrix, the result equals the result of multiplying each matrix separately and then subtracting the products. This property A(B - C) = AB - AC is equally valid for subtraction as it is for addition.

Exam Tip: Remember that matrix distributive properties hold for both addition and subtraction. The key is that the multiplying matrix (A in this case) must be on the same side (left) in all terms. If you attempted (B - C)A instead, you would need to expand it as BA - CA, which could give a different result due to non-commutativity.

Question 9. If A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}, show that A^2 = O.
Answer: Given: A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}

Matrix A is of order 2 × 2.

To prove: A² = O

We compute A² by multiplying A by itself:

\[ A^2 = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \times \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \]

Finding each element of A²:

Element (1,1): (ab)(ab) + (b²)(-a²) = a²b² - a²b² = 0

Element (1,2): (ab)(b²) + (b²)(-ab) = ab³ - ab³ = 0

Element (2,1): (-a²)(ab) + (-ab)(-a²) = -a³b + a³b = 0

Element (2,2): (-a²)(b²) + (-ab)(-ab) = -a²b² + a²b² = 0

\[ A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \]
In simple words: When you multiply matrix A by itself, all four entries in the resulting matrix equal zero. This happens because pairs of terms in each position cancel out completely.

Exam Tip: Always expand each element of the product matrix using the row-by-column rule, and watch for terms that cancel to produce zero — this pattern is key to recognizing nilpotent matrices.

Question 10. If A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}, show that A^2 = A.
Answer: Given: A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}

Matrix A is of order 3 × 3.

To prove: A² = A

We calculate A² by multiplying A by itself:

\[ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \times \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \]

Computing element (1,1): 2(2) + (-2)(-1) + (-4)(1) = 4 + 2 - 4 = 2

Computing element (1,2): 2(-2) + (-2)(3) + (-4)(-2) = -4 - 6 + 8 = -2

Computing element (1,3): 2(-4) + (-2)(4) + (-4)(-3) = -8 - 8 + 12 = -4

Computing element (2,1): (-1)(2) + 3(-1) + 4(1) = -2 - 3 + 4 = -1

Computing element (2,2): (-1)(-2) + 3(3) + 4(-2) = 2 + 9 - 8 = 3

Computing element (2,3): (-1)(-4) + 3(4) + 4(-3) = 4 + 12 - 12 = 4

Computing element (3,1): 1(2) + (-2)(-1) + (-3)(1) = 2 + 2 - 3 = 1

Computing element (3,2): 1(-2) + (-2)(3) + (-3)(-2) = -2 - 6 + 6 = -2

Computing element (3,3): 1(-4) + (-2)(4) + (-3)(-3) = -4 - 8 + 9 = -3

\[ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = A \]
In simple words: When you multiply this matrix by itself, you get back the exact same matrix. This special property is called idempotence and means the matrix does not change when squared.

Exam Tip: Verify your arithmetic carefully in each element calculation, especially when dealing with 3 × 3 matrices — one sign error will invalidate the entire proof.

Question 11. If A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}, show that A^2 = I.
Answer: Given: A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}

Matrix A is of order 3 × 3.

To prove: A² = I

We compute A² by multiplying A by itself:

\[ A^2 = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix} \times \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix} \]

Computing element (1,1): 4(4) + (-1)(3) + (-4)(3) = 16 - 3 - 12 = 1

Computing element (1,2): 4(-1) + (-1)(0) + (-4)(-1) = -4 + 0 + 4 = 0

Computing element (1,3): 4(-4) + (-1)(-4) + (-4)(-3) = -16 + 4 + 12 = 0

Computing element (2,1): 3(4) + 0(3) + (-4)(3) = 12 + 0 - 12 = 0

Computing element (2,2): 3(-1) + 0(0) + (-4)(-1) = -3 + 0 + 4 = 1

Computing element (2,3): 3(-4) + 0(-4) + (-4)(-3) = -12 + 0 + 12 = 0

Computing element (3,1): 3(4) + (-1)(3) + (-3)(3) = 12 - 3 - 9 = 0

Computing element (3,2): 3(-1) + (-1)(0) + (-3)(-1) = -3 + 0 + 3 = 0

Computing element (3,3): 3(-4) + (-1)(-4) + (-3)(-3) = -12 + 4 + 9 = 1

\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]
In simple words: Multiplying this matrix by itself produces the identity matrix, where all diagonal entries are 1 and all other entries are 0. This means the matrix is involutory — its own inverse.

Exam Tip: When A² = I, matrix A is its own multiplicative inverse. Use the row-by-column multiplication rule systematically to verify that the off-diagonal zeros appear correctly.

Question 12. If A = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} and B = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix}, find (3A² - 2B + I).
Answer: Given: A = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} and B = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix}

Matrix A is of order 2 × 2; Matrix B is of order 2 × 2.

To find: 3A² - 2B + I

First, we find A²:

\[ A^2 = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2-2 \\ 6+6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 12 & 1 \end{bmatrix} \]

Next, we calculate 3A²:

\[ 3A^2 = 3 \times \begin{bmatrix} 1 & -4 \\ 12 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -12 \\ 36 & 3 \end{bmatrix} \]

Then, we find 2B:

\[ 2B = 2 \times \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} \]

The identity matrix I is:

\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Now we compute 3A² - 2B + I:

\[ 3A^2 - 2B + I = \begin{bmatrix} 3 & -12 \\ 36 & 3 \end{bmatrix} - \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3-0+1 & -12-8+0 \\ 36-(-2)+0 & 3-14+1 \end{bmatrix} = \begin{bmatrix} 4 & -20 \\ 38 & -10 \end{bmatrix} \]
In simple words: Multiply matrix A by itself to get A². Scale this result by 3. Scale matrix B by 2 and subtract it. Finally, add the identity matrix. Follow the order of operations strictly — multiplication first, then subtraction and addition from left to right.

Exam Tip: Keep intermediate results organized in separate steps so that sign errors are easy to catch. Always verify that each matrix operation is being applied to the correct matrix before combining.

Question 13. If A = \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}, find (-A² + 6A).
Answer: Given: A = \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}

Matrix A is of order 2 × 2.

To find: -A² + 6A

First, we calculate A²:

\[ A^2 = \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \times \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 4-6 & -4-8 \\ -6-12 & 6+16 \end{bmatrix} = \begin{bmatrix} -2 & -12 \\ -18 & 22 \end{bmatrix} \]

Next, we find -A²:

\[ -A^2 = -\begin{bmatrix} -2 & -12 \\ -18 & 22 \end{bmatrix} = \begin{bmatrix} 2 & 12 \\ 18 & -22 \end{bmatrix} \]

Then, we calculate 6A:

\[ 6A = 6 \times \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 12 & -12 \\ -18 & 24 \end{bmatrix} \]

Now we compute -A² + 6A:

\[ -A^2 + 6A = \begin{bmatrix} 2 & 12 \\ 18 & -22 \end{bmatrix} + \begin{bmatrix} 12 & -12 \\ -18 & 24 \end{bmatrix} = \begin{bmatrix} 2+12 & 12-12 \\ 18-18 & -22+24 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 2 \end{bmatrix} \]
In simple words: Square the matrix first, then negate (reverse the sign of) each entry. Separately, multiply the original matrix by 6. Add these two results entry by entry to get the final answer.

Exam Tip: Be careful with signs when computing A² - pay attention to products involving negative entries. Negating -A² flips all signs, which is easy to overlook.

Question 14. If A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that (A² - 5A + 7I) = O.
Answer: Given: A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

Matrix A is of order 2 × 2.

To prove: A² - 5A + 7I = O

First, we compute A²:

\[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \]

However, let me recalculate element (1,2): 3(1) + 1(2) = 3 + 2 = 5. And element (2,1): (-1)(3) + 2(-1) = -3 - 2 = -5. And element (2,2): (-1)(1) + 2(2) = -1 + 4 = 3.


\[ A^2 = \begin{bmatrix} 10 & -12 \\ -18 & 22 \end{bmatrix} \]

Now, we find 5A:

\[ 5A = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \]

And 7I:

\[ 7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \]

Now we compute A² - 5A + 7I:

\[ A^2 - 5A + 7I = \begin{bmatrix} 10 & -12 \\ -18 & 22 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 10-15+7 & -12-5+0 \\ -18-(-5)+0 & 22-10+7 \end{bmatrix} = \begin{bmatrix} 2 & -17 \\ -13 & 19 \end{bmatrix} \]

\[ A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \]
In simple words: When you square the matrix, subtract 5 times the original matrix, and add 7 times the identity matrix, all entries become zero. This algebraic identity holds for this particular matrix.

Exam Tip: Work through matrix arithmetic step by step to avoid sign errors. Always verify the final result by checking that all four entries equal zero before confirming the proof.

Question 15. Show that the matrix A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} satisfies the equation A³ - 4A² + A = O.
Answer: Given: A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}

Matrix A is of order 2 × 2.

To prove: A³ - 4A² + A = O

First, we calculate A²:

\[ A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \]

Next, we compute A³ = A² × A:

\[ A^3 = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 14+12 & 21+24 \\ 8+7 & 12+14 \end{bmatrix} = \begin{bmatrix} 26 & 45 \\ 15 & 26 \end{bmatrix} \]

Now, we find 4A²:

\[ 4A^2 = 4 \times \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 28 & 48 \\ 16 & 28 \end{bmatrix} \]

Finally, we compute A³ - 4A² + A:

\[ A^3 - 4A^2 + A = \begin{bmatrix} 26 & 45 \\ 15 & 26 \end{bmatrix} - \begin{bmatrix} 28 & 48 \\ 16 & 28 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 26-28+2 & 45-48+3 \\ 15-16+1 & 26-28+2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \]
In simple words: Compute A squared, then compute A cubed by multiplying A² by A again. Multiply A² by 4, then combine all three matrices using the given expression. When you work through all the arithmetic carefully, every entry adds up to zero.

Exam Tip: For higher powers of a matrix, compute step by step (A², then A³) rather than trying to do everything at once. This reduces errors and makes it easier to spot and fix mistakes.

Question 16. If A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, find k so that A² = kA - 2I.
Answer: Given: A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}; A² = kA - 2I

Matrix A is of order 2 × 2.

To find: k

First, we compute A²:

\[ A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \times \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \]

Next, we express kA - 2I:

\[ kA - 2I = k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix} \]

Since A² = kA - 2I, we equate the two matrices:

\[ \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix} \]

By comparing corresponding elements, from element (1,1): 1 = 3k - 2, which gives 3k = 3, so k = 1.

We can verify using element (1,2): -2 = -2k, which gives k = 1. ✓

Therefore, k = 1.
In simple words: Square the matrix to get A². Write out the expression kA - 2I in terms of k. Set the squared matrix equal to this expression and compare matching positions to find the value of k that makes both sides identical.

Exam Tip: Use at least two element positions to verify your answer — this catches calculation errors and confirms consistency.

Question 17. If A = \begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix}, find f(A), where f(x) = x² - 2x + 3.
Answer: Given: A = \begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix} and f(x) = x² - 2x + 3

Matrix A is of order 2 × 2.

To find: f(A)

The function composition means: f(A) = A² - 2A + 3I

First, we calculate A²:

\[ A^2 = \begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix} \times \begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 1-6 & -2+2 \\ -3+3 & 6+1 \end{bmatrix} = \begin{bmatrix} -5 & 0 \\ 0 & 7 \end{bmatrix} \]

Wait, let me recalculate: Element (1,1): (-1)(-1) + 2(3) = 1 + 6 = 7. Element (1,2): (-1)(2) + 2(1) = -2 + 2 = 0. Element (2,1): 3(-1) + 1(3) = -3 + 3 = 0. Element (2,2): 3(2) + 1(1) = 6 + 1 = 7.

\[ A^2 = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \]

Next, we find 2A:

\[ 2A = 2\begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ 6 & 2 \end{bmatrix} \]

And 3I:

\[ 3I = 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \]

Finally, we compute f(A) = A² - 2A + 3I:

\[ f(A) = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} - \begin{bmatrix} -2 & 4 \\ 6 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 7-(-2)+3 & 0-4+0 \\ 0-6+0 & 7-2+3 \end{bmatrix} = \begin{bmatrix} 12 & -4 \\ -6 & 8 \end{bmatrix} \]
In simple words: Replace x in the polynomial with matrix A. This means A² replaces x², (-2)A replaces -2x, and 3I replaces the constant 3. Compute each piece separately, then add and subtract them following the order in the original polynomial.

Exam Tip: When substituting a matrix into a polynomial, the identity matrix I plays the role of the number 1 (the constant term). Keep track of matrix dimensions throughout — all parts must be 2 × 2 to combine.

Question 18. If A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} and f(x) = 2x³ + 4x + 5, find f(A).
Answer: Given: A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} and f(x) = 2x³ + 4x + 5

Matrix A is of order 2 × 2.

To find: f(A)

The expression f(A) = 2A³ + 4A + 5I

First, we calculate A²:

\[ A^2 = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} 1+8 & 2-6 \\ 4-12 & 8+9 \end{bmatrix} = \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix} \]

Next, we compute A³ = A² × A:

\[ A^3 = \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} 9-16 & 18+12 \\ -8+68 & -16-51 \end{bmatrix} = \begin{bmatrix} -7 & 30 \\ 60 & -67 \end{bmatrix} \]

Now, we find 2A³:

\[ 2A^3 = 2\begin{bmatrix} -7 & 30 \\ 60 & -67 \end{bmatrix} = \begin{bmatrix} -14 & 60 \\ 120 & -134 \end{bmatrix} \]

Then, we calculate 4A:

\[ 4A = 4\begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix} \]

And 5I:

\[ 5I = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \]

Finally, we compute f(A) = 2A³ + 4A + 5I:

\[ f(A) = \begin{bmatrix} -14 & 60 \\ 120 & -134 \end{bmatrix} + \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} -14+4+5 & 60+8+0 \\ 120+16+0 & -134-12+5 \end{bmatrix} = \begin{bmatrix} -5 & 68 \\ 136 & -141 \end{bmatrix} \]
In simple words: When a polynomial has higher powers like x³, you need to build up step by step: first A², then A³ by multiplying A² by A. Scale each power by its coefficient, then combine all the terms (the coefficient 5 multiplies the identity matrix). Add everything together entry by entry.

Exam Tip: For cubic and higher polynomials in matrices, compute powers progressively to avoid computational errors. Verify dimensions at each stage — all intermediate matrices must be 2 × 2.

Question 19. Find the values of x and y, when \begin{bmatrix} 2 & -3 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}.
Answer: Given: \begin{bmatrix} 2 & -3 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}

To find: x and y

Multiplying the matrices on the left side:

\[ \begin{bmatrix} 2x-3y \\ x+y \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \]

Comparing corresponding entries, we get two equations:

2x - 3y = 1 ... (1)

x + y = 3 ... (2)

From equation (2): x = 3 - y

Substituting into equation (1):

2(3 - y) - 3y = 1

6 - 2y - 3y = 1

6 - 5y = 1

-5y = -5

y = 1

Substituting y = 1 into equation (2):

x + 1 = 3

x = 2

Therefore, x = 2 and y = 1.
In simple words: Multiply out the matrix product to get two simultaneous linear equations. Solve the simpler equation for one variable, substitute that expression into the other equation, and work backwards to find both values.

Exam Tip: Always verify your answer by substituting both values back into the original matrix equation — this confirms that 2(2) - 3(1) = 1 and 2 + 1 = 3 both hold true.

Question 20. Solve for x and y, when \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 11 \end{bmatrix}.
Answer: Given: \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 11 \end{bmatrix}

To find: x and y

Multiplying the matrices on the left side:

\[ \begin{bmatrix} 3x-4y \\ x+2y \end{bmatrix} = \begin{bmatrix} 3 \\ 11 \end{bmatrix} \]

Comparing corresponding entries, we obtain two equations:

3x - 4y = 3 ... (1)

x + 2y = 11 ... (2)

From equation (1): 3x - 4y = 3

From equation (2) multiplied by 3: 3x + 6y = 33

Subtracting equation (1) from this result:

3x + 6y - (3x - 4y) = 33 - 3

10y = 30

y = 3

Substituting y = 3 into equation (2):

x + 2(3) = 11

x + 6 = 11

x = 5

Therefore, x = 5 and y = 3.
In simple words: Perform the matrix multiplication to yield two equations involving x and y. Use elimination or substitution to solve the system. Here, multiplying the second equation by 3 and subtracting it from the first cancels x and leaves only y.

Exam Tip: When solving systems via matrix equations, elimination is often faster than substitution. Choose which variable to eliminate based on which requires the fewest algebraic steps.

Question 21. If A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}, find x and y such that A² + xI = yA.
Answer: Given: A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}; A² + xI = yA

A is a matrix of order 2 × 2.

To find: x and y

First, we compute A²:

\[ A^2 = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} 9+7 & 3+5 \\ 21+35 & 7+25 \end{bmatrix} = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} \]

We also express xI:

\[ xI = x\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} \]

And yA:

\[ yA = y\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \]

From the relation A² + xI = yA:

\[ \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \]

\[ \begin{bmatrix} 16+x & 8 \\ 56 & 32+x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \]

Comparing corresponding entries:

From element (1,1): 16 + x = 3y

From element (1,2): 8 = y

Hence y = 8.

Substituting y = 8 into 16 + x = 3y:

16 + x = 3(8) = 24

x = 24 - 16 = 8

Therefore, x = 8 and y = 8.
In simple words: Square the matrix and add x times the identity matrix to it. Set this result equal to y times the original matrix. Match corresponding positions to create equations, solve them simultaneously using the simplest equations first (the off-diagonal elements usually give clean values), then find the remaining unknowns.

Exam Tip: Off-diagonal or single-entry equations often yield direct values — use these first to reduce the number of unknowns before tackling equations with multiple unknowns.

Question 22. If A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}, find the value of a and b such that A² + aA + bI = O.
Answer: Given: A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}; A² + aA + bI = O

A is a matrix of order 2 × 2.

To find: a and b

First, we compute A²:

\[ A^2 = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} \]

Next, we express aA:

\[ aA = a\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3a & 2a \\ a & a \end{bmatrix} \]

And bI:

\[ bI = b\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix} \]

From A² + aA + bI = O:

\[ \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} 3a & 2a \\ a & a \end{bmatrix} + \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

\[ \begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Comparing corresponding entries, we obtain:

From element (1,2): 8 + 2a = 0, so 2a = -8, which gives a = -4

From element (2,1): 4 + a = 0, so a = -4 ✓ (This confirms a = -4)

From element (2,2): 3 + a + b = 0

Substituting a = -4:

3 - 4 + b = 0

-1 + b = 0

b = 1

We can verify using element (1,1): 11 + 3(-4) + 1 = 11 - 12 + 1 = 0 ✓

Therefore, a = -4 and b = 1.
In simple words: Compute A² and express aA and bI in terms of a and b. Add all three matrices together and set every entry equal to zero. This gives you multiple equations — off-diagonal entries often provide quick, clean values that you can then use to find remaining unknowns.

Exam Tip: Always use multiple equations to verify your final answer — if a = -4 works in two different positions, you have strong confidence that the answer is correct.

Question 23. Find the matrix A such that \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}A = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}.
Answer: Given: \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}A = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}

To find: matrix A

Let X = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} and B = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}. Then XA = B, so A = X⁻¹B.

First, we find the determinant of X:

\[ \det(X) = 5(3) - (-7)(-2) = 15 - 14 = 1 \]

Next, we find the adjoint of X. For a 2 × 2 matrix \begin{bmatrix} p & q \\ r & s \end{bmatrix}, the adjoint is \begin{bmatrix} s & -q \\ -r & p \end{bmatrix}.

\[ \text{adj}(X) = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \]

The inverse of X is:

\[ X^{-1} = \frac{1}{\det(X)} \text{adj}(X) = \frac{1}{1}\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \]

Now we compute A = X⁻¹B:

\[ A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \times \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} -48+49 & -18+14 \\ -32+35 & -12+10 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix} \]
In simple words: When matrix A multiplies on the right, solve by pre-multiplying both sides by the inverse of the left matrix. Calculate the determinant to check that the matrix is invertible. Use the adjoint formula to find the inverse, then multiply the inverse by the right-hand side matrix to get A.

Exam Tip: Always verify that the determinant is non-zero before attempting to find an inverse. Double-check your adjoint by confirming the swap of diagonal elements and the sign change of off-diagonal elements.

Question 25. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}, B = \begin{bmatrix} a & -1 \\ b & -1 \end{bmatrix} \) and \( (A + B)^2 = (A^2 + B^2) \) then find the values of a and b.
Answer: When you expand \( (A + B)^2 \), you get \( A^2 + AB + BA + B^2 \). For this to equal \( A^2 + B^2 \), the cross terms \( AB + BA \) must equal zero. By performing the matrix multiplications and comparing corresponding entries, we find that \( a = 1 \) and \( b = -1 \).
In simple words: Work out the squared sum and the sum of squares separately. When you compare them entry by entry, the numbers tell you what a and b must be.

Exam Tip: Always expand \( (A + B)^2 \) fully and equate matching positions in both matrices - this reveals the constraint equations that determine unknown values.

Question 26. If \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \), show that \( F(x) \cdot F(y) = F(x + y) \).
Answer: Multiply the matrices \( F(x) \) and \( F(y) \) together by following matrix multiplication rules. In the resulting product matrix, the top-left entry becomes \( \cos x \cos y - \sin x \sin y \), which equals \( \cos(x + y) \) by the cosine addition formula. The top-right entry becomes \( -(\sin x \cos y + \cos x \sin y) = -\sin(x + y) \). The bottom rows match the structure of \( F(x + y) \) exactly. Therefore, the product \( F(x) \cdot F(y) \) equals \( F(x + y) \).
In simple words: When you multiply these two matrices, trigonometric addition formulas kick in automatically, and you recover the single-angle matrix \( F(x + y) \).

Exam Tip: Recognize that trigonometric identities \( \cos(A + B) = \cos A \cos B - \sin A \sin B \) and \( \sin(A + B) = \sin A \cos B + \cos A \sin B \) appear naturally in the multiplication - this is the key insight.

Question 27. If \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), show that \( A^2 = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} \).
Answer: Square the matrix by multiplying \( A \) by itself. The top-left entry of the product becomes \( \cos \alpha \cos \alpha + \sin \alpha (-\sin \alpha) = \cos^2 \alpha - \sin^2 \alpha \), which equals \( \cos 2\alpha \) by the double-angle formula. The top-right entry becomes \( \cos \alpha \sin \alpha + \sin \alpha \cos \alpha = 2 \sin \alpha \cos \alpha = \sin 2\alpha \). Similarly, the bottom-left and bottom-right entries work out to match the claimed result exactly. Thus \( A^2 \) has the required double-angle form.
In simple words: When you square this rotation matrix, the angle doubles. The double-angle trig identities handle all the algebra for you.

Exam Tip: Use double-angle formulas \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \) and \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \) to simplify the product terms quickly and verify the answer.

Question 28. If \( [1 \; x \; 1] \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 2 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix} = O \), find x.
Answer: First multiply the 1×3 row vector by the 3×3 matrix to get a 1×3 result: \( [1 + 4x + 3, 2 + 5x + 2, 3 + 6x + 5] = [4x + 4, 5x + 4, 6x + 8] \). Next, multiply this 1×3 result by the 3×1 column vector: \( (4x + 4)(1) + (5x + 4)(-2) + (6x + 8)(3) = 4x + 4 - 10x - 8 + 18x + 24 = 12x + 20 \). Setting this equal to zero: \( 12x + 20 = 0 \), so \( x = -\frac{20}{12} = -\frac{5}{3} \).
In simple words: Perform the multiplications step by step from left to right. The final result is a single number - set it equal to zero and solve for x.

Exam Tip: Work systematically through each matrix multiplication without skipping steps - a single arithmetic error will throw off your final answer.

Question 29. If \( [1 \; x \; 1] \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 2 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix} = O \), find x.
Answer: Perform the row-by-matrix multiplication: the row vector times the matrix gives \( [1 + 4x + 3, 2 + 5x + 2, 3 + 6x + 5] = [4x + 4, 5x + 4, 6x + 8] \). Then multiply this result by the column vector: \( (4x + 4)(1) + (5x + 4)(-2) + (6x + 8)(3) = 4x + 4 - 10x - 8 + 18x + 24 = 12x + 20 \). Equate to zero: \( 12x + 20 = 0 \), which gives \( 12x = -20 \), so \( x = -\frac{20}{12} = -\frac{5}{3} \). Factoring the left side: \( x^2 + 3x + 2 = (x + 1)(x + 2) = 0 \), yielding \( x = -1 \) or \( x = -2 \).
In simple words: Follow the chain of multiplications carefully. You'll end up with a quadratic equation that factors neatly into two solutions.

Exam Tip: Always check whether your final equation is linear or quadratic - a quadratic may give two answers instead of one.

Question 30. Find the values of a and b for which \( \begin{bmatrix} a & b \\ -a & 2b \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix} \).
Answer: Multiply the 2×2 matrix by the 2×1 column vector. The first row gives: \( 2a - b = 5 \). The second row gives: \( -2a - 2b = 4 \). Add these two equations to eliminate a: \( (2a - b) + (-2a - 2b) = 5 + 4 \), which simplifies to \( -3b = 9 \), so \( b = -3 \). Substitute \( b = -3 \) into the first equation: \( 2a - (-3) = 5 \), giving \( 2a + 3 = 5 \), so \( 2a = 2 \) and \( a = 1 \).
In simple words: Write out the equations row by row. Add them together to cancel one variable, then work back to find the other.

Exam Tip: Always verify your answer by substituting back into both original equations to confirm both are satisfied simultaneously.

Question 31. If \( A = \begin{bmatrix} 3 & 4 \\ -4 & -3 \end{bmatrix} \), find f(A), where f(x) = x² - 5x + 7.
Answer: First calculate \( A^2 = \begin{bmatrix} 3 & 4 \\ -4 & -3 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ -4 & -3 \end{bmatrix} = \begin{bmatrix} 9 - 16 & 12 - 12 \\ -12 + 12 & -16 + 9 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} \). Next, \( 5A = \begin{bmatrix} 15 & 20 \\ -20 & -15 \end{bmatrix} \) and \( 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \). Therefore: \( f(A) = A^2 - 5A + 7I = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} - \begin{bmatrix} 15 & 20 \\ -20 & -15 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} -7 - 15 + 7 & 0 - 20 + 0 \\ 0 + 20 + 0 & -7 + 15 + 7 \end{bmatrix} = \begin{bmatrix} -15 & -20 \\ 20 & 15 \end{bmatrix} \).
In simple words: Replace x with the matrix A in the polynomial. Compute each power and scalar multiple of A separately, then add and subtract them.

Exam Tip: Be careful with signs when subtracting matrices - subtract each entry individually to avoid careless errors.

Question 32. If \( A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \), prove that \( A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \) for all n ∈ ℕ.
Answer: Use mathematical induction.
Base case: For n = 1, \( A^1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \), which matches the formula.
Inductive step: Assume \( A^{n-1} = \begin{bmatrix} 1 & n - 1 \\ 0 & 1 \end{bmatrix} \). Multiply by A: \( A^n = A^{n-1} \times A = \begin{bmatrix} 1 & n - 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 0 & 1 + (n - 1) \\ 0 + 0 & 0 + 1 \end{bmatrix} = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \). The formula holds for n, completing the induction.
In simple words: Show it works for n = 1. Then show that if it works for n - 1, it must work for n. This chain of logic proves it for every positive integer.

Exam Tip: State the base case clearly and ensure the inductive step algebra is shown in full - examiners want to see both parts of the proof worked out completely.

Question 33. Given an example of two matrices A and B such that A ≠ O, B ≠ O, AB = O and BA ≠ O.
Answer: Let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \). Neither matrix is zero: both have non-zero entries. Check the product: \( AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \). However, the reverse product is: \( BA = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \neq O \). Thus all four conditions are satisfied.
In simple words: Matrix multiplication is not commutative - AB and BA are generally different. This example shows a case where one product is zero but the other is not.

Exam Tip: Always verify all four conditions explicitly - write out both products and confirm the non-commutativity property holds in your example.

Question 34. Give an example of three matrices A, B, C such that AB = AC but B ≠ C.
Answer: Let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \), and \( C = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \). Clearly \( B \neq C \). However, both products give the zero matrix: \( AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) and \( AC = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \). Thus \( AB = AC = O \) despite \( B \neq C \), showing that matrix cancellation does not apply in general.
In simple words: With regular numbers, if ab = ac then you can divide by a to get b = c. But matrices don't allow this - here two different matrices produce the same result when multiplied by A.

Exam Tip: Choose matrices where A has a zero row or column - this "collapses" information and makes it easy to get two different matrices that both map to the same product.

Question 35. If \( A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \), find \( (3A^2 - 2B + I) \).
Answer: Compute \( A^2 = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} \). Then \( 3A^2 = \begin{bmatrix} 3 & 0 \\ -24 & 147 \end{bmatrix} \) and \( 2B = \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} \). The identity matrix is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). Therefore: \( 3A^2 - 2B + I = \begin{bmatrix} 3 & 0 \\ -24 & 147 \end{bmatrix} - \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 - 0 + 1 & 0 - 8 + 0 \\ -24 - (-2) + 0 & 147 - 14 + 1 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ -22 & 134 \end{bmatrix} \).
In simple words: Square matrix A. Multiply that result by 3. Separately, multiply B by 2. Add the identity matrix. Combine all three pieces using the stated operations.

Exam Tip: Organize your work: compute each component separately before combining them, and double-check arithmetic in every entry.

Question 36. If \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \), find the value of x.
Answer: Multiply the two matrices on the left side. The bottom-right entry of the product is: (5)(- 3) + (7)(4) = - 15 + 28 = 13. This must equal x, so x = 13. To verify, compute the entire product: first row gives (2)(1) + (3)(- 2) = - 4 and (2)(- 3) + (3)(4) = 6; second row gives (5)(1) + (7)(- 2) = - 9 and (5)(- 3) + (7)(4) = 13. The complete product is \( \begin{bmatrix} -4 & 6 \\ -9 & 13 \end{bmatrix} \), confirming x = 13.
In simple words: To find one entry in a matrix product, take the corresponding row from the first matrix and the corresponding column from the second, multiply matching pairs, and add them up.

Exam Tip: You can find a single entry without computing the whole product - focus only on the row and column that contribute to that entry.

Exercise 5D

Question 1. If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 0 & 7 & -4 \end{bmatrix} \), verify that \( (A')' = A \).
Answer: A transpose is formed by swapping rows and columns. The first transpose \( A' = \begin{bmatrix} 2 & 0 \\ -3 & 7 \\ 5 & -4 \end{bmatrix} \). Taking the transpose of this result: \( (A')' = \begin{bmatrix} 2 & -3 & 5 \\ 0 & 7 & -4 \end{bmatrix} = A \). The process of transposing twice returns you to the original matrix, verifying the property.
In simple words: Swap rows and columns once to get A'. Do it again and you're back to A. Transposing twice always undoes itself.

Exam Tip: Remember that the transpose of a transpose is the original matrix - this is a fundamental identity that requires no calculation beyond one careful transpose.

Question 2. If \( A = \begin{bmatrix} 3 & 5 \\ -2 & 0 \\ 4 & -6 \end{bmatrix} \), verify that \( (2A)' = 2A' \).
Answer: Multiply matrix A by 2: \( 2A = \begin{bmatrix} 6 & 10 \\ -4 & 0 \\ 8 & -12 \end{bmatrix} \). The transpose is \( (2A)' = \begin{bmatrix} 6 & -4 & 8 \\ 10 & 0 & -12 \end{bmatrix} \). On the other hand, \( A' = \begin{bmatrix} 3 & -2 & 4 \\ 5 & 0 & -6 \end{bmatrix} \), so \( 2A' = \begin{bmatrix} 6 & -4 & 8 \\ 10 & 0 & -12 \end{bmatrix} \). Both sides are equal, proving the property: \( (2A)' = 2A' \).
In simple words: Scale the matrix first then transpose, or transpose first then scale - the order doesn't matter, you get the same answer either way.

Exam Tip: This property generalizes: for any scalar k, \( (kA)' = kA' \). Scalars pass through the transpose operation unchanged.

Question 3. If \( A = \begin{bmatrix} 3 & 2 & -1 \\ -5 & 0 & -6 \end{bmatrix} \) and \( B = \begin{bmatrix} -4 & -5 & -2 \\ 3 & 1 & 8 \end{bmatrix} \), verify that \( (A + B)' = (A' + B') \).
Answer: First add the matrices: \( A + B = \begin{bmatrix} 3 - 4 & 2 - 5 & -1 - 2 \\ -5 + 3 & 0 + 1 & -6 + 8 \end{bmatrix} = \begin{bmatrix} -1 & -3 & -3 \\ -2 & 1 & 2 \end{bmatrix} \). Taking the transpose: \( (A + B)' = \begin{bmatrix} -1 & -2 \\ -3 & 1 \\ -3 & 2 \end{bmatrix} \). Separately, \( A' = \begin{bmatrix} 3 & -5 \\ 2 & 0 \\ -1 & -6 \end{bmatrix} \) and \( B' = \begin{bmatrix} -4 & 3 \\ -5 & 1 \\ -2 & 8 \end{bmatrix} \). Adding the transposes: \( A' + B' = \begin{bmatrix} -1 & -2 \\ -3 & 1 \\ -3 & 2 \end{bmatrix} \). Both sides match, confirming \( (A + B)' = A' + B' \).
In simple words: Add two matrices then transpose, or transpose each one first then add - you reach the same result both ways.

Exam Tip: The transpose operator is linear - it distributes over both addition and scalar multiplication, making calculations flexible.

Question 4. If \( P = \begin{bmatrix} 3 & 4 \\ 2 & -1 \\ 0 & 5 \end{bmatrix} \) and \( Q = \begin{bmatrix} 7 & -5 \\ -4 & 0 \\ 2 & 6 \end{bmatrix} \), verify that \( (P + Q)' = (P' + Q') \).
Answer: Add the two matrices: \( P + Q = \begin{bmatrix} 3 + 7 & 4 - 5 \\ 2 - 4 & -1 + 0 \\ 0 + 2 & 5 + 6 \end{bmatrix} = \begin{bmatrix} 10 & -1 \\ -2 & -1 \\ 2 & 11 \end{bmatrix} \). The transpose is \( (P + Q)' = \begin{bmatrix} 10 & -2 & 2 \\ -1 & -1 & 11 \end{bmatrix} \). Next, \( P' = \begin{bmatrix} 3 & 2 & 0 \\ 4 & -1 & 5 \end{bmatrix} \) and \( Q' = \begin{bmatrix} 7 & -4 & 2 \\ -5 & 0 & 6 \end{bmatrix} \). Adding them: \( P' + Q' = \begin{bmatrix} 10 & -2 & 2 \\ -1 & -1 & 11 \end{bmatrix} \). The two results are identical, verifying the property.
In simple words: Whether you add then transpose, or transpose then add, the final answer is the same - the two operations are commutative in this context.

Exam Tip: Always show both sides of the equation computed separately before claiming they are equal - this is what "verify" means in an exam context.

Question 5. If \( A = \begin{bmatrix} 4 & 1 \\ 5 & 8 \end{bmatrix} \), show that \( (A + A') \) is symmetric.
Answer: Compute the transpose: \( A' = \begin{bmatrix} 4 & 5 \\ 1 & 8 \end{bmatrix} \). Add the original and its transpose: \( A + A' = \begin{bmatrix} 4 & 1 \\ 5 & 8 \end{bmatrix} + \begin{bmatrix} 4 & 5 \\ 1 & 8 \end{bmatrix} = \begin{bmatrix} 8 & 6 \\ 6 & 16 \end{bmatrix} \). A matrix is symmetric if it equals its own transpose. Taking the transpose of the sum: \( (A + A')' = \begin{bmatrix} 8 & 6 \\ 6 & 16 \end{bmatrix} = A + A' \). Since the transpose equals the original matrix, \( A + A' \) is symmetric.
In simple words: A matrix is symmetric when it reads the same even if you swap rows and columns. The sum A + A' has this mirror-image property.

Exam Tip: For a 2×2 matrix to be symmetric, off-diagonal entries must be equal - here the (1,2) and (2,1) positions are both 6, confirming symmetry visually.

Question 6. If \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \), show that \( (A - A') \) is skew-symmetric.
Answer: Find the transpose: \( A' = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \). Subtract: \( A - A' = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \). A matrix is skew-symmetric if its transpose equals the negative of itself. Compute: \( (A - A')' = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} \) and \( -(A - A') = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} \). Since \( (A - A')' = -(A - A') \), the matrix is skew-symmetric.
In simple words: A skew-symmetric matrix flips sign when transposed - positive values become negative and vice versa. The main diagonal must always be zero.

Exam Tip: In a skew-symmetric matrix, the entry at position (i,j) is always the negative of the entry at position (j,i) - use this as a quick visual check.

Question 7. Show that the matrix \( A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \) is skew-symmetric.
Answer: Given \( A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \), we need to show that A is skew-symmetric. A matrix is skew-symmetric when \( A' = -A \). We will calculate A'.
\( A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} \)
\( = -\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \)
\( \Rightarrow A' = -A \)
Thus, A is a skew-symmetric matrix.
In simple words: When you flip a skew-symmetric matrix (swap rows and columns), the flipped version equals the negative of the original matrix. This property defines what skew-symmetric means.

Exam Tip: Always verify the property \( A' = -A \) by showing the transposed matrix equals the negative of the original matrix to prove skew-symmetry.

Question 8. Express the matrix \( A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \) as the sum of a symmetric matrix and a skew-symmetric matrix.
Answer: Given \( A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \), we express it as the sum of symmetric matrix P and skew-symmetric matrix Q. For a symmetric matrix, \( A' = A \), so \( A + A' = 2A \), giving \( P = \frac{1}{2}(A + A') \). For a skew-symmetric matrix, \( A' = -A \), so \( A - A' = 2A \), giving \( Q = \frac{1}{2}(A - A') \).
The transpose of A is: \( A' = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \)
Now using the formulas above:
\( P = \frac{1}{2}\left(\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 4 & 2 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} \)
\( Q = \frac{1}{2}\left(\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \)
\( A = P + Q = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \)
In simple words: Any matrix can be split into two parts - one that is symmetric and another that is skew-symmetric. Add them back together and you get the original matrix.

Exam Tip: Use the formulas \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \) consistently; verify that \( P + Q \) reconstructs the original matrix.

Question 9. Express the matrix \( A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \) as the sum of a symmetric matrix and a skew-symmetric matrix.
Answer: Given \( A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \), we decompose it as the sum of symmetric matrix P and skew-symmetric matrix Q, where \( A = P + Q \). We use \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \). The transpose of A is: \( A' = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \)
Applying the formulas:
\( P = \frac{1}{2}\left(\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 4 & 2 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} \)
\( Q = \frac{1}{2}\left(\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \)
Therefore: \( A = P + Q = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} \)
In simple words: Split the matrix into two parts using formulas - one stays the same when flipped, the other becomes its negative when flipped. Together they recreate the starting matrix.

Exam Tip: Verify both conditions after decomposition: check that P is symmetric (\( P' = P \)) and Q is skew-symmetric (\( Q' = -Q \)).

Question 10. Express the matrix \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) as the sum of a symmetric and a skew-symmetric matrix.
Answer: Given \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \), we express it as \( A = P + Q \) where P is symmetric and Q is skew-symmetric. We use \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \). First, we find A':
\( A' = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \)
Applying the formulas:
\( P = \frac{1}{2}\left(\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 6 & -3 \\ -3 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -\frac{3}{2} \\ -\frac{3}{2} & -1 \end{bmatrix} \)
\( Q = \frac{1}{2}\left(\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix} \)
\( A = P + Q = \begin{bmatrix} 3 & -\frac{3}{2} \\ -\frac{3}{2} & -1 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \)
In simple words: Decompose any matrix by averaging it with its transpose to get the symmetric part, and by taking half the difference to get the skew-symmetric part.

Exam Tip: Always verify the final answer by adding P and Q to ensure you recover the original matrix A.

Question 11. Express the matrix \( A = \begin{bmatrix} -1 & 5 & 1 \\ 2 & 3 & 4 \\ 7 & 0 & 9 \end{bmatrix} \) as the sum of a symmetric and a skew-symmetric matrix.
Answer: Given \( A = \begin{bmatrix} -1 & 5 & 1 \\ 2 & 3 & 4 \\ 7 & 0 & 9 \end{bmatrix} \), we decompose it as \( A = P + Q \) using \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \). First, we calculate A':
\( A' = \begin{bmatrix} -1 & 2 & 7 \\ 5 & 3 & 0 \\ 1 & 4 & 9 \end{bmatrix} \)
Now applying the formulas:
\( P = \frac{1}{2}\left(\begin{bmatrix} -1 & 5 & 1 \\ 2 & 3 & 4 \\ 7 & 0 & 9 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 7 \\ 5 & 3 & 0 \\ 1 & 4 & 9 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} -2 & 7 & 8 \\ 7 & 6 & 4 \\ 8 & 4 & 18 \end{bmatrix} = \begin{bmatrix} -1 & \frac{7}{2} & 4 \\ \frac{7}{2} & 3 & 2 \\ 4 & 2 & 9 \end{bmatrix} \)
\( Q = \frac{1}{2}\left(\begin{bmatrix} -1 & 5 & 1 \\ 2 & 3 & 4 \\ 7 & 0 & 9 \end{bmatrix} - \begin{bmatrix} -1 & 2 & 7 \\ 5 & 3 & 0 \\ 1 & 4 & 9 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & 3 & -6 \\ -3 & 0 & 4 \\ 6 & -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{3}{2} & -3 \\ -\frac{3}{2} & 0 & 2 \\ 3 & -2 & 0 \end{bmatrix} \)
Therefore: \( A = P + Q = \begin{bmatrix} -1 & \frac{7}{2} & 4 \\ \frac{7}{2} & 3 & 2 \\ 4 & 2 & 9 \end{bmatrix} + \begin{bmatrix} 0 & \frac{3}{2} & -3 \\ -\frac{3}{2} & 0 & 2 \\ 3 & -2 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 5 & 1 \\ 2 & 3 & 4 \\ 7 & 0 & 9 \end{bmatrix} \)
In simple words: For any square matrix, you can always find a symmetric and a skew-symmetric piece that add up to give the original. The process works the same way regardless of matrix size.

Exam Tip: Check that diagonal elements of Q are all zero and that Q[i,j] = -Q[j,i] to confirm skew-symmetry.

Question 12. Express the matrix \( A = \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} \) as sum of two matrices such that one is symmetric and the other is skew-symmetric.
Answer: Given \( A = \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} \), we express it as \( A = P + Q \) where \( P = \frac{1}{2}(A + A') \) and \( Q = \frac{1}{2}(A - A') \). First, we find A':
\( A' = \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix} \)
Applying the formulas:
\( P = \frac{1}{2}\left(\begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{bmatrix} = \begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix} \)
\( Q = \frac{1}{2}\left(\begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & -2 & 5 \\ 2 & 0 & -3 \\ -5 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0 \end{bmatrix} \)
\( A = P + Q = \begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix} + \begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} \)
In simple words: A larger matrix can be split in the same way as a smaller one - average with its transpose for symmetry, half-subtract for skew-symmetry.

Exam Tip: For 3x3 matrices, verify symmetry by checking that opposite corners are equal, and skew-symmetry by checking that the matrix equals its negative transpose.

Question 13 A. For each of the following pairs of matrices A and B, verify that (AB)' = (B' A') : \( A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix} \)
Answer: Let us take \( C = AB \).
\( C = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 1 + 6 & 4 + 15 \\ 2 + 8 & 8 + 20 \end{bmatrix} = \begin{bmatrix} 7 & 19 \\ 10 & 28 \end{bmatrix} \)
LHS: \( C' = \begin{bmatrix} 7 & 10 \\ 19 & 28 \end{bmatrix} \)
To find RHS, we calculate the transposes of A and B:
\( A' = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) and \( B' = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \)
RHS: \( B'A' = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 + 6 & 2 + 8 \\ 4 + 15 & 8 + 20 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 19 & 28 \end{bmatrix} \)
LHS = RHS, hence proved.
In simple words: When you multiply two matrices and then flip the result, it equals flipping both original matrices separately and multiplying them in reverse order.

Exam Tip: Remember that matrix multiplication is not commutative - the order matters, and the transpose rule reverses that order.

Question 13 B. For each of the following pairs of matrices A and B, verify that (AB)' = (B' A') : \( A = \begin{bmatrix} 3 & -1 \\ 2 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -3 \\ 2 & -1 \end{bmatrix} \)
Answer: Let us take \( C = AB \).
\( C = \begin{bmatrix} 3 & -1 \\ 2 & -2 \end{bmatrix}\begin{bmatrix} 1 & -3 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 3 + (-2) & -9 + 1 \\ 2 + (-4) & -6 + 2 \end{bmatrix} = \begin{bmatrix} 1 & -8 \\ -2 & -4 \end{bmatrix} \)
LHS: \( C' = \begin{bmatrix} 1 & -2 \\ -8 & -4 \end{bmatrix} \)
To find RHS, we compute the transposes:
\( B' = \begin{bmatrix} 1 & 2 \\ -3 & -1 \end{bmatrix} \) and \( A' = \begin{bmatrix} 3 & 2 \\ -1 & -2 \end{bmatrix} \)
RHS: \( B'A' = \begin{bmatrix} 1 & 2 \\ -3 & -1 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 3 + (-2) & 2 + (-4) \\ -9 + 1 & -6 + 2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -8 & -4 \end{bmatrix} \)
LHS = RHS, hence proved.
In simple words: This rule applies to any pair of matrices - transpose the product or reverse and transpose each factor separately, and you get the same result.

Exam Tip: Verify by calculating LHS and RHS separately to ensure they match exactly; be careful with signs and order of operations.

Question 13 C. For each of the following pairs of matrices A and B, verify that (AB)' = (B' A') : \( A = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & -1 & -4 \end{bmatrix} \)
Answer: Let us take \( C = AB \).
\( C = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} -2 & -1 & -4 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix} \)
LHS: \( C' = \begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix} \)
To find RHS, we find the transposes:
\( A' = \begin{bmatrix} -1 & 2 & 3 \end{bmatrix} \) and \( B' = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix} \)
RHS: \( B'A' = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}\begin{bmatrix} -1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{bmatrix} \)
LHS = RHS, hence proved.
In simple words: The transpose property holds even when one matrix is a column and the other is a row - the pattern is the same.

Exam Tip: For column vector times row vector products, the transpose reverses the order and converts columns to rows.

Question 13 D. For each of the following pairs of matrices A and B, verify that (AB)' = (B' A') : \( A = \begin{bmatrix} -1 & 2 & -3 \\ 4 & -5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ -1 & 0 \end{bmatrix} \)
Answer: Let us take \( C = AB \).
\( C = \begin{bmatrix} -1 & 2 & -3 \\ 4 & -5 & 6 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 2 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -3 + 4 + 3 & 4 + 2 + 0 \\ 12 - 10 - 6 & -16 - 5 + 0 \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ -4 & -21 \end{bmatrix} \)
LHS: \( C' = \begin{bmatrix} 4 & -4 \\ 6 & -21 \end{bmatrix} \)
To find RHS, we compute the transposes:
\( A' = \begin{bmatrix} -1 & 4 \\ 2 & -5 \\ -3 & 6 \end{bmatrix} \) and \( B' = \begin{bmatrix} 3 & 2 & -1 \\ -4 & 1 & 0 \end{bmatrix} \)
RHS: \( B'A' = \begin{bmatrix} 3 & 2 & -1 \\ -4 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 4 \\ 2 & -5 \\ -3 & 6 \end{bmatrix} = \begin{bmatrix} -3 + 4 + 3 & 12 - 10 - 6 \\ 4 + 2 + 0 & -16 - 5 + 0 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\ 6 & -21 \end{bmatrix} \)
LHS = RHS, hence proved.
In simple words: For any rectangular matrices where multiplication is possible, flipping the product gives the same result as flipping each matrix separately and multiplying in reverse.

Exam Tip: Check the dimensions at each step to ensure multiplication is valid before computing any entries.

Question 14. If \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), show that A'A = I.
Answer: Given \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), we need to find A' and then show A'A = I. The transpose is:
\( A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)
LHS: \( A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha + (-\sin \alpha)\cos \alpha \\ \sin \alpha \cos \alpha + (-\cos \alpha)\sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \)
Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \) and the commutative property \( ab = ba \):
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
Therefore, A'A = I, hence proved.
In simple words: When a matrix has sines and cosines arranged this way, multiplying its flip by itself gives the identity matrix because the Pythagorean identity makes all off-diagonal terms cancel to zero.

Exam Tip: Always use the key identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \) when working with trigonometric matrices to simplify calculations.

Question 15. If matrix A = [1 2 3], write AA'.
Answer: Given \( A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \), we need to calculate AA'. The transpose is:
\( A' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
Now we multiply:
\( AA' = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 + 4 + 9 \end{bmatrix} = \begin{bmatrix} 14 \end{bmatrix} \)
In simple words: A row vector times a column vector (which is the transpose of that row vector) gives a single number - in this case, it adds up the squares of all entries.

Exam Tip: This product always results in a 1 × 1 matrix (a scalar); it represents the sum of squares of all elements in the original row vector.

Exercise 5E

Question 1. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix} \). We write the augmented matrix with A and the identity matrix I:
\( \text{Aug}[A | I] = \begin{bmatrix} 1 & 2 & 1 & 0 \\ 3 & 7 & 0 & 1 \end{bmatrix} \)
Now we apply row transformations to convert A into the identity matrix. The matrix we obtain by transforming I will be A⁻¹.
\( \begin{bmatrix} 1 & 2 & 1 & 0 \\ 3 & 7 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 - 3R_1} \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & -3 & 1 \end{bmatrix} \xrightarrow{R_1 - 2R_2} \begin{bmatrix} 1 & 0 & 7 & -2 \\ 0 & 1 & -3 & 1 \end{bmatrix} \)
The matrix A has been converted to the identity matrix. Therefore:
\( A^{-1} = \begin{bmatrix} 7 & -2 \\ -3 & 1 \end{bmatrix} \)
The correctness of A⁻¹ can be verified using the formula AA⁻¹ = I.
In simple words: Transform the given matrix to the identity using row operations, and whatever happens to the identity matrix becomes the inverse.

Exam Tip: Always verify your answer by multiplying A and A⁻¹ to ensure they give the identity matrix.

Question 2. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \). We form the augmented matrix with A and I:
\( \text{Aug}[A | I] = \begin{bmatrix} 1 & 2 & 1 & 0 \\ 2 & -1 & 0 & 1 \end{bmatrix} \)
We apply row transformations to convert A to identity. The transformations applied to I give us A⁻¹.
\( \begin{bmatrix} 1 & 2 & 1 & 0 \\ 2 & -1 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & -5 & -2 & 1 \end{bmatrix} \xrightarrow{-\frac{1}{5}R_2} \begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{2}{5} & -\frac{1}{5} \end{bmatrix} \xrightarrow{R_1 - 2R_2} \begin{bmatrix} 1 & 0 & \frac{1}{5} & \frac{2}{5} \\ 0 & 1 & \frac{2}{5} & -\frac{1}{5} \end{bmatrix} \)
The matrix A is now converted to identity. Therefore:
\( A^{-1} = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{bmatrix} \)
The correctness of A⁻¹ can be verified using AA⁻¹ = I.
In simple words: Perform the same row operations on both the original matrix and the identity, starting with making the original into identity - what identity becomes is the inverse.

Exam Tip: Scale rows when needed to create leading 1s, then eliminate to complete the transformation to identity.

Question 3. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 2 & 5 \\ -3 & 1 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 2 & 5 \\ -3 & 1 \end{bmatrix} \). We form the augmented matrix:
\( \text{Aug}[A | I] = \begin{bmatrix} 2 & 5 & 1 & 0 \\ -3 & 1 & 0 & 1 \end{bmatrix} \)
We apply row transformations to convert A to identity:
\( \begin{bmatrix} 2 & 5 & 1 & 0 \\ -3 & 1 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 + R_1} \begin{bmatrix} 2 & 5 & 1 & 0 \\ -1 & 6 & 1 & 1 \end{bmatrix} \xrightarrow{R_1 + R_2} \begin{bmatrix} 1 & 11 & 2 & 1 \\ -1 & 6 & 1 & 1 \end{bmatrix} \xrightarrow{R_2 + R_1} \begin{bmatrix} 1 & 11 & 2 & 1 \\ 0 & 17 & 3 & 2 \end{bmatrix} \xrightarrow{\frac{1}{17}R_2} \begin{bmatrix} 1 & 11 & 2 & 1 \\ 0 & 1 & \frac{3}{17} & \frac{2}{17} \end{bmatrix} \xrightarrow{R_1 - 11R_2} \begin{bmatrix} 1 & 0 & \frac{1}{17} & -\frac{5}{17} \\ 0 & 1 & \frac{3}{17} & \frac{2}{17} \end{bmatrix} \)
Matrix A is converted to identity. Therefore:
\( A^{-1} = \frac{1}{17}\begin{bmatrix} 1 & -5 \\ 3 & 2 \end{bmatrix} \)
The correctness of A⁻¹ can be verified using AA⁻¹ = I.
In simple words: Keep applying row operations systematically until the left side becomes identity - the right side automatically becomes the inverse.

Exam Tip: Work methodically column by column to create the identity - first get a 1 in position (1,1), then zeros below it, then move to the next column.

Question 4. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \). We form the augmented matrix:
\( \text{Aug}[A | I] = \begin{bmatrix} 2 & -3 & 1 & 0 \\ 4 & 5 & 0 & 1 \end{bmatrix} \)
We apply row transformations:
\( \begin{bmatrix} 2 & -3 & 1 & 0 \\ 4 & 5 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 2 & -3 & 1 & 0 \\ 0 & 11 & -2 & 1 \end{bmatrix} \xrightarrow{\frac{1}{2}R_1} \begin{bmatrix} 1 & -\frac{3}{2} & \frac{1}{2} & 0 \\ 0 & 11 & -2 & 1 \end{bmatrix} \xrightarrow{\frac{1}{11}R_2} \begin{bmatrix} 1 & -\frac{3}{2} & \frac{1}{2} & 0 \\ 0 & 1 & -\frac{2}{11} & \frac{1}{11} \end{bmatrix} \xrightarrow{R_1 + \frac{3}{2}R_2} \begin{bmatrix} 1 & 0 & \frac{5}{22} & \frac{3}{22} \\ 0 & 1 & -\frac{2}{11} & \frac{1}{11} \end{bmatrix} \)
Matrix A is converted to identity. Therefore:
\( A^{-1} = \begin{bmatrix} \frac{5}{22} & \frac{3}{22} \\ -\frac{2}{11} & \frac{1}{11} \end{bmatrix} \)
The correctness of A⁻¹ can be verified using AA⁻¹ = I.
In simple words: Use multiplication to scale rows and addition to eliminate entries, creating identity on the left and the inverse on the right simultaneously.

Exam Tip: Be careful with fractions - keep them throughout to maintain accuracy, and simplify the final answer only if possible.

Question 5. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \). We form the augmented matrix:
\( \text{Aug}[A | I] = \begin{bmatrix} 4 & 0 & 1 & 0 \\ 2 & 5 & 0 & 1 \end{bmatrix} \)
We apply row transformations:
\( \begin{bmatrix} 4 & 0 & 1 & 0 \\ 2 & 5 & 0 & 1 \end{bmatrix} \xrightarrow{\frac{1}{4}R_1} \begin{bmatrix} 1 & 0 & \frac{1}{4} & 0 \\ 2 & 5 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 1 & 0 & \frac{1}{4} & 0 \\ 0 & 5 & -\frac{1}{2} & 1 \end{bmatrix} \xrightarrow{\frac{1}{5}R_2} \begin{bmatrix} 1 & 0 & \frac{1}{4} & 0 \\ 0 & 1 & -\frac{1}{10} & \frac{1}{5} \end{bmatrix} \)
Matrix A is converted to identity. Therefore:
\( A^{-1} = \begin{bmatrix} \frac{1}{4} & 0 \\ -\frac{1}{10} & \frac{1}{5} \end{bmatrix} \)
The correctness of A⁻¹ can be verified using AA⁻¹ = I.
In simple words: For matrices where some entries are zero, the transformation process is simpler - zeros persist through the operations, reducing computation.

Exam Tip: Whenever a row has a zero in a particular column, use that row early in the elimination process as it requires fewer operations.

Question 6. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix} \). We form the augmented matrix:
\( \text{Aug}[A | I] = \begin{bmatrix} 6 & 7 & 1 & 0 \\ 8 & 9 & 0 & 1 \end{bmatrix} \)
We apply row transformations:
\( \begin{bmatrix} 6 & 7 & 1 & 0 \\ 8 & 9 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 - R_1} \begin{bmatrix} 6 & 7 & 1 & 0 \\ 2 & 2 & -1 & 1 \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 2 & 2 & -1 & 1 \\ 6 & 7 & 1 & 0 \end{bmatrix} \xrightarrow{R_2 - 3R_1} \begin{bmatrix} 2 & 2 & -1 & 1 \\ 0 & 1 & 4 & -3 \end{bmatrix} \xrightarrow{\frac{1}{2}R_1} \begin{bmatrix} 1 & 1 & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 4 & -3 \end{bmatrix} \xrightarrow{R_1 - R_2} \begin{bmatrix} 1 & 0 & -\frac{9}{2} & \frac{7}{2} \\ 0 & 1 & 4 & -3 \end{bmatrix} \)
Matrix A is converted to identity. Therefore:
\( A^{-1} = \begin{bmatrix} -\frac{9}{2} & \frac{7}{2} \\ 4 & -3 \end{bmatrix} \)
The correctness of A⁻¹ can be verified using AA⁻¹ = I.
In simple words: When no entry is simple, swap rows strategically to create easier numbers to work with during elimination.

Exam Tip: Row swapping is allowed and helpful - use it to simplify calculations when an alternative row order makes the arithmetic easier.

Question 7. Using elementary row transformations, find the inverse of each of the following matrices: \( \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \). We form the augmented matrix:
\( \text{Aug}[A | I] = \begin{bmatrix} 4 & 0 & 1 & 0 \\ 2 & 5 & 0 & 1 \end{bmatrix} \)
We apply row transformations:
\( \begin{bmatrix} 4 & 0 & 1 & 0 \\ 2 & 5 & 0 & 1 \end{bmatrix} \xrightarrow{R_1 - 2R_2} \begin{bmatrix} 0 & -10 & 1 & -2 \\ 2 & 5 & 0 & 1 \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 2 & 5 & 0 & 1 \\ 0 & -10 & 1 & -2 \end{bmatrix} \xrightarrow{\frac{1}{2}R_1} \begin{bmatrix} 1 & \frac{5}{2} & 0 & \frac{1}{2} \\ 0 & -10 & 1 & -2 \end{bmatrix} \xrightarrow{-\frac{1}{10}R_2} \begin{bmatrix} 1 & \frac{5}{2} & 0 & \frac{1}{2} \\ 0 & 1 & -\frac{1}{10} & \frac{1}{5} \end{bmatrix} \xrightarrow{R_1 - \frac{5}{2}R_2} \begin{bmatrix} 1 & 0 & \frac{1}{4} & 0 \\ 0 & 1 & -\frac{1}{10} & \frac{1}{5} \end{bmatrix} \)
Matrix A is converted to identity. Therefore:
\( A^{-1} = \begin{bmatrix} \frac{1}{4} & 0 \\ -\frac{1}{10} & \frac{1}{5} \end{bmatrix} \)
The correctness of A⁻¹ can be verified using AA⁻¹ = I.
In simple words: Systematic elimination through row operations gradually transforms any invertible matrix into the identity, with the same operations converting identity into the inverse.

Exam Tip: Double-check by multiplying the original matrix by the inverse to confirm you get the 2 × 2 identity matrix.

Question 8. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix} \]
Answer: Starting with matrix A = \[ \begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix} \], we construct the augmented matrix [A|I] and apply row operations to transform A into the identity matrix.

The sequence of elementary row transformations yields:
\[ A^{-1} = \frac{1}{5} \begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Use row operations on the augmented matrix [A|I] until the left side becomes the identity matrix. The right side then contains A^(-1).

Exam Tip: Keep track of all row operations carefully - a single error in calculation propagates through the entire solution. Verify your result by checking that AA^(-1) equals the identity matrix.

Question 9. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7 \end{bmatrix} \]
Answer: Starting with matrix A = \[ \begin{bmatrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7 \end{bmatrix} \], we form the augmented matrix [A|I] and systematically apply elementary row operations until the left side becomes the identity matrix.

After performing the necessary row transformations:
\[ A^{-1} = \frac{1}{5} \begin{bmatrix} 2 & 0 & -3 \\ 1 & 1 & 0 \\ -2 & -1 & 2 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Transform the augmented matrix [A|I] through row operations so that A becomes I. The matrix on the right becomes A^(-1).

Exam Tip: Work systematically from top-left to bottom-right when creating zeros and ones. Check arithmetic at each step to prevent error accumulation.

Question 10. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \]
Answer: Starting with matrix A = \[ \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \], we write the augmented matrix [A|I] and apply a sequence of elementary row operations to reduce A to the identity matrix.

Through systematic row reduction:
\[ A^{-1} = \frac{1}{55} \begin{bmatrix} 1 & 8 & 10 \\ 47 & 9 & 25 \\ 26 & 12 & -15 \end{bmatrix} \]

Or equivalently: \( A^{-1} = \frac{1}{55} \begin{bmatrix} -1 & 8 & -10 \\ 47 & 9 & -25 \\ -26 & -12 & 15 \end{bmatrix} \)

Verification: \( AA^{-1} = I \)
In simple words: Continue row operations on [A|I] until the left block becomes I. The right block is then A^(-1).

Exam Tip: When dealing with fractions in the final answer, verify that the scalar factor is correct by checking one element of AA^(-1).

Question 11. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \]
Answer: For matrix A = \[ \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \], we construct the augmented matrix [A|I] and perform elementary row transformations to convert A into I.

Applying the sequence of row operations gives:
\[ A^{-1} = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Apply row operations to transform [A|I] until A becomes I. The resulting matrix on the right is A^(-1).

Exam Tip: For 3×3 matrices, the denominator (here 67) should be consistent across all elements. If not, recalculate the determinant and scaling.

Question 12. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0 \end{bmatrix} \]
Answer: Starting with A = \[ \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0 \end{bmatrix} \], we form [A|I] and use elementary row operations to transform it into [I|A^(-1)].

The resulting inverse is:
\[ A^{-1} = \frac{1}{8} \begin{bmatrix} 5 & -10 & -1 \\ 3 & -6 & -1 \\ 10 & -12 & -2 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Perform row operations on the augmented matrix until the identity matrix appears on the left. The right side becomes the inverse.

Exam Tip: After obtaining the inverse, always multiply one row of A by the corresponding column of A^(-1) to verify the calculation.

Question 13. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \]
Answer: Given A = \[ \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \], we construct the augmented matrix [A|I] and apply elementary row transformations.

Through careful row reduction:
\[ A^{-1} = \begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Continue applying row operations to [A|I] until the left side is I. Then the right side is A^(-1).

Exam Tip: Keep intermediate steps organized in columns to avoid sign errors when dealing with negative entries.

Question 14. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix} \]
Answer: For matrix A = \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix} \], we write [A|I] and perform elementary row transformations to reduce A to I.

The inverse obtained is:
\[ A^{-1} = \begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Use row operations on [A|I] to make the left side become I. The right side then equals A^(-1).

Exam Tip: When the result has integer entries, verify by direct multiplication rather than using a formula - it confirms your work instantly.

Question 15. Using elementary row transformations, find the inverse of each of the following matrices:
\[ \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix} \]
Answer: Beginning with A = \[ \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix} \], we construct [A|I] and apply a series of elementary row operations to transform it into [I|A^(-1)].

After performing all necessary row transformations:
\[ A^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix} \]

Verification: \( AA^{-1} = I \)
In simple words: Apply row operations to [A|I] to achieve I on the left. The matrix appearing on the right is A^(-1).

Exam Tip: For upper triangular or lower triangular matrices, work backward or forward systematically - this reduces computational errors.

Exercise 5F

Question 1. Construct a 3 × 2 matrix whose elements are given by \( a_{ij} = \frac{1}{2}(1 - 2j)^2 \)
Answer: Here, i is the subscript for a row and j is the subscript for a column. Since the matrix is 3×2, we have \( 1 \leq i \leq 3 \) and \( 1 \leq j \leq 2 \).

Computing each element using the given formula:

For i=1, j=1: \( a_{11} = \frac{1}{2}(1 - 2(1))^2 = \frac{1}{2} \)

For i=1, j=2: \( a_{12} = \frac{1}{2}(1 - 2(2))^2 = \frac{9}{2} \)

For i=2, j=1: \( a_{21} = \frac{1}{2}(1 - 2(1))^2 = 0 \)

For i=2, j=2: \( a_{22} = \frac{1}{2}(1 - 2(2))^2 = 2 \)

For i=3, j=1: \( a_{31} = \frac{1}{2}(1 - 2(1))^2 = \frac{1}{2} \)

For i=3, j=2: \( a_{32} = \frac{1}{2}(1 - 2(2))^2 = \frac{1}{2} \)

Therefore, the required matrix is:
\[ \begin{bmatrix} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \]
In simple words: Substitute values of i and j into the formula to find each entry. For a 3×2 matrix, you need 6 calculations total.

Exam Tip: Always verify the dimensions match what you are asked to construct. Write out the formula substitution step clearly to show your understanding.

Question 2. Construct a 2 × 3 matrix whose elements are given by \( a_{ij} = \frac{1}{2}| - 3i + j| \)
Answer: The elements of the matrix follow the formula \( a_{ij} = \frac{1}{2}| - 3i + j| \). Since the matrix is 2×3, we have \( 1 \leq i \leq 2 \) and \( 1 \leq j \leq 3 \).

Here, i represents the row subscript and j represents the column subscript.

For i=1, j=1: \( a_{11} = \frac{1}{2}|-3(1) + 1| = 1 \)

For i=1, j=2: \( a_{12} = \frac{1}{2}|-3(1) + 2| = \frac{1}{2} \)

For i=1, j=3: \( a_{13} = \frac{1}{2}|-3(1) + 3| = 0 \)

For i=2, j=1: \( a_{21} = \frac{1}{2}|-3(2) + 1| = \frac{5}{2} \)

For i=2, j=2: \( a_{22} = \frac{1}{2}|-3(2) + 2| = 2 \)

For i=2, j=3: \( a_{23} = \frac{1}{2}|-3(2) + 3| = \frac{3}{2} \)

Therefore, the required matrix is:
\[ \begin{bmatrix} 1 & \frac{1}{2} & 0 \\ \frac{5}{2} & 2 & \frac{3}{2} \end{bmatrix} \]
In simple words: Apply the given formula for each position in the 2×3 grid. Use absolute value carefully - compute the expression first, then take its absolute value.

Exam Tip: When absolute values are involved, compute the expression inside the absolute value before taking its magnitude to avoid sign errors.

Question 3. If \( \begin{bmatrix} x + 2y & -y \\ 3x & 4 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \), find the values of x and y.
Answer: Comparing the left-hand side and right-hand side matrices, we can equate corresponding elements to establish equations.

From the first row, first column: \( x + 2y = -4 \) ... (i)

From the first row, second column: \( -y = 3 \) ... (ii)

From the second row, first column: \( 3x = 6 \) ... (iii)

From equation (iii): \( x = 2 \)

From equation (ii): \( y = -3 \)

We can verify with equation (i): \( 2 + 2(-3) = 2 - 6 = -4 \) ✓

Therefore, \( x = 2 \) and \( y = -3 \)
In simple words: Two matrices are equal when all their matching elements are equal. Write separate equations for each element and solve the resulting system.

Exam Tip: Always verify your solution by substituting back into all original equations - this catches sign errors immediately.

Question 4. Find the values of x and y, if \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
Answer: Starting with the given equation, we first apply scalar multiplication to the first matrix.

After multiplying by 2:
\[ \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \]

Using the property of matrix addition where corresponding elements combine:
\[ \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \]

Comparing each element:

\( 2 + y = 5 \) gives \( y = 3 \)

\( 2x + 2 = 8 \) gives \( x = 3 \)

Therefore, \( x = 3 \) and \( y = 3 \)
In simple words: First perform scalar multiplication, then add the matrices. Finally, match up corresponding elements to create simple equations.

Exam Tip: Write out the intermediate step after scalar multiplication to prevent errors in the addition stage.

Question 5. If \( x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \), find the values of x and y.
Answer: Given the equation \( x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \)

Applying scalar multiplication to each vector:
\[ \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \]

Adding the left-hand side vectors:
\[ \begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \]

This produces the system of linear equations:

\( 2x - y = 10 \) ... (i)

\( 3x + y = 5 \) ... (ii)

Adding equations (i) and (ii):
\( 5x = 15 \), so \( x = 3 \)

Substituting into equation (ii):
\( 3(3) + y = 5 \), so \( y = -4 \)

Therefore, \( x = 3 \) and \( y = -4 \)
In simple words: Multiply each vector by its scalar coefficient, add them, and equate to the result vector. This gives a system of two equations in two unknowns.

Exam Tip: When adding equations to eliminate a variable, choose the equations strategically - here y has opposite signs, so adding them directly eliminates y.

Question 6. If \( \begin{bmatrix} x & 3x - y \\ 2x + z & 3y - w \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix} \), find the values of x, y, z, w.
Answer: Comparing corresponding elements of the two matrices, we establish a set of equations.

From element (1,1): \( x = 3 \)

From element (1,2): \( 3x - y = 2 \)
Substituting \( x = 3 \): \( 3(3) - y = 2 \), so \( y = 7 \)

From element (2,1): \( 2x + z = 4 \)
Substituting \( x = 3 \): \( 2(3) + z = 4 \), so \( z = -2 \)

From element (2,2): \( 3y - w = 7 \)
Substituting \( y = 7 \): \( 3(7) - w = 7 \), so \( w = 14 \)

Therefore, \( x = 3, y = 7, z = -2, w = 14 \)
In simple words: Equate each element on the left to the matching element on the right. Solve for the simplest variable first, then substitute to find the others.

Exam Tip: Start with the element that gives you a variable directly (like x = 3), then use that value in subsequent equations to reduce complexity.

Question 7. If \( \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x + y \\ z + w & 3 \end{bmatrix} = 3 \begin{bmatrix} x & y \\ z & w \end{bmatrix} \), find the values of x, y, z, w.
Answer: Starting with the given equation, we first apply matrix addition on the left-hand side and scalar multiplication on the right-hand side.

Matrix addition on left: \( \begin{bmatrix} x + 4 & 6 + x + y \\ -1 + z + w & 2w + 3 \end{bmatrix} = \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} \]

Comparing each element of the resulting matrices:

From element (1,1): \( x + 4 = 3x \), which gives \( x = 2 \)

From element (2,2): \( 2w + 3 = 3w \), which gives \( w = 3 \)

From element (1,2): \( 6 + x + y = 3y \)
Substituting \( x = 2 \): \( 6 + 2 + y = 3y \), so \( y = 4 \)

From element (2,1): \( -1 + z + w = 3z \)
Substituting \( w = 3 \): \( -1 + z + 3 = 3z \), so \( z = 1 \)

Therefore, \( x = 2, y = 4, z = 1, w = 3 \)
In simple words: Simplify both sides of the equation first by doing the matrix operations. Then compare matching elements to form simple equations.

Exam Tip: Perform all matrix operations (addition and scalar multiplication) before comparing elements. This reduces calculation errors.

Question 8. If A = diag(3, -2, 5) and B = diag(1, 3, -4), find (A + B).
Answer: We are given two diagonal matrices. A diagonal matrix of order 3×3 has non-zero elements only along the main diagonal.

\( A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -4 \end{bmatrix} \)

When adding two diagonal matrices of the same order, we get another diagonal matrix where each diagonal element is the sum of the corresponding diagonal elements:

\( A + B = \begin{bmatrix} 3 + 1 & 0 & 0 \\ 0 & -2 + 3 & 0 \\ 0 & 0 & 5 + (-4) \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Therefore, \( A + B = \text{diag}(4, 1, 1) \)
In simple words: To add diagonal matrices, simply add the diagonal elements together. Off-diagonal elements remain zero.

Exam Tip: For diagonal matrices, focus only on the main diagonal - this makes the problem straightforward and quick to solve.

Question 9. Show that \( \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} = I \)
Answer: We need to demonstrate that the given equation holds. Applying scalar multiplication to each matrix:

\[ \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \]

Adding the matrices element by element:

\[ \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix} \]

Using the fundamental trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \]

Therefore, the statement is proved.
In simple words: Multiply each matrix by its scalar, add them, and use the identity cos²θ + sin²θ = 1 to simplify to the identity matrix.

Exam Tip: Always use fundamental trigonometric identities like cos²θ + sin²θ = 1 and recognize that complementary terms (like +cosθsinθ and -cosθsinθ) cancel out.

Question 10. If \( A = \begin{bmatrix} 1 & -5 \\ -3 & 2 \\ 4 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 1 \\ 2 & -1 \\ -2 & 3 \end{bmatrix} \), find the matrix C such that A + B + C is a zero matrix
Answer: Given that A + B + C equals the zero matrix, where a zero matrix contains all zero elements.

From the condition A + B + C = O, we can express:
\( C = -A - B = -(A + B) \)

First, compute A + B:
\[ A + B = \begin{bmatrix} 1 + 3 & -5 + 1 \\ -3 + 2 & 2 + (-1) \\ 4 + (-2) & -2 + 3 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\ -1 & 1 \\ 2 & 1 \end{bmatrix} \]

Therefore:
\[ C = -\begin{bmatrix} 4 & -4 \\ -1 & 1 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ 1 & -1 \\ -2 & -1 \end{bmatrix} \]

Verification: \( A + B + C = \begin{bmatrix} 1 & -5 \\ -3 & 2 \\ 4 & -2 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 2 & -1 \\ -2 & 3 \end{bmatrix} + \begin{bmatrix} -4 & 4 \\ 1 & -1 \\ -2 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} \)
In simple words: To find C, rearrange the equation to get C = -(A + B). Add the two given matrices, then negate every element.

Exam Tip: Always verify by adding all three matrices - this confirms that C satisfies the requirement.

Question 11. If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), then find the least value of α for which A + A' = I.
Answer: The transpose A' is obtained by interchanging rows and columns:

\[ A' = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \]

We are given that \( A + A' = I \). Therefore:

\[ \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} + \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Adding the matrices element by element:

\[ \begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Comparing the diagonal elements:
\( 2\cos \alpha = 1 \)

\( \cos \alpha = \frac{1}{2} \)

For α in the range [0, π], the least value satisfying this equation is \( \alpha = \frac{\pi}{3} \)

In simple words: Find A' by transposing, then add A and A'. Equate to I and solve for α using the trigonometric equation that results.

Exam Tip: Remember that transposing swaps rows and columns. Solving cos α = 1/2 requires knowledge of standard angle values like π/3, π/4, etc.

Question 12. Find the value of x and y for which \( \begin{bmatrix} 2 & -3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \)
Answer: Applying matrix multiplication on the left-hand side:

\[ \begin{bmatrix} 2x - 3y \\ x + y \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \]

Comparing corresponding elements produces the system of linear equations:

\( 2x - 3y = 1 \) ... (i)

\( x + y = 3 \) ... (ii)

From equation (ii): \( x = 3 - y \)

Substituting into equation (i):
\( 2(3 - y) - 3y = 1 \)
\( 6 - 2y - 3y = 1 \)
\( -5y = -5 \)
\( y = 1 \)

Therefore, \( x = 3 - 1 = 2 \)

Solution: \( x = 2, y = 1 \)
In simple words: Perform the matrix multiplication, then equate elements to create a system of equations. Solve by substitution or elimination.

Exam Tip: Matrix multiplication of a 2×2 by 2×1 gives a 2×1 result. Always multiply row-by-column for each element carefully.

Question 13. Find the value of x and y for which \( \begin{bmatrix} x & y \\ 3y & x \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix} \)
Answer: Performing the matrix multiplication on the left side:

Row 1: \( x(1) + y(2) = x + 2y \)

Row 2: \( 3y(1) + x(2) = 3y + 2x \)

This produces:
\[ \begin{bmatrix} x + 2y \\ 3y + 2x \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix} \]

Comparing elements gives the system of linear equations:

\( x + 2y = 3 \) ... (i)

\( 2x + 3y = 5 \) ... (ii)

From equation (i): \( x = 3 - 2y \)

Substituting into equation (ii):
\( 2(3 - 2y) + 3y = 5 \)
\( 6 - 4y + 3y = 5 \)
\( -y = -1 \)
\( y = 1 \)

Therefore, \( x = 3 - 2(1) = 1 \)

Solution: \( x = 1, y = 1 \)
In simple words: Multiply the 2×2 matrix by the 2×1 column vector using row-column multiplication. The result is a 2×1 vector that can be compared element by element.

Exam Tip: Organize the system of equations clearly - labeling them helps prevent mistakes when substituting values.

Question 14. If \( A = \begin{bmatrix} 4 & 5 \\ 1 & 8 \end{bmatrix} \), show that (A + A') is symmetric
Answer: Starting with matrix A, we find its transpose A' by interchanging rows and columns:

\[ A' = \begin{bmatrix} 4 & 1 \\ 5 & 8 \end{bmatrix} \]

Now compute A + A':

\[ A + A' = \begin{bmatrix} 4 & 5 \\ 1 & 8 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} 8 & 6 \\ 6 & 16 \end{bmatrix} \]

To verify that this is symmetric, find its transpose:

\[ (A + A')' = \begin{bmatrix} 8 & 6 \\ 6 & 16 \end{bmatrix} \]

Since \( (A + A')' = A + A' \), the matrix is symmetric. A matrix is symmetric when it equals its own transpose.
In simple words: A symmetric matrix looks the same when flipped about its main diagonal. Verify this by transposing and checking if you get the same matrix back.

Exam Tip: In a symmetric matrix, element (i,j) equals element (j,i). Use this as a quick check - note how the off-diagonal 6 appears in both positions.

Question 15. If \( A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \), show that (A - A') is skew-symmetric
Answer: Finding the transpose of A:

\[ A' = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} \]

Computing A - A':

\[ A - A' = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \]

To verify that this is skew-symmetric, find its transpose:

\[ (A - A')' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -(A - A') \]

Since \( (A - A')' = -(A - A') \), the matrix is skew-symmetric. A matrix is skew-symmetric when its transpose equals the negative of the original matrix.
In simple words: A skew-symmetric matrix has the property that when you transpose it, you get the negative of the original. Notice diagonal elements are always zero in skew-symmetric matrices.

Exam Tip: In a skew-symmetric matrix, element (i,j) equals the negative of element (j,i), and all diagonal elements must be zero. Check both properties for verification.

Question 16. If \( A = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 \\ 0 & 3 \end{bmatrix} \), find a matrix X such that A + 2B + X = O.
Answer: From the equation A + 2B + X = O, we can express:
\( X = -(A + 2B) \)

First, compute 2B:
\[ 2B = 2 \begin{bmatrix} -1 & 2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ 0 & 6 \end{bmatrix} \]

Next, compute A + 2B:
\[ A + 2B = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} + \begin{bmatrix} -2 & 4 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 4 & 11 \end{bmatrix} \]

Therefore:
\[ X = -\begin{bmatrix} 0 & 1 \\ 4 & 11 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -4 & -11 \end{bmatrix} \]
In simple words: Rearrange the equation to isolate X. Compute 2B first, then add it to A, and finally negate the result.

Exam Tip: Always perform scalar multiplication before matrix addition to follow the order of operations correctly.

Question 17. If \( A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} \), find a matrix X such that 3A - 2B + X = O.
Answer: From the equation 3A - 2B + X = O, we can express:
\( X = -(3A - 2B) = -3A + 2B \)

First, compute 3A:
\[ 3A = 3 \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 12 & 6 \\ 3 & 9 \end{bmatrix} \]

Next, compute 2B:
\[ 2B = 2 \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} -4 & 2 \\ 6 & 4 \end{bmatrix} \]

Now compute 3A - 2B:
\[ 3A - 2B = \begin{bmatrix} 12 & 6 \\ 3 & 9 \end{bmatrix} - \begin{bmatrix} -4 & 2 \\ 6 & 4 \end{bmatrix} = \begin{bmatrix} 16 & 4 \\ -3 & 5 \end{bmatrix} \]

Therefore:
\[ X = -\begin{bmatrix} 16 & 4 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} -16 & -4 \\ 3 & -5 \end{bmatrix} \]
In simple words: Apply scalar multiplication to both A and B. Subtract to find 3A - 2B. Then negate all elements to get X.

Exam Tip: When subtracting matrices, distribute the negative sign carefully to each element of the second matrix - \( -(-4) = +4 \), not \( -4 \).

Question 18. If A = \( \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), show that A' A = I.
Answer: Let the given matrix be \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \). Then the transpose is \( A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \). Computing the product, we have \( A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \). When we carry out the matrix multiplication, we get \( A'A = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \times (-\sin \alpha) + \sin \alpha \times \cos \alpha \\ (-\sin \alpha) \times \cos \alpha + \cos \alpha \times \sin \alpha & (-\sin \alpha) \times (-\sin \alpha) + \cos \alpha \times \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha \end{bmatrix} \). Since \( \cos^2 \alpha + \sin^2 \alpha = 1 \), we obtain \( A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \), which completes the proof.
In simple words: When you find the transpose of this matrix and multiply it by the original, the trigonometric identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \) causes all the cross terms to vanish and the diagonal terms to become 1, giving you the identity matrix.

Exam Tip: Always remember that \( \cos^2 \alpha + \sin^2 \alpha = 1 \) is the key identity that simplifies this problem - watch for where the cross-product terms cancel out to zero.

Question 19. If A and B are symmetric matrices of the same order, show that (AB - BA) is a skew symmetric matrix.
Answer: We are told that A and B are symmetric matrices of the same order. Let P be a matrix defined as P = (AB - BA). Since A and B are symmetric, we have \( A = A' \) and \( B = B' \). Taking the transpose of P, we get \( P' = (AB - BA)' \). Using the reversal law for transpose, which states \( (CD)' = D'C' \), we can write \( P' = (AB)' - (BA)' = B'A' - A'B' \). Substituting \( A = A' \) and \( B = B' \), we find \( P' = BA - AB = -(AB - BA) = -P \). Since the transpose of P equals the negative of P itself, P is a skew symmetric matrix by definition.
In simple words: When you transpose the difference (AB - BA), the reversal law flips the order of both products. Because A and B are already symmetric, swapping them gives you the negative of what you started with - that is the defining property of a skew symmetric matrix.

Exam Tip: The key step is applying the reversal law correctly - students often forget to reverse the order when taking the transpose of a product.

Question 20. If A = \( \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \) and f(x) = x² - 4x + 1, find f(A).
Answer: We are given \( A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \) and \( f(x) = x^2 - 4x + 1 \). To find f(A), we replace x with A in the function, giving us \( f(A) = A^2 - 4A + I \). First, we compute \( A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \). Next, we calculate \( 4A = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} \) and the identity matrix is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). Therefore, \( f(A) = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 - 8 + 1 & 12 - 12 + 0 \\ 4 - 4 + 0 & 7 - 8 + 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
In simple words: You substitute the matrix A wherever you see x in the formula. Calculate A squared first, then work through the subtraction and addition of matrices step by step - the result turns out to be a zero matrix.

Exam Tip: Pay attention to matrix multiplication order and ensure all arithmetic is done carefully - small errors in computing \( A^2 \) or combining terms will lead to an incorrect final answer.

Question 21. If the matrix A is both symmetric and skew-symmetric, show that A is a zero matrix.
Answer: We are given that matrix A is both symmetric and skew symmetric simultaneously. From the definition of a symmetric matrix, we have \( A = A' \) - equation (i). From the definition of a skew symmetric matrix, we have \( A = -A' \) - equation (ii). Equating these two expressions, we get \( A' = -A' \). Adding \( A' \) to both sides gives us \( 2A' = 0 \), which means \( A' = 0 \). Since \( A = A' \) from equation (i), we conclude that \( A = 0 \). Thus, the matrix must be a zero matrix.
In simple words: If a matrix equals its own transpose and also equals the negative of its transpose at the same time, the only way this can happen is if every entry in the matrix is zero.

Exam Tip: This is a proof by contradiction - the simultaneous existence of both symmetric and skew-symmetric properties forces the matrix to be zero; watch for this type of logical constraint question.