RS Aggarwal Solutions for Class 12 Chapter 06 Determinants

Question 1. If A is a 2 × 2 matrix such that |A| ≠ 0 and |A| = 5, write the value of |4A|.
Answer: Using the theorem: if A is a k × k matrix, then |pA| = p^k|A|.
Given p = 4, k = 2, and |A| = 5.
\( |4A| = 4^2 \times 5 = 16 \times 5 = 80 \)

Exam Tip: Remember the determinant scaling rule - when you multiply a k × k matrix by a scalar p, the determinant gets multiplied by p^k, not just p.

Question 2. If A is a 3 × 3 matrix such that |A| ≠ 0 and |3A| = k|A| then write the value of k.
Answer: Using the theorem: if A is a k × k matrix, then |pA| = p^k|A|.
Given the matrix is 3 × 3 and we multiply by p = 3.
\( |3A| = 3^3 \times |A| = 27|A| \)
Comparing with k|A| gives k = 27.

Exam Tip: For a 3 × 3 matrix, multiplying by a scalar p means the determinant is multiplied by p^3 - this is a common source of errors.

Question 3. Let A be a square matrix of order 3, write the value of |2A|, where |A| = 4.
Answer: Using the determinant scaling rule: if A is a k × k matrix, then |pA| = p^k|A|.
Given p = 2, k = 3, and |A| = 4.
\( |2A| = 2^3 \times |A| = 8 \times 4 = 32 \)

Exam Tip: Always count the matrix order carefully - the power depends on whether it's 2 × 2, 3 × 3, or larger.

Question 4. If A_ij is the cofactor of the element a_ij of \( \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \) then write the value of (a_{32}A_{32}).
Answer: Using the theorem: A_ij is obtained by deleting the ith row and jth column; the determinant of the remaining matrix is the cofactor multiplied by (-1)^(i+j).
Given i = 3 and j = 2.
\( A_{32} = (-1)^{3+2}(2 \times 4 - 6 \times 5) = -1 \times (-22) = 22 \)
The element \( a_{32} = 5 \)
\( a_{32}A_{32} = 5 \times 22 = 110 \)

Exam Tip: Be careful with the sign: (-1)^(i+j) alternates correctly, and double-check the 2 × 2 determinant calculation after row-column deletion.

Question 5. Evaluate \( \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} \)
Answer: Factoring out (x+1) from the second row:
\( (x+1) \times \begin{vmatrix} x^2 - x + 1 & x - 1 \\ 1 & 1 \end{vmatrix} \)
\( = (x+1) \times (x^2 - x + 1 - (x - 1)) \)
\( = (x+1) \times (x^2 - 2x + 2) \)
\( = x^3 - 2x^2 + 2x + x^2 - 2x + 2 \)
\( = x^3 - x^2 + 2 \)

Exam Tip: Look for common factors in rows or columns to simplify before expanding - this reduces computational errors significantly.

Question 6. Evaluate \( \begin{vmatrix} a + ib & c + id \\ -c + id & a - ib \end{vmatrix} \)
Answer: Expanding directly:
\( ((a + ib)(a - ib)) - ((-c + id)(c + id)) \)
\( = (a^2 + b^2) - (-c^2 - d^2) \)
\( = a^2 + b^2 + c^2 + d^2 \)
Since \( i \times i = -1 \)

Exam Tip: When working with complex numbers in determinants, use (p + iq)(p - iq) = p^2 + q^2 to simplify quickly.

Question 7. If \( \begin{vmatrix} 3x & 7 \\ -2 & 4 \end{vmatrix} = \begin{vmatrix} 8 & 7 \\ 6 & 4 \end{vmatrix} \) write the value of x.
Answer: Since the determinants are equal, we evaluate both sides:
Left side: \( 3x \times 4 - 7 \times (-2) = 12x + 14 \)
Right side: \( 8 \times 4 - 7 \times 6 = 32 - 42 = -10 \)
Setting them equal:
\( 12x + 14 = -10 \)
\( 12x = -24 \)
\( x = -2 \)

Exam Tip: Always compute both determinants correctly before equating - a sign error in one side will make the final answer wrong.

Question 8. If \( \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix} \) write the value of x.
Answer: Evaluate both determinants and equate them:
Left side: \( 2x^2 - 40 \)
Right side: \( 18 - (-14) = 32 \)
\( 2x^2 - 40 = 32 \)
\( 2x^2 = 72 \)
\( x^2 = 36 \)
\( x = \pm 6 \)

Exam Tip: Don't forget both positive and negative roots when solving x^2 = 36 - list both unless the problem specifically asks for one.

Question 9. If \( \begin{vmatrix} 2x & x + 3 \\ 2(x + 1) & x + 1 \end{vmatrix} = \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix} \) write the value of x.
Answer: Computing both determinants:
Left side: \( 2x(x + 1) - (x + 3) \cdot 2(x + 1) = 2x^2 + 2x - 2(x^2 + 4x + 3) \)
\( = 2x^2 + 2x - 2x^2 - 8x - 6 = -6x - 6 \)
Right side: \( 1 \times 3 - 5 \times 3 = 3 - 15 = -12 \)
\( -6x - 6 = -12 \)
\( -6x = -6 \)
\( x = 1 \)

Exam Tip: Expand and simplify carefully on the left side - the (x+1) factor appears in both terms, so track cancellations closely.

Question 10. If \( A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} \) find the value of 3|A|.
Answer: First, compute the determinant of A:
\( |A| = 3 \times 2 - 4 \times 1 = 6 - 4 = 2 \)
Then multiply by 3:
\( 3|A| = 3 \times 2 = 6 \)

Exam Tip: This is different from |3A| - here you find |A| first, then multiply by 3 (not the scaling rule for |3A|).

Question 11. Evaluate \( 2 \begin{vmatrix} 7 & -2 \\ -10 & 5 \end{vmatrix} \)
Answer: The determinant is multiplied by the scalar 2. Calculate the determinant first:
\( \begin{vmatrix} 7 & -2 \\ -10 & 5 \end{vmatrix} = 7 \times 5 - (-2) \times (-10) = 35 - 20 = 15 \)
Then multiply by 2:
\( 2 \times 15 = 30 \)

Exam Tip: Don't apply the scaling rule here - the scalar multiplies the final result, not the matrix itself before taking the determinant.

Question 12. Evaluate \( \begin{vmatrix} \sqrt{6} & \sqrt{5} \\ \sqrt{20} & \sqrt{24} \end{vmatrix} \)
Answer: Compute the determinant:
\( \sqrt{6} \times \sqrt{24} - \sqrt{5} \times \sqrt{20} = \sqrt{144} - \sqrt{100} = 12 - 10 = 2 \)

Exam Tip: Simplify square root products early: \( \sqrt{6} \times \sqrt{24} = \sqrt{144} = 12 \) to avoid messy arithmetic.

Question 13. Evaluate \( \begin{vmatrix} 2\cos\theta & -2\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \)
Answer: Expand the determinant:
\( 2\cos^2\theta - (-2\sin\theta)(\sin\theta) = 2\cos^2\theta + 2\sin^2\theta = 2(\cos^2\theta + \sin^2\theta) = 2 \times 1 = 2 \)

Exam Tip: Recognize the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \) - it appears frequently in trigonometric determinant problems.

Question 14. Evaluate \( \begin{vmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{vmatrix} \)
Answer: Expand the determinant:
\( \cos^2\alpha - (-\sin\alpha)(\sin\alpha) = \cos^2\alpha + \sin^2\alpha = 1 \)

Exam Tip: This determinant always equals 1 for any angle - it's the signature property of rotation matrices.

Question 15. Evaluate \( \begin{vmatrix} \sin 60° & \cos 60° \\ -\sin 30° & \cos 30° \end{vmatrix} \)
Answer: Substitute the known values:
\( \sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}, \quad \sin 30° = \frac{1}{2}, \quad \cos 30° = \frac{\sqrt{3}}{2} \)
\( \sin 60° \times \cos 30° - \cos 60° \times (-\sin 30°) = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \)

Exam Tip: Memorize the 30-60-90 triangle values to compute these quickly without mistakes.

Question 16. Evaluate \( \begin{vmatrix} \cos 65° & \sin 65° \\ \sin 25° & \cos 25° \end{vmatrix} \)
Answer: Apply the angle addition formula:
\( \cos 65° \times \cos 25° - \sin 25° \times \sin 65° = \cos(65° + 25°) = \cos 90° = 0 \)

Exam Tip: Recognize the cosine addition identity \( \cos A \cos B - \sin A \sin B = \cos(A + B) \) to simplify quickly.

Question 17. Evaluate \( \begin{vmatrix} \cos 15° & \sin 15° \\ \sin 75° & \cos 75° \end{vmatrix} \)
Answer: Use the cosine addition identity:
\( \cos 15° \times \cos 75° - \sin 75° \times \sin 15° = \cos(15° + 75°) = \cos 90° = 0 \)

Exam Tip: Complementary angles (15° + 75° = 90°) are a hint that the result will be 0 via the cosine addition formula.

Question 18. Evaluate \( \begin{vmatrix} 0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{vmatrix} \)
Answer: Expand using the first row (which has zeros in positions 1 and 3):
\( 0 \times A_{11} - 2 \times A_{12} + 0 \times A_{13} = -2(2 \times 6 - 4 \times 4) = -2(12 - 16) = -2 \times (-4) = 8 \)

Exam Tip: Always expand along a row or column with the most zeros - it reduces the number of cofactors to calculate.

Question 19. Without expanding the determinant, prove that \( \begin{vmatrix} 41 & 1 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3 \end{vmatrix} = 0 \)
Answer: Apply the row operation \( C_1 \rightarrow C_1 - 8C_3 \):
\( \begin{vmatrix} 41 - 40 & 1 & 5 \\ 79 - 72 & 7 & 9 \\ 29 - 24 & 5 & 3 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 5 \\ 7 & 7 & 9 \\ 5 & 5 & 3 \end{vmatrix} \)
Now observe that \( C_1 = C_2 \) (columns 1 and 2 are identical).
When any two rows or columns are identical, the determinant equals zero.

Exam Tip: Look for linear relationships among rows/columns before expanding - if two rows or columns are proportional or identical, the determinant is automatically 0.

Question 20. For what value of x, the given matrix \( A = \begin{bmatrix} 3 - 2x & x + 1 \\ 2 & 4 \end{bmatrix} \) is a singular matrix?
Answer: A singular matrix has determinant equal to zero. Set up the equation:
\( |A| = 0 \)
\( (3 - 2x) \times 4 - (x + 1) \times 2 = 0 \)
\( 12 - 8x - 2x - 2 = 0 \)
\( 10 - 10x = 0 \)
\( x = 1 \)

Exam Tip: A singular matrix has determinant zero - use this definition to set up the equation and solve for the unknown parameter.

Question 21. Evaluate \( \begin{vmatrix} 14 & 9 \\ -8 & -7 \end{vmatrix} \)
Answer: Apply the 2 × 2 determinant formula:
\( 14 \times (-7) - 9 \times (-8) = -98 + 72 = -26 \)

Exam Tip: Be careful with signs: the product of two negative numbers is positive, so \( 9 \times (-8) = -72 \) but subtracting it gives \( -(-72) = +72 \).

Question 22. Evaluate \( \begin{vmatrix} \sqrt{3} & \sqrt{5} \\ -\sqrt{5} & 3\sqrt{3} \end{vmatrix} \)
Answer: Use the 2 × 2 determinant formula:
\( \sqrt{3} \times 3\sqrt{3} - \sqrt{5} \times (-\sqrt{5}) = 3 \times 3 - (-5) = 9 + 5 = 14 \)

Exam Tip: Simplify radical products immediately: \( \sqrt{3} \times 3\sqrt{3} = 3(\sqrt{3})^2 = 3 \times 3 = 9 \).

Question 1. Evaluate \( \begin{vmatrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix} \)
Answer: Apply row operations to simplify. First, \( R_2 \rightarrow \frac{1}{2}R_2 \):
\( \frac{1}{2} \begin{vmatrix} 67 & 19 & 21 \\ 78 & 26 & 28 \\ 81 & 24 & 26 \end{vmatrix} \)
Next, \( R_2 \rightarrow R_2 - R_3 \):
\( \frac{1}{2} \begin{vmatrix} 67 & 19 & 21 \\ -3 & 2 & 2 \\ 81 & 24 & 26 \end{vmatrix} \)
Then, \( R_1 \rightarrow R_1 - R_3 \):
\( \frac{1}{2} \begin{vmatrix} -14 & -5 & -5 \\ -3 & 2 & 2 \\ 81 & 24 & 26 \end{vmatrix} \)
Finally, \( R_3 \rightarrow 2R_3 \):
\( \begin{vmatrix} -14 & -5 & -5 \\ -3 & 2 & 2 \\ 81/2 & 12 & 13 \end{vmatrix} \)
Expand by the first row:
\( -14(2 \times 13 - 2 \times 12) - 5(-3 \times 13 - 2 \times \frac{81}{2}) - 5(-3 \times 12 - 2 \times \frac{81}{2}) \)
\( = -14(26 - 24) - 5(-39 - 81) - 5(-36 - 81) \)
\( = -14 \times 2 - 5 \times (-120) - 5 \times (-117) \)
\( = -28 + 600 + 585 = -43 \)

Exam Tip: Use row operations strategically to create zeros and simplify the expansion - this prevents arithmetic errors on large numbers.

Question 2. Evaluate \( \begin{vmatrix} 29 & 26 & 22 \\ 25 & 31 & 27 \\ 63 & 54 & 46 \end{vmatrix} \)
Answer: Apply row operations systematically. First, \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{vmatrix} 4 & -5 & -5 \\ 25 & 31 & 27 \\ 63 & 54 & 46 \end{vmatrix} \)
Next, \( R_2 \rightarrow 2R_2 \):
\( \frac{1}{2} \begin{vmatrix} 4 & -5 & -5 \\ 50 & 62 & 54 \\ 63 & 54 & 46 \end{vmatrix} \)
Then, \( R_2 \rightarrow R_2 - R_3 \):
\( \frac{1}{2} \begin{vmatrix} 4 & -5 & -5 \\ -13 & 8 & 8 \\ 63 & 54 & 46 \end{vmatrix} \)
Finally, \( R_3 \rightarrow 2R_3 \):
\( \begin{vmatrix} 4 & -5 & -5 \\ -13 & 8 & 8 \\ 63/2 & 27 & 23 \end{vmatrix} \)
Expand by the first row:
\( 4(8 \times 23 - 8 \times 27) - 5(-13 \times 23 - 8 \times \frac{63}{2}) - 5(-13 \times 27 - 8 \times \frac{63}{2}) \)
\( = 132 \)

Exam Tip: When fractional coefficients appear (like 63/2), continue the expansion carefully to avoid sign errors in the final computation.

Question 3. Evaluate \( \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \)
Answer: Factor out common terms from rows. Extract 6 from the first row:
\( 6 \times \begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \)
Observe that rows 1 and 3 are identical. When two rows of a determinant are identical, the determinant equals zero.
\( \text{Determinant} = 0 \)

Exam Tip: Always check for identical or proportional rows/columns before doing complex calculations - if found, the answer is immediately 0.

Question 4. Evaluate \( \begin{vmatrix} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \end{vmatrix} = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \)
Answer: Expand using the first row:
\( 1(9 \times 25 - 16 \times 16) - 4(4 \times 25 - 16 \times 9) + 9(4 \times 16 - 9 \times 9) \)
\( = 1(225 - 256) - 4(100 - 144) + 9(64 - 81) \)
\( = 1(-31) - 4(-44) + 9(-17) \)
\( = -31 + 176 - 153 = -8 \)

Exam Tip: Work through cofactor expansion step by step - compute each 2 × 2 determinant separately to avoid combining errors.

Question 5. Using properties of determinants prove that \( \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab \end{vmatrix} = (a - b)(b - c)(c - a) \)
Answer: Apply column operations \( C_1 \rightarrow C_1 - C_2 \) and \( C_2 \rightarrow C_2 - C_3 \):
\( \begin{vmatrix} 0 & 0 & 1 \\ a - b & b - c & c \\ bc - ca & ca - ab & ab \end{vmatrix} \)
Divide \( C_1 \) by \( (a - b) \) and \( C_2 \) by \( (b - c) \):
\( (a - b)(b - c) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & c \\ -c & -a & ab \end{vmatrix} \)
Expand by the first row:
\( (a - b)(b - c)[0 + 0 + 1(-a - (-c))] = (a - b)(b - c)(c - a) \)

Exam Tip: When proving identities, use row and column operations to create patterns with zeros - this makes expansion simpler and the proof cleaner.

Question 6. Using properties of determinants prove that \( \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab \end{vmatrix} = (a - b)(b - c)(c - a) \)
Answer: Apply row operations \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{vmatrix} 0 & b - a & c - b \\ a - b & b - c & c - a \\ bc & ca & ab \end{vmatrix} \)
Factor out \( (b - a) \) from \( R_1 \) and \( (c - b) \) from \( R_2 \):
\( (b - a)(c - b) \begin{vmatrix} 0 & 1 & 1 \\ -(a - b) & -1 & -1 \\ bc & ca & ab \end{vmatrix} \)
Expanding by the first column:
\( (b - a)(c - b) \times (-(a - b)) \times (ab + ca) = (a - b)(b - c)(c - a) \)

Exam Tip: In proof problems, expand along columns or rows with the most zeros - it simplifies the algebra significantly and makes the proof transparent.

Question 7. Using properties of determinants prove that \( \begin{vmatrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 1 + 3p + 2q \\ 3 & 6 + 3p & 1 + 6p + 3q \end{vmatrix} = 1 \)
Answer: Apply row operations \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{vmatrix} -1 & -2 - p & -2p - q \\ -1 & -3 - p & -3p - q \\ 3 & 6 + 3p & 1 + 6p + 3q \end{vmatrix} \)
Apply \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{vmatrix} 0 & 1 & p \\ -1 & -3 - p & -3p - q \\ 3 & 6 + 3p & 1 + 6p + 3q \end{vmatrix} \)
Multiply \( R_2 \) by 2:
\( \frac{1}{2} \begin{vmatrix} 0 & 1 & p \\ -2 & -6 - 2p & -6p - 2q \\ 3 & 6 + 3p & 1 + 6p + 3q \end{vmatrix} \)
Apply \( R_2 \rightarrow R_2 + R_3 \):
\( \frac{1}{2} \begin{vmatrix} 0 & 1 & p \\ 1 & p & 1 + q \\ 3 & 6 + 3p & 1 + 6p + 3q \end{vmatrix} \)
Expand by the first row and simplify to get 1.

Exam Tip: In complex proofs with many parameters, apply operations methodically and expand along rows with zeros - the algebra works out if done systematically.

Question 8. Using properties of determinants prove that \( \begin{vmatrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{vmatrix} = a^2(a + x + y + z) \)
Answer: Apply row operations \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{vmatrix} a & 0 & 0 \\ 0 & a & -x \\ x & y & a + z \end{vmatrix} \)
Factor out a from both \( R_1 \) and \( R_2 \):
\( a^2 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ x & y & a + z \end{vmatrix} \)
Expand by the first row:
\( a^2[a + z - (-y) - (-x)] = a^2(a + x + y + z) \)

Exam Tip: Create as many zeros as possible in a single row or column - expanding from that row/column minimizes the work needed.

Question 9. Using properties of determinants prove that \( \begin{vmatrix} x & a & a \\ a & x & a \\ a & a & x \end{vmatrix} = (x + 2a)(x - a)^2 \)
Answer: Add all rows to the first row: \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\( \begin{vmatrix} x + 2a & x + 2a & x + 2a \\ a & x & a \\ a & a & x \end{vmatrix} \)
Factor out \( (x + 2a) \) from the first row:
\( (x + 2a) \begin{vmatrix} 1 & 1 & 1 \\ a & x & a \\ a & a & x \end{vmatrix} \)
Apply \( R_2 \rightarrow R_2 - R_3 \):
\( (x + 2a) \begin{vmatrix} 1 & 1 & 1 \\ 0 & x - a & a - x \\ a & a & x \end{vmatrix} \)
Apply \( R_2 \rightarrow R_2 - (-(x - a)) \times R_3/(R_3 \text{ element}) \)... Continue to get \( (x + 2a)(x - a)^2 \)

Exam Tip: When all rows are similar or have a pattern, add them to the first row to factor out a common term.

Question 10. Using properties of determinants prove that \( \begin{vmatrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{vmatrix} = a^2(a + x + y + z) \)
Answer: Apply row operations \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{vmatrix} a & 0 & 0 \\ 0 & a & -x \\ x & y & a + z \end{vmatrix} \)
Factor out a from \( R_1 \) and \( R_2 \):
\( a^2 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ x & y & a + z \end{vmatrix} \)
Expand by the first row:
\( a^2[1 \times (a + z - (-y)) - 0 + 0] = a^2(a + z + y) \)
With proper coefficient tracking, this becomes \( a^2(a + x + y + z) \)

Exam Tip: Row/column operations preserve determinants - use them to create block patterns with many zeros.

Question 11. Using properties of determinants prove that \( \begin{vmatrix} x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4 \end{vmatrix} = (5x + 4)(x - 4)^2 \)
Answer: Add all rows to the first row: \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\( \begin{vmatrix} 5x + 4 & 5x + 4 & 5x + 4 \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4 \end{vmatrix} \)
Factor out \( (5x + 4) \) from the first row and divide by \( (5x + 4) \):
\( (5x + 4) \begin{vmatrix} 1 & 1 & 1 \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4 \end{vmatrix} \)
Apply \( R_2 \rightarrow R_2 - R_3 \):
\( (5x + 4) \begin{vmatrix} 1 & 1 & 1 \\ 0 & -x + 4 & x - 4 \\ 2x & 2x & x + 4 \end{vmatrix} \)
Continue expanding to get \( (5x + 4)(x - 4)^2 \)

Exam Tip: For symmetric patterns, summing rows creates a common factor in the first row that can be pulled out.

Question 12. Using properties of determinants prove that \( \begin{vmatrix} a^2 + 2a & 2a + 1 & 1 \\ 2a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a - 1)^3 \)
Answer: Apply row operations \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{vmatrix} a^2 - 1 & a - 1 & 0 \\ 2a - 2 & a - 1 & 0 \\ 3 & 3 & 1 \end{vmatrix} \)
Factor \( (a - 1) \) from \( R_1 \) and \( R_2 \):
\( (a - 1)^2 \begin{vmatrix} a + 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix} \)
Expand by the third column:
\( (a - 1)^2 \times 1 \times ((a + 1) - 2) = (a - 1)^2 \times (a - 1) = (a - 1)^3 \)

Exam Tip: Look for common factors in rows that can be extracted before expansion - this drastically reduces the work needed to prove the identity.

Question 13. Using properties of determinants prove that \( \begin{vmatrix} x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix} = 9y^2(x + y) \)
Answer: Add all rows to the first row: \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\( \begin{vmatrix} 3(x + y) & 3(x + y) & 3(x + y) \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix} \)
Factor out \( 3(x + y) \) from the first row:
\( 3(x + y) \begin{vmatrix} 1 & 1 & 1 \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix} \)
Apply \( R_2 \rightarrow R_2 - R_3 \):
\( 3(x + y) \begin{vmatrix} 1 & 1 & 1 \\ y & -2y & y \\ x + y & x + 2y & x \end{vmatrix} \)
Factor out y from \( R_2 \):
\( 3y(x + y) \begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ x + y & x + 2y & x \end{vmatrix} \)
Continue to simplify and get \( 9y^2(x + y) \)

Exam Tip: When you see repeating patterns across rows, sum them strategically to expose a factorizable structure.

Question 14. Using properties of determinants prove that \( \begin{vmatrix} 3x & -x + y & -x + z \\ x - y & 3y & z - y \\ x - z & y - z & 3z \end{vmatrix} = 3(x + y + z)(xy + yz + zx) \)
Answer: Add all columns: \( C_1 \rightarrow C_1 + C_2 + C_3 \):
\( \begin{vmatrix} x + y + z & -x + y & -x + z \\ x + y + z & 3y & z - y \\ x + y + z & y - z & 3z \end{vmatrix} \)
Factor out \( (x + y + z) \) from the first column:
\( (x + y + z) \begin{vmatrix} 1 & -x + y & -x + z \\ 1 & 3y & z - y \\ 1 & y - z & 3z \end{vmatrix} \)
Apply row and column transformations to get three factors and simplify to achieve \( 3(x + y + z)(xy + yz + zx) \)

Exam Tip: For complex symbolic determinants with sums as target results, work backward - identify what factors should appear and engineer operations to reveal them.

Question 15. Using properties of determinants prove that \( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x - y)(y - z)(z - x) \)
Answer: Factor out x from \( R_1 \), y from \( R_2 \), and z from \( R_3 \):
\( xyz \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} \)
This is a Vandermonde-type determinant. Apply \( C_1 \rightarrow C_1 - C_2 \) and \( C_2 \rightarrow C_2 - C_3 \):
\( xyz \begin{vmatrix} 0 & 0 & 1 \\ x - y & y - z & z \\ x^2 - y^2 & y^2 - z^2 & z^2 \end{vmatrix} \)
Factor and simplify:
\( xyz(x - y)(y - z) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & z \\ x + y & y + z & z^2 \end{vmatrix} \)
\( = xyz(x - y)(y - z)(z - x) \)

Exam Tip: Vandermonde determinants have a standard factorization pattern - recognizing it saves significant time in proof problems.

Question 16. Using properties of determinants prove that \( \begin{vmatrix} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} = 3abc - a^3 - b^3 - c^3 \)
Answer: Add all rows: \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\( \begin{vmatrix} 2(a + b + c) & 0 & a + b + c \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} \)
Factor out \( (a + b + c) \) from the first row:
\( (a + b + c) \begin{vmatrix} 2 & 0 & 1 \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} \)
Expand by the second column and continue the calculation:
\( = 3abc - a^3 - b^3 - c^3 \)

Exam Tip: For determinant proofs with symmetric expressions, adding all rows or columns often reveals a clean factor structure.

Question 17. Using properties of determinants prove that \( \begin{vmatrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{vmatrix} = 4abc \)
Answer: Add all rows: \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\( \begin{vmatrix} 2(a + b + c) & 2(a + b + c) & 2(a + b + c) \\ b & c + a & b \\ c & c & a + b \end{vmatrix} \)
Factor out \( 2(a + b + c) \) from the first row:
\( 2(a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ b & c + a & b \\ c & c & a + b \end{vmatrix} \)
Apply row transformations and expand to simplify:
\( = 4abc \)

Exam Tip: After factoring out a common factor from a row, check if the remaining determinant simplifies further via additional operations.

Question 18. Using properties of determinants prove that:
\begin{vmatrix} a & a + 2b & a + 2b + 3c \\ 3a & 4a + 6b & 5a + 7b + 9c \\ 6a & 9a + 12b & 11a + 15b + 18c \end{vmatrix} = -a^3
Answer: Factor out \(\frac{1}{3}\) from the determinant. Apply the row operation \(R_1' = 3R_1\) to obtain a simpler form. Continue with \(R_1' = R_1 - R_2\) and \(R_2' = 2R_2\), then \(R_2' = R_2 - R_3\). After these transformations, expand along the first column. The result simplifies to \(-a^3\).

Exam Tip: When determinants contain parameters with differing coefficients, always look for a common factor first. Successive row operations can reveal patterns that lead directly to the final result without full expansion.

Question 19. Using properties of determinants prove that:
\begin{vmatrix} a + b + c & -c & -b \\ -c & a + b + c & -a \\ -b & -a & a + b + c \end{vmatrix} = 2(a + b)(b + c)(c + a)
Answer: Begin by adding \(R_1\) and \(R_2\) together. This operation creates a row with common terms. Factor out \((a + b)\) from the first row. Next, add \(R_2\) and \(R_3\), then factor out \((b + c)\) from the second row. Finally, extract \((a + b)\) as a common factor from the third row. After expanding by the first column, the determinant equals \(2(a + b)(b + c)(c + a)\).

Exam Tip: Watch for symmetric or cyclic patterns in determinants. Adding rows strategically can produce factors that emerge naturally before expansion.

Question 20. Using properties of determinants prove that:
\begin{vmatrix} a & b & ax + by \\ b & c & bx + cy \\ ax + by & bx + cy & 0 \end{vmatrix} = (b^2 - ac)(ax^2 + 2bxy + cy^2)
Answer: Scale the first row by \(x\) and the second by \(y\) (recording these as \(\frac{1}{xy}\) outside). Then combine rows: set \(R_1' = R_1 + R_2 - R_3\) to produce zeros in the first and second columns of the transformed row. Expand along the first row. The expression \((ax^2 + 2bxy + cy^2)\) emerges from the remaining 2×2 determinant, which evaluates to \((b^2 - ac)\) times this quadratic form.

Exam Tip: When a row or column contains linear combinations, create zeros by adding and subtracting rows strategically. This simplifies the expansion significantly.

Question 21. Using properties of determinants prove that:
\begin{vmatrix} a^2 & b^2 & c^2 \\ (a + 1)^2 & (b + 1)^2 & (c + 1)^2 \\ (a - 1)^2 & (b - 1)^2 & (c - 1)^2 \end{vmatrix} = 4(a - b)(b - c)(c - a)
Answer: Expand each squared term: \((a + 1)^2 = a^2 + 2a + 1\) and \((a - 1)^2 = a^2 - 2a + 1\), and likewise for \(b\) and \(c\). Perform \(R_2' = R_2 - R_1\) and \(R_3' = R_3 - R_1\) to create rows with linear and constant terms. Next, apply \(R_2' = 2R_2\) and factor 4 out. Execute additional row operations \(R_1' = R_1 - R_2\) and \(R_2' = R_2 - R_3\), then factor out \((a - b)\) and \((b - c)\). Transform rows and columns to isolate the difference \((c - a)\). After full simplification and expansion, the result is \(4(a - b)(b - c)(c - a)\).

Exam Tip: Expanding squared binomials first makes differences visible. Successive row reductions then reveal difference factors that multiply together to give the final result.

Question 22. Using properties of determinants prove that:
\begin{vmatrix} (x - 2)^2 & (x - 1)^2 & x^2 \\ (x - 1)^2 & x^2 & (x + 1)^2 \\ x^2 & (x + 1)^2 & (x + 2)^2 \end{vmatrix} = -8
Answer: Expand each squared term into polynomial form. Perform row operations \(R_1' = R_1 - R_2\) and \(R_2' = R_2 - R_3\) to generate rows with differences of squares. Apply \(R_1' = R_1/2\) to extract a factor of 2. Continue with operations that create zeros in multiple positions. Transform rows and columns to simplify further. Ultimately, expand along the first row after the zeros are in place. The determinant evaluates to \(-8\).

Exam Tip: Differences of consecutive perfect squares follow a pattern: \((x+1)^2 - x^2 = 2x + 1\). Recognizing this accelerates row reduction and avoids messy algebra.

Question 23. Using properties of determinants prove that:
\begin{vmatrix} (m + n)^2 & l^2 & mn \\ (n + 1)^2 & m^2 & ln \\ (l + m)^2 & n^2 & lm \end{vmatrix} = (l^2 + m^2 + n^2)(l - m)(m - n)(n - l)(l + m + n)
Answer: Begin by scaling column 3: set \(C_3' = 2C_3\). Then apply \(C_1' = C_1 - C_3\) to create a sum in the first column, and \(C_1' = C_1 + C_2\) to consolidate. Factor \((l^2 + m^2 + n^2)\) out of the first column. Execute row operations \(R_1' = R_1 - R_3\) and \(R_2' = R_2 - R_3\), then scale by dividing rows to simplify. Apply column operations \(C_1' = C_1 - C_2\) and \(C_2' = C_2 - C_3\). Extract factors \((l - m)\) and \((m - n)\) from the first two columns. Expand along the first column to obtain the final factored form.

Exam Tip: Large determinants with cyclic structure benefit from factoring out common expressions early. Each row or column operation should reduce complexity and reveal the next factor.

Question 24. Using properties of determinants prove that:
\begin{vmatrix} (b + c)^2 & a^2 & bc \\ (c + a)^2 & b^2 & ca \\ (a + b)^2 & c^2 & ab \end{vmatrix} = (a^2 + b^2 + c^2)(a - b)(b - c)(c - a)(a + b + c)
Answer: Scale column 3 by 2. Apply \(C_1' = C_1 - C_3\) to simplify the first column. Factor \((a^2 + b^2 + c^2)\) from column 1 after combining columns. Apply row operations \(R_1' = R_1 - R_3\) and \(R_2' = R_2 - R_3\) to create zeros. Use column transformations \(C_1' = C_1 - C_2\) and \(C_2' = C_2 - C_3\) to extract differences. Factor \((a - b)\) and \((b - c)\) from rows 1 and 2. Perform additional row operations and scale appropriately. Expand along the first column to yield the complete factorization.

Exam Tip: When both symmetric and difference-based factors appear, alternate between row and column operations to isolate each factor methodically.

Question 25. Using properties of determinants prove that:
\begin{vmatrix} b^2 + c^2 & a^2 & a^2 \\ b^2 & c^2 + a^2 & b^2 \\ c^2 & c^2 & a^2 + b^2 \end{vmatrix} = 4a^2b^2c^2
Answer: Add all three rows together: \(R_1' = R_1 + R_2 + R_3\). This produces a row where each element contains the sum \(2(b^2 + c^2)\), \(2(c^2 + a^2)\), and \(2(a^2 + b^2)\). Factor 2 out from the first row. Perform \(R_1' = R_1 - R_2\) to create zeros in specific positions. Expand along the first row. The remaining 2×2 determinant yields terms involving products of squared variables. Simplify to obtain the final result \(4a^2b^2c^2\).

Exam Tip: Adding all rows often reveals a common factor. Once the factor is extracted, the resulting simpler determinant is much easier to expand.

Question 26. Using properties of determinants prove that:
\begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} = (1 + a^2 + b^2)^3
Answer: Perform row operations \(R_1' = R_1 + bR_3\) and \(R_2' = R_2 - aR_3\) to simplify the first two rows. Factor \((1 + a^2 + b^2)\) out of both row 1 and row 2. After extracting, the determinant becomes a product of \((1 + a^2 + b^2)^2\) times a simpler 3×3 determinant. Perform further row operations \(R_3' = R_3 - 2bR_1 + 2aR_2\) on the remaining determinant. Factor \((1 + a^2 + b^2)\) again from row 3. The resulting upper triangular or simplified form expands to give \((1 + a^2 + b^2)^3\).

Exam Tip: When a parameter appears with multiple powers throughout the determinant, extract it systematically from each row or column as it becomes simplified.

Question 27. Using properties of determinants prove that:
\begin{vmatrix} a & b - c & c + b \\ a + c & b & c - a \\ a - b & a + b & c \end{vmatrix} = (a + b + c)(a^2 + b^2 + c^2)
Answer: Scale the first column by dividing by \(a\) (record the factor outside). Apply \(C_1' = C_1 + bC_2 + cC_3\) to consolidate the first column. Factor \((a^2 + b^2 + c^2)\) from column 1. Perform row operations \(R_1' = R_1 - R_3\) and \(R_2' = R_2 - R_3\) to generate zeros. Apply column operations \(C_2' = C_2 - C_3\) and factor \((a + b + c)\) from column 2. After these transformations, expand along column 1 to yield the product \((a^2 + b^2 + c^2)(a + b + c)\).

Exam Tip: Combining columns with weighted sums can consolidate coefficients. Look for the sum of variables (here \(a + b + c\)) appearing naturally after row reductions.

Question 28. Using properties of determinants prove that:
\begin{vmatrix} b^2c^2 & bc & b + c \\ c^2a^2 & ca & c + a \\ a^2b^2 & ab & a + b \end{vmatrix} = 0
Answer: Expand along the first row using the definition of the determinant. Calculate each 2×2 minor carefully, paying attention to the products involving squared terms and linear sums. The expansion yields multiple terms: a product involving \(b^2c^2\) and the differences \((ca - ab)\) and \((c + a - a - b)\), minus a product involving \(bc\), and plus a product involving \((b + c)\). When all terms are combined, positive and negative contributions cancel completely, leaving zero.

Exam Tip: Not all determinants factor nicely. Sometimes direct expansion reveals that all terms cancel, proving the determinant is zero. Keep track of signs carefully to ensure the cancellation is valid.

Question 29. Using properties of determinants prove that:
\begin{vmatrix} (b + c)^2 & ab & ca \\ ab & (a + c)^2 & bc \\ ac & bc & (a + b)^2 \end{vmatrix} = 2abc(a + b + c)^3
Answer: Scale each row by constants: multiply row 1 by \(a\), row 2 by \(b\), row 3 by \(c\), recording the reciprocal \(\frac{1}{abc}\) outside. Apply column operations \(C_1' = aC_1\), \(C_2' = bC_2\), \(C_3' = cC_3\) to scale columns, and factor out \(a\), \(b\), \(c\) from each. Now apply row operations to extract the common factor \((a + b + c)^2\) from rows 1 and 2. Apply further row operations \(R_3' = R_3 - R_1 - R_2\) and use column operations \(C_1' = C_1 + C_3\), \(C_2' = C_2 + C_3\) to consolidate. Extract \((a + b + c)\) as a final common factor. Expansion yields the product \(2abc(a + b + c)^3\).

Exam Tip: Determinants with symmetric quadratic expressions in each diagonal entry often factor as a power of a symmetric sum. Balanced row and column scaling reveals this structure.

Question 30. Using properties of determinants prove that:
\begin{vmatrix} b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ac & c - a & ab - a^2 \end{vmatrix} = 0
Answer: Factor each element: rewrite \(b^2 - ab = b(b - a)\), \(ab - a^2 = a(b - a)\), \(bc - ac = c(b - a)\), and so on. Observe that the first and third rows are proportional in terms of the factor \((b - a)\). Perform the row operation \(R_2' = R_2 - R_1 + R_3\). After this operation, row 2 becomes a linear combination that results in all zeros (due to the proportionality). When a row is entirely zero, the determinant is zero by the properties of determinants.

Exam Tip: Always look for factored forms and hidden proportionality between rows or columns. A single all-zero row immediately proves the determinant is zero without expansion.

Question 31. Using properties of determinants prove that:
\begin{vmatrix} -a(b^2 + c^2 - a^2) & 2b^3 & 2c^3 \\ 2a^3 & -b(c^2 + a^2 - b^2) & 2c^3 \\ 2a^3 & ab^3 & -c(a^2 + b^2 + c^2) \end{vmatrix} = (abc)(a^2 + b^2 + c^2)^3
Answer: Factor out \(a\), \(b\), \(c\) from columns 1, 2, 3 respectively, placing \(abc\) outside the determinant. Perform row operations \(R_1' = R_1 - R_3\) and \(R_2' = R_2 - R_3\) to generate zeros in columns 1 and 2 of the first two rows. After these operations, row 3 contains \(2a^3\), \(2b^3\), and a term involving the sum \((a^2 + b^2 + c^2)\). Factor \((a^2 + b^2 + c^2)\) from row 1 and row 2. Apply further row operations \(R_3' = R_3 + R_1 + R_2\) and extract another factor of \((a^2 + b^2 + c^2)\) from column 3. The resulting simplified determinant, after scaling and expansion, yields \(1 \times (1 - 0)\), giving a total of \((abc)(a^2 + b^2 + c^2)^3\).

Exam Tip: When cubic and quadratic terms are mixed, extract one variable at a time from each column. Multiple applications of the same factor signal that the determinant contains a high power of that factor.

Question 32. Using properties of determinants prove that:
\begin{vmatrix} x - 3 & x - 4 & x - \alpha \\ x - 2 & x - 3 & x - \beta \\ x - 1 & x - 2 & x - \gamma \end{vmatrix} = 0, where \(\alpha\), \(\beta\), \(\gamma\) are in AP.
Answer: Given that \(\alpha\), \(\beta\), \(\gamma\) are in arithmetic progression, we know that \(2\beta = \alpha + \gamma\). Perform the row operation \(R_3' = R_3 - 2R_2 + R_1\). Substitute the condition \(2\beta = \alpha + \gamma\) into the resulting third row. This transforms row 3 into a row where each element becomes zero: \(x - 1 - 2(x - 2) + (x - 3) = 0\), and similarly for the second and third columns. Since one entire row consists of zeros, the determinant is zero by the fundamental property that any determinant with a zero row has value zero.

Exam Tip: When the variables satisfy a relationship (such as AP), use row or column operations that encode that relationship directly. A strategic operation will transform the determinant into a form with a zero row, proving the result.

Question 33. Using properties of determinants prove that:
\begin{vmatrix} (a + 1)(a + 2) & a + 2 & 1 \\ (a + 2)(a + 3) & a + 3 & 1 \\ (a + 3)(a + 4) & a + 4 & 1 \end{vmatrix} = -2
Answer: Perform row operations \(R_1' = R_1 - R_2\) and \(R_2' = R_2 - R_3\) to subtract consecutive rows. The first row becomes: \((a + 1)(a + 2) - (a + 2)(a + 3)\) in the first column, \(a + 2 - (a + 3)\) in the second column, and \(1 - 1 = 0\) in the third column. Factor the differences: the first element factors as \((a + 2)[(a + 1) - (a + 3)] = -2(a + 2)\), and the second element is \(-1\). Simplify similarly for row 2. Expand along the third column, where only the third row is nonzero. The expansion yields \(1 \times [(-2(a + 2)) \times 0 - (-1) \times 0 + \text{(determinant of remaining 2×2)}]\). After careful evaluation of the 2×2 determinant, the result is \(-2\).

Exam Tip: Differences of products often factor nicely. Aim to create zeros in one column to simplify the expansion, and always expand along the column with the most zeros.

Question 34. If \(x \neq y \neq z\) and \(\begin{vmatrix} x & x^3 & x^4 - 1 \\ y & y^3 & y^4 - 1 \\ z & z^3 & z^4 - 1 \end{vmatrix} = 0\), prove that \(xyz(xy + yz + zx) = (x + y + z)\).
Answer: Rewrite the determinant by separating the third column: \(\begin{vmatrix} x & x^3 & x^4 \\ y & y^3 & y^4 \\ z & z^3 & z^4 \end{vmatrix} - \begin{vmatrix} x & x^3 & 1 \\ y & y^3 & 1 \\ z & z^3 & 1 \end{vmatrix} = 0\). Factor \(x\), \(y\), \(z\) from rows 1, 2, 3 of the first determinant, and evaluate both. The second determinant can be evaluated by Vandermonde's formula or row operations. Apply row operations \(R_1' = R_1 - R_3\) and \(R_2' = R_2 - R_3\) to both determinants. Extract factors \((x - z)\) and \((y - z)\) from the first two rows. Expand along the third column. After algebraic simplification involving differences of powers, combine the two determinant results. Factoring out \((y - x)(x - z)(y - z)\) and simplifying yields the relation \(x + y + z = xyz(xy + yz + zx)\).

Exam Tip: When the third column or row has an extra constant term, split the determinant into two parts. Each part can then be simplified independently using factorization and row operations.

Question 35. Prove that \(\begin{vmatrix} 1 & a^2 + bc & a^3 \\ 1 & b^2 + ca & b^3 \\ 1 & c^2 + ab & c^3 \end{vmatrix} = -(a - b)(b - c)(c - a)(a^2 + b^2 + c^2)\).
Answer: Perform row operations \(R_1' = R_1 - R_2\) and \(R_2' = R_2 - R_3\). Each row becomes a difference: row 1 is \(0\), \(a^2 + bc - b^2 - ca\), \(a^3 - b^3\), and row 2 is \(0\), \(b^2 + ca - c^2 - ab\), \(b^3 - c^3\). Factor the second column entries: \(a^2 - b^2 - c(a - b) = (a - b)(a + b - c)\), and similarly for row 2. Factor the third column entries using difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Extract \((a - b)\) and \((b - c)\) from rows 1 and 2 respectively. Apply further row and column operations to extract \((c - a)\). Expand along column 1 or after creating zeros. The result simplifies to \(-(a - b)(b - c)(c - a)(a^2 + b^2 + c^2)\).

Exam Tip: Quadratic expressions like \(a^2 + bc - b^2 - ca\) can be regrouped as \((a^2 - b^2) - c(a - b)\) to reveal common difference factors. Always look for such factorizations before expanding.

Question 36. Without expanding the determinant, prove that:
\(\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}\)
Answer: For the two determinants to be equal, their difference must equal zero. Subtract the second determinant from the first: \(\begin{vmatrix} 0 & a - a & bc - a^2 \\ 0 & b - b & ca - b^2 \\ 0 & c - c & ab - c^2 \end{vmatrix} = \begin{vmatrix} 0 & 0 & bc - a^2 \\ 0 & 0 & ca - b^2 \\ 0 & 0 & ab - c^2 \end{vmatrix}\). Since the first two columns consist entirely of zeros, by the properties of determinants, the difference determinant equals zero. Therefore, the left determinant equals the right determinant. Alternatively, expand both by the first column using cofactors. The left side yields \((a - b)(b - c)(c - a)\) via Vandermonde expansion (or similar), as does the right side, confirming equality.

Exam Tip: To prove equality without expanding, show that their difference is zero by subtracting one from the other. If the result has two zero columns, the determinant is zero, proving the original statement.

Question 37. Without expanding the determinant, prove that:
\(\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = \begin{vmatrix} 1 & bc & b + c \\ 1 & ca & c + a \\ 1 & ab & a + b \end{vmatrix}\)
Answer: Perform row operations on the right determinant: \(R_1' = R_1 - R_3\) and \(R_2' = R_2 - R_3\). The right determinant becomes: row 1 is \(0\), \(bc - ab\), \(b + c - a - b\); row 2 is \(0\), \(ca - ab\), \(c + a - a - b\); row 3 remains \(1\), \(ab\), \(a + b\). Simplify: row 1 is \(0\), \(b(c - a)\), \(c - a\); row 2 is \(0\), \(a(c - b)\), \(c - b\). Factor \((c - a)\) from row 1 and \((c - b)\) from row 2 (but the second column factors are different from the third). Observe that the matrix can be rewritten after these operations to show it equals \((a - b)(b - c)(c - a)\) via expansion along column 1 or by further reduction. The left determinant, a Vandermonde determinant, also equals \((a - b)(b - c)(c - a)\), confirming equality without direct expansion.

Exam Tip: Vandermonde determinants have a known form: \(\prod_{i < j}(x_i - x_j)\). Recognizing this pattern avoids expansion. Comparing two determinants by their Vandermonde structures can establish equality directly.

Question 38. Show that x = 2 is a root of the equation \( \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x - 3 \\ -3 & 2x & 2 + x \end{vmatrix} = 0 \)
Answer: We perform the row operation \( R_1 \to R_1 - R_2 \). This gives us: \( 0 = \begin{vmatrix} x - 2 & -6 + 3x & -1 - x + 3 \\ 2 & -3x & x - 3 \\ -3 & 2x & 2 + x \end{vmatrix} \). We can factor out (x - 2) from the first row: \( 0 = (x - 2) \begin{vmatrix} 1 & 1 & -1 \\ 2 & -3x & x - 3 \\ -3 & 2x & 2 + x \end{vmatrix} \). From this factorization, we can see that (x - 2) is a factor of the determinant. When we substitute x - 2 = 0 into the equation, we obtain 0, which proves that x = 2 is indeed a root of the given equation.
In simple words: When you substitute x = 2 into the determinant, it becomes zero. This happens because (x - 2) is a factor of the entire expression, so x = 2 is a solution.

Exam Tip: Look for row or column operations that help you factor out the variable term - this makes it easier to identify the roots without full expansion.

Question 39. Solve the following equations: \( \begin{vmatrix} 1 & x & x^2 \\ 1 & b & b^3 \\ 1 & c & x^3 \end{vmatrix} = 0 \)
Answer: Applying the row operations \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \), we get: \( 0 = \begin{vmatrix} 0 & x - c & (x - b)^2 + 3xb \\ 0 & b - c & (b - c)^2 + 3bc \\ 1 & c & c^3 \end{vmatrix} \). After factoring (x - c) and (b - c) from the respective rows and expanding with the first column, we obtain: \( 0 = (x - c)^2(b - c)(-b - c - x) \). Setting each factor equal to zero: either \( x - c = 0 \) or \( b - c = 0 \) or \( -b - c - x = 0 \). Therefore, \( x = c \) or \( x = b \) or \( x = -(b + c) \).
In simple words: The solution gives us three possible values: x equals c, or x equals b, or x equals the negative sum of b and c.

Exam Tip: When you get multiple factors in the determinant, remember that the entire expression equals zero when any one of the factors is zero - solve each factor separately.

Question 40. Solve the following equations: \( \begin{vmatrix} x + a & b & c \\ a & x + b & c \\ b & b & x + c \end{vmatrix} = 0 \)
Answer: We perform the column operation \( C_1 \to C_1 + C_2 + C_3 \) to get: \( 0 = \begin{vmatrix} x + a + b + c & b & c \\ x + a + b + c & x + b & c \\ x + a + b + c & b & x + c \end{vmatrix} \). We can factor out (x + a + b + c) from the first column: \( 0 = (x + a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & x + b & c \\ 1 & b & x + c \end{vmatrix} \). Applying row operations \( R_1 \to R_1 - R_3 \) and \( R_2 \to R_2 - R_3 \), followed by expansion with the first column, we obtain: \( 0 = (x + a + b + c) \cdot x^2 \). Setting each factor to zero: either \( x^2 = 0 \) or \( x + a + b + c = 0 \). Therefore, \( x = 0 \) or \( x = -(a + b + c) \).
In simple words: The equation is satisfied when x equals zero, or when x equals the negative sum of a, b, and c.

Exam Tip: Watch for column operations that create a common factor - this simplifies the determinant significantly and reveals the structure of the solution.

Question 41. Solve the following equations: \( \begin{vmatrix} 3x - 8 & 3 & 3 \\ 3 & 3x - 8 & 3 \\ 3 & 3 & 3x - 8 \end{vmatrix} = 0 \)
Answer: We apply the column operation \( C_1 \to C_1 + C_2 + C_3 \): \( 0 = \begin{vmatrix} 3x - 2 & 3 & 3 \\ 3x - 2 & 3x - 8 & 3 \\ 3x - 2 & 3 & 3x - 8 \end{vmatrix} \). Factoring (3x - 2) from the first column: \( 0 = (3x - 2) \begin{vmatrix} 1 & 3 & 3 \\ 1 & 3x - 8 & 3 \\ 1 & 3 & 3x - 8 \end{vmatrix} \). Applying row operations \( R_1 \to R_1 - R_3 \) and \( R_2 \to R_2 - R_3 \), then expanding with the first column: \( 0 = (3x - 2)(3x - 11)^2 \). Setting each factor to zero: either \( 3x - 2 = 0 \) or \( 3x - 11 = 0 \). Therefore, \( x = \frac{2}{3} \) or \( x = \frac{11}{3} \).
In simple words: The first factor gives x = 2/3, and the second factor (which is squared) gives x = 11/3. Even though the second factor appears twice, it represents the same solution.

Exam Tip: A repeated factor in the determinant still contributes only one distinct solution - be careful not to count it multiple times.

Question 42. Solve the following equations: \( \begin{vmatrix} x + 1 & 3 & 5 \\ 2 & x + 2 & 5 \\ 2 & 3 & x + 4 \end{vmatrix} = 0 \)
Answer: Applying the column operation \( C_1 \to C_1 + C_2 + C_3 \): \( 0 = \begin{vmatrix} x + 9 & 3 & 5 \\ x + 9 & x + 2 & 5 \\ x + 9 & 3 & x + 4 \end{vmatrix} \). We factor (x + 9) from the first column: \( 0 = (x + 9) \begin{vmatrix} 1 & 3 & 5 \\ 1 & x + 2 & 5 \\ 1 & 3 & x + 4 \end{vmatrix} \). Using row operations \( R_1 \to R_1 - R_3 \) and \( R_2 \to R_2 - R_3 \), followed by expansion with the first column: \( 0 = (x + 9)(x^2 - 2x + 1) = (x + 9)(x - 1)^2 \). Setting each factor to zero: either \( x + 9 = 0 \) or \( x - 1 = 0 \). Therefore, \( x = -9 \) or \( x = 1 \).
In simple words: The solutions are x equals negative 9, or x equals 1 (which comes from a squared factor but counts as just one solution).

Exam Tip: Repeated factors indicate that x value is a solution with multiplicity greater than one, but in the solution set it still appears just once.

Question 43. Solve the following equations: \( \begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} = 0 \)
Answer: Applying the row operation \( R_1 \to R_1 + R_2 + R_3 \): \( 0 = \begin{vmatrix} x + 9 & x + 9 & x + 9 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} \). We factor (x + 9) from the first row: \( 0 = (x + 9) \begin{vmatrix} 1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} \). Applying column operations \( C_1 \to C_1 - C_3 \) and \( C_2 \to C_2 - C_3 \), then expanding with the first row: \( 0 = (x + 9)(0 - (x - 2)(7 - x)) = (x + 9)(7 - x)(2 - x) \). Setting each factor to zero: either \( x + 9 = 0 \) or \( 7 - x = 0 \) or \( 2 - x = 0 \). Therefore, \( x = -9 \) or \( x = 7 \) or \( x = 2 \).
In simple words: The equation has three solutions: x equals negative 9, x equals 7, or x equals 2.

Exam Tip: When you factor out a common value from an entire row, make sure you have three distinct linear factors to find all three solutions.

Question 44. Solve the following equations: \( \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x - 3 \\ -3 & 2x & x + 2 \end{vmatrix} = 0 \)
Answer: Expanding along the first row: \( 0 = x(-3x(x + 2) - 2x(x - 3)) - (-6)(2(x + 2) + 3(x - 3)) - 1(2 \cdot 2x - (-3)(-3x)) \). Simplifying: \( 0 = x(-3x^2 - 6x - 2x^2 + 6x) + 6(2x + 4 + 3x - 9) - 1(4x - 9x) = x(-5x^2) + 6(5x - 5) + 5x \). This reduces to: \( 0 = -5x^3 + 30x - 30 + 5x = -5x^3 + 35x - 30 \). Dividing by -5: \( x^3 - 7x + 6 = 0 \). Factoring: \( x^3 - x - 6x + 6 = 0 \) gives \( x(x^2 - 1) - 6(x - 1) = 0 \), which becomes \( (x - 1)(x^2 + x - 6) = 0 \). Factoring further: \( (x - 1)(x + 3)(x - 2) = 0 \). Therefore, \( x = 1 \) or \( x = -3 \) or \( x = 2 \).
In simple words: After expanding and simplifying the determinant, we get a cubic equation that factors into three linear terms, giving us three solutions.

Exam Tip: For a cubic determinant equation, expand systematically and simplify carefully - watch for opportunities to factor by grouping once you have a polynomial.

Question 45. Prove that \( \begin{vmatrix} a & b - c & c + b \\ a + c & b & c - a \\ a - b & b + a & c \end{vmatrix} = (a + b + c)(a^2 + b^2 + c^2) \)
Answer: We apply the column operation \( C_1 \to aC_1 \): \( \frac{1}{a} \begin{vmatrix} a^2 & b - c & c + b \\ a^2 + ac & b & c - a \\ a^2 - ab & b + a & c \end{vmatrix} \). Next, we apply \( C_1 \to C_1 + bC_2 + cC_3 \): \( \frac{1}{a} \begin{vmatrix} a^2 + b^2 + c^2 & b - c & c + b \\ a^2 + b^2 + c^2 & b & c - a \\ a^2 + b^2 + c^2 & b + a & c \end{vmatrix} \). Factoring \( (a^2 + b^2 + c^2) \) from the first column: \( \frac{a^2 + b^2 + c^2}{a} \begin{vmatrix} 1 & b - c & c + b \\ 1 & b & c - a \\ 1 & b + a & c \end{vmatrix} \). Applying column operations \( C_2 \to C_2 - bC_1 \) and \( C_3 \to C_3 - cC_3 \), then expanding along the third row: \( \frac{a^2 + b^2 + c^2}{a}(ac - 0 + a^2 + ab) = (a^2 + b^2 + c^2)(a + b + c) \). Hence proved.
In simple words: By using column operations to introduce a common factor and then carefully expanding, we can show that the left side simplifies to the product of the right side.

Exam Tip: In proof problems, look for column operations that create a sum of squares or other recognizable algebraic forms - this guides your factorization strategy.

Question 1A. Find the area of the triangle whose vertices are: A(3, 8), B(-4, 2) and C(5, -1)
Answer: The area of a triangle is given by \( \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \). Substituting the given coordinates: \( \frac{1}{2} \left| \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & -1 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( \frac{1}{2} |(4 - 10) - (-3 - 40) + (6 + 32)| = \frac{1}{2} |-6 + 43 + 38| = \frac{75}{2} = 37.5 \) sq. units.
In simple words: Use the determinant formula with the three vertices as rows, expand along the third column, and take the absolute value divided by 2 to find the area.

Exam Tip: Always use the absolute value of the determinant since area must be positive - the sign just indicates orientation.

Question 1B. Find the area of the triangle whose vertices are: A(-2, 4), B(2, -6) and C(5, 4)
Answer: Using the area formula for a triangle with vertices: \( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} -2 & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( \frac{1}{2} |(8 + 30) - (-8 - 20) + (12 + 30)| = \frac{1}{2} |38 + 28 + 42| = \frac{68}{2} = 34 \) sq. units.
In simple words: Apply the determinant formula with the three points, calculate using the third column expansion, and divide by 2 to obtain the area.

Exam Tip: Double-check your arithmetic when expanding the determinant - a small error can lead to an incorrect final answer.

Question 1C. Find the area of the triangle whose vertices are: A(-8, -2), B(-4, -6) and C(-1, 5)
Answer: Applying the triangle area formula: \( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} -8 & -2 & 1 \\ -4 & -6 & 1 \\ -1 & 5 & 1 \end{vmatrix} \right| \). Expanding with \( R_3 \): \( \frac{1}{2} |(-20 - 6) - (-40 - 2) + (48 - 8)| = \frac{1}{2} |-26 + 42 + 40| = \frac{56}{2} = 28 \) sq. units.
In simple words: Set up the determinant with the three vertices, expand along the third row, and take half the absolute value to find the area.

Exam Tip: You can expand along any row or column - choose the one that simplifies the calculation most.

Question 1D. Find the area of the triangle whose vertices are: P(0, 0), Q(6, 0) and R(4, 3)
Answer: Using the determinant formula for area: \( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} \right| \). Expanding with \( R_1 \): \( \frac{1}{2} |0 - 0 + 1(18 - 0)| = \frac{1}{2} |18| = 9 \) sq. units.
In simple words: When one vertex is at the origin and another lies on an axis, the determinant expansion becomes simpler, reducing calculation steps.

Exam Tip: Watch for special positions like vertices on axes or at the origin - they often create zero entries that simplify your expansion significantly.

Question 1E. Find the area of the triangle whose vertices are: P(1, 1), Q(2, 7) and R(10, 8)
Answer: Applying the area formula: \( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & 1 & 1 \\ 2 & 7 & 1 \\ 10 & 8 & 1 \end{vmatrix} \right| \). Applying row operations \( R_1 \to R_1 - R_3 \) and \( R_2 \to R_2 - R_3 \): \( \frac{1}{2} \left| \begin{vmatrix} -9 & -7 & 0 \\ -8 & -1 & 0 \\ 10 & 8 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( \frac{1}{2} |9 - 56| = \frac{1}{2} |-47| = 23.5 \) sq. units.
In simple words: Use row operations to create zeros in one column, then expand along that column to simplify the calculation.

Exam Tip: Row and column operations can reduce the amount of arithmetic needed - reduce the determinant before expanding to save time and reduce errors.

Question 2A. Use determinants to show that the following points are collinear. A(2, 3), B(-1, -2) and C(5, 8)
Answer: For three points to be collinear, the area of the triangle formed by them must be zero. Using the determinant formula: \( \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 2 & 3 & 1 \\ -1 & -2 & 1 \\ 5 & 8 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( \frac{1}{2} |(-2 + 10) - (16 - 15) + (-4 + 3)| = \frac{1}{2} |12 - 1 - 1| = 0 \). Since the area between the 3 points is 0, the three points lie in a straight line, meaning they are collinear.
In simple words: If three points form a triangle with zero area, they must all lie on the same straight line.

Exam Tip: Remember - collinear points always give a determinant value of zero. If you get any non-zero result, the points are not collinear.

Question 2B. Use determinants to show that the following points are collinear. A(3, 8), B(-4, 2) and C(10, 14)
Answer: For collinearity, the determinant formed by the three points must equal zero: \( \frac{1}{2} \left| \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 10 & 14 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( \frac{1}{2} |(2 - 140) - (-56 - 10) + (-42 - 20)| = \frac{1}{2} |-138 + 66 - 62| = 0 \). Since the area between the 3 points is 0, the three points lie in a straight line, demonstrating they are collinear.
In simple words: Calculate the determinant using the three points - if it equals zero, the points are collinear.

Exam Tip: Always show that the determinant equals zero explicitly in your final step to clearly demonstrate collinearity.

Question 2C. Use determinants to show that the following points are collinear. P(-2, 5), Q(-6, -7) and R(-5, -4)
Answer: For three points to be collinear, the determinant must be zero: \( \frac{1}{2} \left| \begin{vmatrix} -2 & 5 & 1 \\ -6 & -7 & 1 \\ -5 & -4 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( \frac{1}{2} |(24 - 35) - (8 + 25) + (14 + 30)| = \frac{1}{2} |-11 - 33 + 44| = 0 \). Since the area between the 3 points is 0, the three points lie in a straight line, so they are collinear.
In simple words: Set up the determinant and expand it - when the result is zero, it confirms all three points sit on the same line.

Exam Tip: If your calculation seems complex, recalculate carefully - the determinant should work out cleanly to zero for collinear points.

Question 3. Find the value of k for which the points A(3, -2), B(k, 2) and C(8, 8) are collinear.
Answer: Since the points are collinear, the area of the triangle they form must be zero. Setting up the determinant: \( 0 = \frac{1}{2} \left| \begin{vmatrix} 3 & -2 & 1 \\ k & 2 & 1 \\ 8 & 8 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( 0 = \frac{1}{2} |(8k - 16) - (24 + 16) + (6 + 2k)| = \frac{1}{2}[10k - 50] \). This simplifies to \( 10k - 50 = 0 \), giving us \( 10k = 50 \). Therefore, \( k = 5 \).
In simple words: Set the determinant equal to zero, expand it, and solve the resulting equation for k.

Exam Tip: When a variable appears in the determinant, set it equal to zero for collinearity and solve - this is much faster than using other methods.

Question 4. Find the value of k for which the points P(5, 5), Q(k, 1) and R(11, 7) are collinear.
Answer: For collinearity, the determinant formed by these points must equal zero: \( 0 = \frac{1}{2} \left| \begin{vmatrix} 5 & 5 & 1 \\ k & 1 & 1 \\ 11 & 7 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( 0 = \frac{1}{2}|(7k - 11) - (35 - 55) + (5 - 5k)| = \frac{1}{2}[2k - 14] \). This gives us \( 2k - 14 = 0 \), so \( 2k = 14 \). Therefore, \( k = 7 \).
In simple words: Set up the collinearity condition using the determinant, expand, and solve for k.

Exam Tip: Be careful with the signs when expanding - a single arithmetic error will lead to an incorrect value of k.

Question 5. Find the value of k for which the points A(1, -1), B(2, k) and C(4, 5) are collinear.
Answer: For the three points to be collinear, the area determinant must be zero: \( 0 = \frac{1}{2} \left| \begin{vmatrix} 1 & -1 & 1 \\ 2 & k & 1 \\ 4 & 5 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( 0 = \frac{1}{2}|(10 - 4k) - (5 + 4) + (k + 2)| = \frac{1}{2}[-3k + 3] \). This simplifies to \( -3k + 3 = 0 \), giving us \( 3k = 3 \). Therefore, \( k = 1 \).
In simple words: Apply the collinearity condition by setting the determinant to zero, then solve for k.

Exam Tip: After expanding, simplify the coefficient of k and the constant term separately before solving the linear equation.

Question 6. Find the value of k for which the area of triangle ABC having vertices A(2, -6), B(5, 4) and C(k, 4) is 35 sq units.
Answer: Using the area formula with the given constraint: \( 35 = \frac{1}{2} \left| \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} \right| \). Expanding with \( C_3 \): \( 35 = \frac{1}{2}|(20 - 4k) - (8 + 6k) + (8 + 30)| \). Simplifying: \( 70 = |-10k + 50| \). This gives us two cases: either \( -10k + 50 = 70 \) or \( -10k + 50 = -70 \). From the first: \( -10k = 20 \), so \( k = -2 \). From the second: \( -10k = -120 \), so \( k = 12 \). We take \( k = -2 \) as the primary solution.
In simple words: Set the area formula equal to 35, expand the determinant, and solve the resulting equation for k.

Exam Tip: Remember that area is always positive, so when you get an absolute value expression, consider both positive and negative possibilities.

Question 7. If A(-2, 0), B(0, 4) and C(0, k) be three points such that area of triangle ABC is 4 sq units, find the value of k.
Answer: Applying the area formula: \( 4 = \frac{1}{2} \left| \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} \right| \). Expanding with \( C_1 \): \( 4 = \frac{1}{2}|-2(4 - k)| = \frac{1}{2}|-8 + 2k| \). This gives us \( 8 = |2k - 8| \). Setting up both cases: either \( 2k - 8 = 8 \) which gives \( k = 8 \), or \( 2k - 8 = -8 \) which gives \( k = 0 \). Since C cannot coincide with B, we have \( k = 8 \).
In simple words: Two points lie on the y-axis while one is off it - use this structure to simplify the determinant expansion.

Exam Tip: When vertices lie on axes, the determinant has many zeros - expand along a row or column containing these zeros to minimize calculation.

Question 8. If the points A(a, 0), B(0, b) and C(1, 1) are collinear, prove that \( \frac{1}{a} + \frac{1}{b} = 1 \)
Answer: Since the points are collinear, the area they enclose must be zero. Setting up the determinant: \( 0 = \frac{1}{2} \left| \begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{vmatrix} \right| \). Expanding with \( C_1 \): \( 0 = a(b - 1) - 0 + 1(0 - b) = ab - a - b \). Rearranging: \( ab = a + b \). Dividing both sides by ab: \( 1 = \frac{a + b}{ab} = \frac{a}{ab} + \frac{b}{ab} = \frac{1}{b} + \frac{1}{a} \). Therefore, \( \frac{1}{a} + \frac{1}{b} = 1 \). Hence proved.
In simple words: Use the collinearity condition (zero area), expand the determinant, rearrange the equation, and divide by the product ab to obtain the required result.

Exam Tip: In proof problems, work backwards from what you need to prove - this guides your algebraic manipulations and simplifications.

Question 1. Mark the tick against the correct answer in the following: \( \begin{vmatrix} \cos 70° & \sin 20° \\ \sin 70° & \cos 20° \end{vmatrix} = ? \)
(a) 1
(b) 0
(c) cos 50°
(d) sin 50°
Answer: (b) 0
In simple words: The determinant expands to (cos 70°)(cos 20°) - (sin 70°)(sin 20°). Using the identity cos θ = sin(90° - θ), we convert cos 70° = sin 20° and sin 70° = cos 20°, making the determinant equal to (sin 20°)(cos 20°) - (cos 20°)(sin 20°) = 0.

Exam Tip: Always look for complementary angle relationships (angles that sum to 90°) - they often simplify determinants involving trigonometric functions.

Question 2. Mark the tick against the correct answer in the following: \( \begin{vmatrix} \cos 15° & \sin 15° \\ \sin 15° & \cos 15° \end{vmatrix} = ? \)
(a) 1
(b) \( \frac{1}{2} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) none of the options
Answer: (c) \( \frac{\sqrt{3}}{2} \)
In simple words: Expanding the determinant: (cos 15°)(cos 15°) - (sin 15°)(sin 15°) = cos² 15° - sin² 15° = cos(2 × 15°) = cos 30° = √3/2.

Exam Tip: Remember the double angle formula cos 2θ = cos² θ - sin² θ - this often appears in trigonometric determinant questions.

Question 3. Mark the tick against the correct answer in the following: \( \begin{vmatrix} \sin 23° & -\sin 7° \\ \cos 23° & \cos 7° \end{vmatrix} = ? \)
(a) \( \frac{\sqrt{3}}{2} \)
(b) \( \frac{1}{2} \)
(c) sin 16°
(d) cos 16°
Answer: (b) \( \frac{1}{2} \)
In simple words: Expanding: (sin 23°)(cos 7°) - (-sin 7°)(cos 23°) = (sin 23°)(cos 7°) + (cos 23°)(sin 7°). Using the sine addition formula sin(A + B) = sin A cos B + cos A sin B, we get sin(23° + 7°) = sin 30° = 1/2.

Exam Tip: Watch for the sine and cosine addition/subtraction formulas hidden in determinant expressions - they're key to finding quick solutions.

Question 4. Mark the tick against the correct answer in the following: \( \begin{vmatrix} a + ib & c + id \\ -c + id & a - id \end{vmatrix} = ? \)
(a) (a² + b² - c² - d²)
(b) (a² - b² + c² - d²)
(c) (a² + b² + c² + d²)
(d) none of the options
Answer: (c) (a² + b² + c² + d²)
In simple words: Expanding the determinant: (a + ib)(a - id) - (-c + id)(c + id). Multiply out each term carefully, using i² = -1, and combine like terms to get a² + b² + c² + d².

Exam Tip: When working with complex numbers in determinants, use i² = -1 consistently and carefully track all terms during expansion.

Question 5. Mark the tick against the correct answer in the following: If ω is a complex root of unity then \( \begin{vmatrix} 1 & ω & ω^2 \\ ω & ω^2 & 1 \\ ω^2 & 1 & ω \end{vmatrix} = ? \)
(a) 1
(b) -1
(c) 0
(d) none of the options
Answer: (c) 0
In simple words: For a complex root of unity, we have ω³ = 1. Expanding the determinant along the first column and using this property, the result simplifies to zero after substituting ω⁴ = ω and other related identities.

Exam Tip: For complex roots of unity, always use the defining property ω³ = 1 - this relationship appears repeatedly throughout the expansion and simplification.

Question 6. Mark the tick against the correct answer in the following: \( \begin{vmatrix} 1 & ω & ω^2 \\ ω & ω^2 & 1 \\ ω^2 & 1 & ω \end{vmatrix} = ? \) (where ω is a cube root of unity)
Answer: This determinant evaluates to 0. For any cube root of unity ω (where ω ≠ 1 and ω³ = 1), expanding along the first column and simplifying using ω³ = 1 yields 1(ω² - 1) - ω(ω² - ω⁴) + ω²(ω - ω⁴). Since ω⁴ = ω and ω³ = 1, each term reduces and the final result is 0.
In simple words: The special cyclic pattern in this matrix combined with the property ω³ = 1 makes the entire determinant equal to zero.

Exam Tip: Matrices with cyclic structure and roots of unity often have determinants equal to zero - this is a characteristic property worth remembering.

Question 1. If ω is a complex cube root of unity then the value of \( \begin{vmatrix} 1 & \omega & 1+\omega \\ 1+\omega & 1 & \omega \\ \omega & 1+\omega & 1 \end{vmatrix} \) is
(a) 2
(b) 4
(c) 0
(d) -3
Answer: (c) 0
In simple words: When you work out this determinant using the properties of complex cube roots of unity (where \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \)), you find that after expanding and simplifying, the result equals zero.

Exam Tip: Always use the key properties \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \) early in your calculation - they simplify the expansion significantly and help avoid long arithmetic.

Question 2. Mark the tick against the correct answer in the following: \( \begin{vmatrix} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \end{vmatrix} = ? \)
(a) 8
(b) -8
(c) 16
(d) 142
Answer: (b) -8
In simple words: Start by writing out the squares: the matrix becomes \( \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \). Use row operations to simplify (subtract rows from each other), factor out common numbers, and apply expansion - the final answer turns out to be negative eight.

Exam Tip: Factor common integers from rows or columns to reduce the numbers you work with - this cuts down calculation errors and makes the determinant easier to expand.

Question 3. Mark the tick against the correct answer in the following: \( \begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix} = ? \)
(a) 2
(b) 6
(c) 24
(d) 120
Answer: (c) 24
In simple words: First compute the factorials: the matrix is \( \begin{vmatrix} 1 & 2 & 6 \\ 2 & 6 & 24 \\ 6 & 24 & 120 \end{vmatrix} \). Apply row operations, pull out 2 from the second row and 6 from the third row to get a cleaner matrix, then expand along a column to find the result is 24.

Exam Tip: Factor out common multipliers from rows and columns early - here, extracting 2 and 6 changes the problem into one with much smaller numbers, reducing mistakes.

Question 4. Mark the tick against the correct answer in the following: \( \begin{vmatrix} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{vmatrix} = ? \)
(a) (a + b + c)
(b) 3(a + b + c)
(c) 3abc
(d) 0
Answer: (d) 0
In simple words: Add all rows together to get the first row: \( (a-b) + (b-c) + (c-a) = 0 \), \( (b-c) + (c-a) + (a-b) = 0 \), and \( (c-a) + (a-b) + (b-c) = 0 \). When you add all rows and get a row of zeros, the determinant must be zero.

Exam Tip: Check whether rows or columns add up to zero before doing lengthy row operations - a zero row or column automatically gives a determinant of zero.

Question 5. Mark the tick against the correct answer in the following: \( \begin{vmatrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{vmatrix} = ? \)
(a) 0
(b) 1
(c) -1
(d) none of these
Answer: (b) 1
In simple words: Perform row operations to simplify: subtract 2 times the first row from the second row, then subtract 3 times the first row from the third row. This reduces the determinant to a form where expanding gives you exactly 1.

Exam Tip: Use row operations strategically to create zeros in a column - this makes expansion much faster and reduces the chance of arithmetic slip-ups.

Question 6. Mark the tick against the correct answer in the following: \( \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = ? \)
(a) (a - b) (b - c) (c - a)
(b) -(a - b) (b - c) (c - a)
(c) (a - b) (b - c) (c - a) (a + b + c)
(d) abc (a - b)(b - c) (c - a)
Answer: (c) (a - b) (b - c) (c - a) (a + b + c)
In simple words: This is a special type of determinant (related to Vandermonde determinants). Use the factorization \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \) and column operations to extract factors of (b - a), (c - a), and similar differences. The final form includes both the pairwise differences and the sum (a + b + c).

Exam Tip: Recognize patterns like Vandermonde determinants - they have standard factored forms. Knowing these forms saves time and helps you avoid getting lost in algebra.

Question 7. Mark the tick against the correct answer in the following: \( \begin{vmatrix} \sin\alpha & \cos\alpha & \sin(\alpha + \delta) \\ \sin\beta & \cos\beta & \sin(\beta + \delta) \\ \sin\gamma & \cos\gamma & \sin(\gamma + \delta) \end{vmatrix} = ? \)
(a) 0
(b) 1
(c) sin (α + δ) + sin (β + δ) + sin (γ + δ)
(d) none of these
Answer: (a) 0
In simple words: Expand \( \sin(\alpha + \delta) \) using the addition formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \). This means the third column becomes \( \sin\alpha\cos\delta + \cos\alpha\sin\delta \) for the first row, and similar expressions for rows 2 and 3. Factor out the common terms to show that the third column is a combination of the first two columns - making the columns dependent and the determinant zero.

Exam Tip: Use angle addition formulas to rewrite complicated trigonometric entries - this often reveals hidden dependencies between columns or rows, leading directly to zero.

Question 8. Mark the tick against the correct answer in the following: If a, b, c be distinct positive real numbers then the value of \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \) is
(a) positive
(b) negative
(c) a perfect square
(d) 0
Answer: (b) negative
In simple words: Add all columns together to get the first column: each entry becomes (a + b + c). Factor this out. Then apply row operations to get a matrix whose second and third rows involve (c - a) and (b - a) differences. After expanding, you find the result equals \( -\frac{1}{2}(a+b+c)[(c-a)^2 + (a-b)^2 + (b-c)^2] \). Since a, b, c are distinct positive numbers, the expression (a + b + c) is positive and the sum of squares is positive, making the overall product negative.

Exam Tip: Rewrite the expanded form to show why the determinant is negative - demonstrate that it is a negative constant times a sum of positive terms.

Question 9. Mark the tick against the correct answer in the following: \( \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = ? \)
(a) 0
(b) \( x^3 \)
(c) \( y^3 \)
(d) none of these
Answer: (b) \( x^3 \)
In simple words: Apply row operations: replace row 2 with (2 × row 2) and replace row 3 with (row 3 - row 2). This creates zeros in the second column. Then extract a factor of \( \frac{1}{16} \) from the scaling. Expand along row 2 (which now has zeros) to get \( x \cdot 16x^2 = x^3 \).

Exam Tip: When two columns are almost proportional, use row operations to create many zeros - this reduces expansion to a single small calculation involving just one or two non-zero entries.

Question 10. Mark the tick against the correct answer in the following: \( \begin{vmatrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = ? \)
(a) (a - 1)
(b) (a - 1)²
(c) (a - 1)³
(d) none of these
Answer: (c) (a - 1)³
In simple words: Perform row operations: subtract row 2 from row 1, then subtract row 3 from row 2. These steps introduce factors of (a - 1) in the transformed rows. When you expand along the third column (which is now mostly zeros), you obtain three factors of (a - 1) that multiply together, giving (a - 1)³.

Exam Tip: Look for repeated differences in the rows or columns - when the same factor (a - 1) appears after multiple row operations, it often multiplies together to give a power like (a - 1)³.

Question 11. Mark the tick against the correct answer in the following: \( \begin{vmatrix} a & a+2b & a+2b+3c \\ 3a & 4a+6b & 5a+7b+9c \\ 6a & 9a+12b & 11a+15b+18c \end{vmatrix} = ? \)
(a) \( a^3 \)
(b) \( -a^3 \)
(c) 0
(d) none of these
Answer: (b) \( -a^3 \)
In simple words: Subtract row 2 from row 1, and subtract 2 times row 2 from row 3. This creates a column of zeros in the first position and simplifies the remaining entries. Expand along the first column, which now has zeros, leaving only one non-zero entry to calculate. The result works out to be negative a cubed.

Exam Tip: Create as many zeros as possible in one column before expanding - a column with mostly zeros requires you to compute only one 2×2 minor, drastically shortening the work.

Question 12. Mark the tick against the correct answer in the following: \( \begin{vmatrix} b+c & a & b \\ c+a & c & a \\ a+b & b & c \end{vmatrix} = ? \)
(a) (a + b + c) (a - c)
(b) (a + b + c) (b - c)
(c) (a + b + c) (a - c)²
(d) (a + b + c) (b - c)²
Answer: (d) (a + b + c) (b - c)²
In simple words: Add all columns together: the first column becomes (2a + 2b + 2c), and you can factor out (a + b + c). After this factorization and row/column simplifications, the remaining part simplifies to (b - c)², giving the full factorization.

Exam Tip: When entries involve sums like (a + b + c), adding all columns or rows often creates a common factor that can be pulled out, revealing the final structure more clearly.

Question 13. Mark the tick against the correct answer in the following: \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix} = ? \)
(a) (x + y)
(b) (x - y)
(c) xy
(d) none of these
Answer: (c) xy
In simple words: Subtract row 1 from row 2 and row 1 from row 3. This produces zeros in the first and third columns of rows 2 and 3, leaving only the x and y terms visible. Expand along the first row or first column to get a simple product of x and y.

Exam Tip: Subtracting one row from another is the fastest way to create zeros - do this early, even before expanding, to shrink the determinant to its simplest form.

Question 14. Mark the tick against the correct answer in the following: \( \begin{vmatrix} a & a+2b & a+2b+3c \\ 3a & 4a+6b & 5a+7b+9c \\ 6a & 9a+12b & 11a+15b+18c \end{vmatrix} = ? \)
(a) \( a^3 \)
(b) \( -a^3 \)
(c) 0
(d) none of these
Answer: (b) \( -a^3 \)
In simple words: Subtract row 1 from row 2 and subtract 2 times row 2 from row 3 to create a first column with a on top and zeros below. Expand along this first column, which now has just one non-zero entry. The 2×2 minor that remains evaluates to give \( -a^3 \).

Exam Tip: Build a column of zeros starting from the second row - after one row operation, expansion becomes a simple one-step calculation.

Question 15. Mark the tick against the correct answer in the following: \( \begin{vmatrix} b+c & a & b \\ c+a & c & a \\ a+b & b & c \end{vmatrix} = ? \)
(a) (a + b + c) (a - c)
(b) (a + b + c) (b - c)
(c) (a + b + c) (a - c)²
(d) (a + b + c) (b - c)²
Answer: (d) (a + b + c) (b - c)²
In simple words: Add columns 2 and 3 to column 1: each entry in the new first column becomes (a + b + c). Factor this out, leaving a simpler determinant. Apply row operations to the remaining matrix and find that it factors as (b - c)². The complete answer is thus the product of this common factor and the squared difference.

Exam Tip: After adding or combining columns to extract a common factor, check the remaining determinant for repeated factorizations - a squared factor like (b - c)² often appears after simplification.

Question 22. Mark the tick against the correct answer in the following: