CBSE Class 12 Mathematics Determinants MCQs Set 05

Answer: (d) x = 0, y = 0

Explanation:
1. Expand along \( R_1 \): 6i[3i × i − (−1)(3)] − (−3i)[4i − (−1)(20)] + 1[4 × 3 − 3i × 20]
2. = 6i(−3 + 3) + 3i(20 + 4i) + (12 − 60i)
3. = 0 + 60i − 12 + 12 − 60i = 0 = 0 + 0i
4. x = 0, y = 0

Teacher's Note:
1. Use \( i^2 = -1 \) inside each minor before multiplying by the outside entry.
2. Once the value is 0, compare real and imaginary parts: both must be zero.

Answer: (b) −1, 2

Explanation:
1. Expand along \( R_1 \): 1(−10x² − 10x) − 4(5x² − 5) + 20(2x + 2)
2. = −10x² − 10x − 20x² + 20 + 40x + 40 = −30x² + 30x + 60
3. Set to zero: −30(x² − x − 2) = 0 ⇒ (x − 2)(x + 1) = 0
4. x = −1, 2

Teacher's Note:
1. Take out the common factor −30 before factorising the quadratic.
2. Check: substituting x = 2 or x = −1 in −30x² + 30x + 60 gives −120 + 60 + 60 = 0 and −30 − 30 + 60 = 0.

Answer: (b) 0, 0, −3

Explanation:
1. Expand along \( R_1 \): (1 + x)[(1 + x)² − 1] − 1[(1 + x) − 1] + 1[1 − (1 + x)]
2. = (1 + x)(x² + 2x) − x − x = x³ + 3x² + 2x − 2x
3. = x³ + 3x² = x²(x + 3) = 0
4. Roots: 0, 0, −3

Teacher's Note:
1. Expand (1 + x)² − 1 as x² + 2x before multiplying, so the cubic comes out cleanly.
2. A cubic equation has three roots counted with repetition, so the double root 0 must be listed twice.

Answer: (a) 2, 7

Explanation:
1. Expand along \( R_1 \): \( x(x^2 - 12) - 3(2x - 14) + 7(12 - 7x) = x^3 - 67x + 126 \)
2. Check the given root: (−9)³ − 67(−9) + 126 = −729 + 603 + 126 = 0
3. Divide by (x + 9): \( x^3 - 67x + 126 = (x + 9)(x^2 - 9x + 14) \)
4. \( x^2 - 9x + 14 = (x - 2)(x - 7) \), so the other roots are 2 and 7

Teacher's Note:
1. The sum of the three roots equals minus the coefficient of \( x^2 \), which is 0 here; so the other two roots add up to 9.
2. Also the product of the roots is −126, so the other two multiply to 14; 2 and 7 fit both checks.

Answer: (d) None of these

Explanation:
1. Expand along \( R_1 \): \( 4(4x - 9) - 3(12 - 9) + 3(9 - 3x) \)
2. = 16x − 36 − 9 + 27 − 9x = 7x − 18
3. 7x − 18 = 0 ⇒ \( x = \frac{18}{7} \)
4. This is not among 2, 3, 4, so the answer is None of these

Teacher's Note:
1. The equation in x is only linear because x appears once, so there is exactly one value to find.
2. Substituting x = 3 gives 3, not 0, which quickly rules out the tempting option (b).

Answer: (a) 1

Explanation:
1. Expand along \( R_1 \): 1[x(x + 3) − 0] − x[(x + 3) − (x + 2)] + (x + 1)[0 − x]
2. = x² + 3x − x − x² − x
3. = x, so the equation is x = 0
4. Number of roots = 1

Teacher's Note:
1. Do not assume a 3 × 3 determinant always gives a cubic; expand fully to see the true degree.
2. Here the x² terms cancel, leaving a linear expression with a single root.

Answer: (d) a, b

Explanation:
1. Expand along \( R_1 \): 1(x² − ab) − 1(ax − ab) + 1(ab − bx)
2. = x² − ax − bx + ab
3. = (x − a)(x − b) = 0 ⇒ x = a, b

Teacher's Note:
1. Expanding along the row of 1s gives the minors directly with no extra multipliers.
2. Check: substituting x = a or x = b in x² − ax − bx + ab gives 0.

Answer: (b) \( \frac{2}{3} \)

Explanation:
1. Expand along \( R_1 \): (3x − 8)[(3x − 8)² − 9] − 3[3(3x − 8) − 9] + 3[9 − 3(3x − 8)]
2. = (3x − 8)(3x − 5)(3x − 11) − 9(3x − 11) − 9(3x − 11) = (3x − 11)[(3x − 8)(3x − 5) − 18]
3. = (3x − 11)(9x² − 39x + 22) = (3x − 11)²(3x − 2) = 0
4. \( x = \frac{2}{3} \) or \( x = \frac{11}{3} \); of the options, \( \frac{2}{3} \) is correct

Teacher's Note:
1. Factor (3x − 8)² − 9 as a difference of squares, (3x − 5)(3x − 11), to spot the common factor quickly.
2. When a single option is asked for, check which listed value is a root rather than expecting all roots in the options.

Answer: (b) −1, 2

Explanation:
1. Expand along \( R_1 \): (x − 1)[(x − 1)² − 1] − 1[(x − 1) − 1] + 1[1 − (x − 1)]
2. = (x − 1)x(x − 2) − (x − 2) − (x − 2) = (x − 2)(x² − x − 2)
3. = (x − 2)(x − 2)(x + 1) = 0
4. Roots: −1, 2

Teacher's Note:
1. Pull out the common factor (x − 2) from all three terms before multiplying out.
2. Check: x² − x − 2 factorises as (x − 2)(x + 1), so 2 is a double root.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): (3 − x)[(3 − x)(−6 − x) − 9] + 6[(−6)(−6 − x) − 9] + 3[−18 − 3(3 − x)]
2. = (3 − x)(x² + 3x − 27) + 6(6x + 27) + 3(3x − 27)
3. = (−x³ + 36x − 81) + (36x + 162) + (9x − 81) = −x³ + 81x = −x(x − 9)(x + 9)
4. Roots are 0, 9, −9; a root among the options is 0

Teacher's Note:
1. Expand each minor fully before multiplying, then collect like terms.
2. Putting x = 0 in the expanded form −x³ + 81x gives 0 immediately, confirming option (c).

Answer: (d) \( \left(\frac{a}{b}\right) \) is one of the cube roots of −1

Explanation:
1. Expand along \( R_1 \): \( a(a^2 - 0) - b(0 - b^2) + 0 = a^3 + b^3 \)
2. \( a^3 + b^3 = 0 \) ⇒ \( \left(\frac{a}{b}\right)^3 = -1 \)
3. So \( \frac{a}{b} \) is one of the cube roots of −1

Teacher's Note:
1. Divide by \( b^3 \) to turn \( a^3 + b^3 = 0 \) into an equation in the ratio a/b.
2. Cube roots of −1 are −1, −ω and \( -\omega^2 \); do not confuse them with the cube roots of unity.

Answer: (c) \( b^3 \)

Explanation:
1. Expand along \( R_1 \): \( a(b^2 - ac) - 2b(3b - 4c) + 2c(3a - 4b) \)
2. = \( ab^2 - a^2c - 6b^2 + 8bc + 6ac - 8bc = (6 - a)(ac - b^2) \)
3. Since a ≠ 6, we need \( ac = b^2 \)
4. Then \( abc = b \cdot ac = b \cdot b^2 = b^3 \)

Teacher's Note:
1. The condition a ≠ 6 is given precisely so that you can cancel the factor (6 − a).
2. The result \( b^2 = ac \) means a, b, c are in geometric progression.

Answer: (d) 2 and −7

Explanation:
1. Expand: \( x(x^2 + 16) - 3(2x + 14) + 7(16 - 7x) = x^3 - 39x + 70 \)
2. x = 5 is a root: 125 − 195 + 70 = 0. Divide: \( x^3 - 39x + 70 = (x - 5)(x^2 + 5x - 14) \)
3. \( x^2 + 5x - 14 = (x + 7)(x - 2) \)
4. Other roots: 2 and −7

Teacher's Note:
1. With no \( x^2 \) term, the roots add up to 0, so the other two must sum to −5; 2 and −7 fit.
2. Their product must be −70 ÷ 5 = −14, which again confirms 2 × (−7).

Answer: (a) x = 1, 2

Explanation:
1. Expand along \( R_1 \): 1[(p + 1)(x + 2) − (p + x)(x + 1)] − 1[(p + 1)(x + 2) − 3(p + x)] + x[(p + 1)(x + 1) − 3(p + 1)]
2. = (p + 2 − x²) − (px − p − 2x + 2) + x(p + 1)(x − 2)
3. = px² − 3px + 2p = p(x − 1)(x − 2) = 0
4. For p ≠ 0: x = 1, 2

Teacher's Note:
1. Simplify each minor on its own line before combining; the x² terms cancel except for px².
2. Check x = 1 in the expanded form: p − 3p + 2p = 0.

Answer: (a) 0, −6

Explanation:
1. Expand along \( R_1 \): (x + 1)[(x + 2)(x + 3) − 6] − 1[2(x + 3) − 6] + 1[6 − 3(x + 2)]
2. = (x + 1)(x² + 5x) − 2x − 3x = x³ + 6x² + 5x − 5x
3. = x³ + 6x² = x²(x + 6) = 0 ⇒ x = 0 or x = −6

Teacher's Note:
1. Each minor simplifies nicely: x² + 5x, 2x and −3x, so expand them first.
2. Check x = −6 in x³ + 6x²: −216 + 216 = 0.

Answer: (d) 1

Explanation:
1. Expand along \( R_1 \): \( (x - 1)[6(x - 3) - 20] - 3(12 - 12) + 0 \)
2. = (x − 1)(6x − 38) = 2(x − 1)(3x − 19)
3. x = 1 or \( x = \frac{19}{3} \); of the options, x = 1

Teacher's Note:
1. The middle minor 2 × 6 − 4 × 3 is zero, which removes a whole term from the expansion.
2. Keep the factor (x − 1) intact instead of multiplying out; it hands you a root immediately.

Answer: (a) 0

Explanation:
1. Expand along \( R_1 \): (a + x)[a(−x) − xa] − a[(a − x)(−x) − x(a − x)] + x[(a − x)a − a(a − x)]
2. = (a + x)(−2ax) − a(−2ax + 2x²) + x(0)
3. = −2a²x − 2ax² + 2a²x − 2ax² = −4ax²
4. −4ax² = 0 ⇒ x = 0 (for a ≠ 0)

Teacher's Note:
1. The third minor is 0, so only two terms need to be multiplied out.
2. The a²x terms cancel, leaving −4ax², whose only root is x = 0.

Answer: (b) −5

Explanation:
1. Expand: \( x(x^2 - 12) - 2(3x - 15) + 5(12 - 5x) = x^3 - 43x + 90 \)
2. x = 5 is an integral root: 125 − 215 + 90 = 0
3. Divide: \( x^3 - 43x + 90 = (x - 5)(x^2 + 5x - 18) \)
4. The non-integral roots satisfy \( x^2 + 5x - 18 = 0 \), so their sum is −5

Teacher's Note:
1. For a quadratic \( x^2 + px + q = 0 \) the sum of roots is −p, so there is no need to solve it.
2. The roots of \( x^2 + 5x - 18 \) are \( \frac{-5 \pm \sqrt{97}}{2} \), which are indeed non-integral.

Answer: (b) 45

Explanation:
1. Given ad − bc = 5
2. \( \begin{vmatrix} 3a & 3b \\ 3c & 3d \end{vmatrix} = (3a)(3d) - (3b)(3c) = 9(ad - bc) \)
3. = 9 × 5 = 45

Teacher's Note:
1. For a 2 × 2 matrix, \( |kA| = k^2|A| \), because each product in ad − bc picks up k twice.
2. A common slip is to multiply by k once, giving 15.

Answer: (d) \( k^3\Delta \)

Explanation:
1. The new determinant is |kA|, where A is the matrix of Δ
2. Expanding along \( R_1 \), each term is (k × entry) × (2 × 2 minor with every entry multiplied by k), i.e. k × k² = k³ times the original term
3. Value = \( k^3\Delta \)

Teacher's Note:
1. For a square matrix of order n, \( |kA| = k^n|A| \); here n = 3.
2. A common slip is to answer kΔ; since every term of the expansion is a product of three entries, the factor is k³.

Answer: (d) 125Δ

Explanation:
1. The new determinant is |5A|, where A is the matrix of the original determinant
2. For a 3 × 3 matrix, \( |kA| = k^3|A| \), since every term in the expansion is a product of three entries
3. New value = \( 5^3\Delta \) = 125Δ

Teacher's Note:
1. Each term of a 3 × 3 expansion is a product of three entries, so each term gets multiplied by 5 × 5 × 5.
2. For matrices this reads \( |kA| = k^n|A| \), a result used often in board exams.

Answer: (a) |A|

Explanation:
1. For any row i, \( a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3} \) is exactly the expansion of the determinant along row i
2. Every such expansion gives the same value
3. So the sum equals |A|

Teacher's Note:
1. Elements of a row multiplied by their own cofactors give |A|, while elements of one row with cofactors of a different row give 0.
2. This pair of facts is the basis of the formula \( A(\text{adj } A) = |A| I \).

Answer: (d) \( a_1A_2 + b_1B_2 + c_1C_2 = \Delta \)

Explanation:
1. (a) is the expansion along \( R_1 \), so it equals Δ
2. (b) is the expansion along the second column, so it equals Δ
3. (c) is the expansion along \( R_3 \), so it equals Δ
4. (d) multiplies elements of \( R_1 \) by cofactors of \( R_2 \); such a sum is 0, not Δ, so (d) is incorrect

Teacher's Note:
1. Elements of one row multiplied by the cofactors of a different row always add up to 0.
2. Expansion can be done along any row or any column, as long as elements and cofactors come from the same line.

Answer: (a) 4, 3, 2, 1

Explanation:
1. Cofactor of 1 (position 1, 1): \( (+1) \times 4 = 4 \)
2. Cofactor of −2 (position 1, 2): \( (-1) \times (-3) = 3 \)
3. Cofactor of −3 (position 2, 1): \( (-1) \times (-2) = 2 \)
4. Cofactor of 4 (position 2, 2): \( (+1) \times 1 = 1 \). So 4, 3, 2, 1

Teacher's Note:
1. In a 2 × 2 determinant the minor of each element is just the diagonally opposite element.
2. Then apply the sign pattern + − / − +; the signs often turn negative entries positive.

Answer: (c) 8

Explanation:
1. The element 2 is in row 2, column 2, so the sign is \( (-1)^{2+2} = +1 \)
2. Minor: delete \( R_2 \) and \( C_2 \) to get \( \begin{vmatrix} 1 & 1 \\ 1 & 9 \end{vmatrix} = 9 - 1 = 8 \)
3. Cofactor = +8

Teacher's Note:
1. For diagonal positions the sign is always +, so the cofactor equals the minor.
2. Delete the full row and column of the element before writing the minor.

Answer: (c) 1

Explanation:
1. 2x is at row 2, column 2, so its cofactor is \( +\begin{vmatrix} x & -2 \\ x - 1 & 0 \end{vmatrix} \)
2. = x · 0 − (−2)(x − 1) = 2(x − 1)
3. 2(x − 1) = 0 ⇒ x = 1

Teacher's Note:
1. Only the four elements not in the row or column of 2x matter for its cofactor.
2. Check: at x = 1 the minor is \( \begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} \), whose zero row makes it 0.

Answer: (d) −2

Explanation:
1. 0 is at row 1, column 2, so the sign is \( (-1)^{1+2} = -1 \)
2. Minor of 0 = the remaining element 2
3. Cofactor = −1 × 2 = −2

Teacher's Note:
1. A zero element still has a non-zero cofactor; the cofactor depends on the other entries, not on the element itself.
2. Position (1, 2) always carries a minus sign.

Answer: (c) −5

Explanation:
1. 0 is at row 3, column 2, so the sign is \( (-1)^{3+2} = -1 \)
2. Minor: delete \( R_3 \) and \( C_2 \) to get \( \begin{vmatrix} -1 & 1 \\ -2 & -3 \end{vmatrix} = 3 + 2 = 5 \)
3. Cofactor = −5

Teacher's Note:
1. Write the sign chessboard + − + / − + − / + − + to read off the sign quickly.
2. Be careful with the double negative inside the minor: (−1)(−3) − (1)(−2) = 3 + 2.

Answer: (b) 7, 11, 2

Explanation:
1. \( M_{11} = \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = 4 + 3 = 7 \)
2. \( M_{12} = \begin{vmatrix} 4 & -3 \\ 1 & 2 \end{vmatrix} = 8 + 3 = 11 \)
3. \( M_{13} = \begin{vmatrix} 4 & 2 \\ 1 & 1 \end{vmatrix} = 4 - 2 = 2 \)
4. Minors: 7, 11, 2

Teacher's Note:
1. Minors carry no sign; signs are attached only when you convert minors into cofactors.
2. So the cofactors here would be 7, −11, 2, which is a different list.

Answer: (d) −5

Explanation:
1. Infinitely many solutions need the coefficient determinant to vanish: \( \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \lambda \end{vmatrix} = 0 \)
2. Expand along \( R_1 \): 3(2λ + 15) + 1(λ + 18) + 4(5 − 12) = 7λ + 35
3. 7λ + 35 = 0 ⇒ λ = −5
4. With λ = −5 the equations are consistent (rank of coefficient and augmented matrices is 2), so there are infinitely many solutions

Teacher's Note:
1. |A| = 0 is necessary for infinitely many solutions, but you must also make sure the system is not inconsistent.
2. Write the equations in the form ax + by + cz = d before forming the matrix, moving constants to the right.

Answer: (b) Inconsistent

Explanation:
1. \( |A| = \begin{vmatrix} 4 & -5 & -2 \\ 5 & -4 & 2 \\ 2 & 2 & 8 \end{vmatrix} \) = 4(−32 − 4) + 5(40 − 4) − 2(10 + 8) = −144 + 180 − 36 = 0
2. Look for a relation among the left sides: −2(4x − 5y − 2z) + 2(5x − 4y + 2z) = 2x + 2y + 8z
3. The same combination of right sides gives −2(2) + 2(3) = 2, but the third equation has right side 1
4. 2 ≠ 1, so the system is inconsistent

Teacher's Note:
1. When |A| = 0, find how the third equation's left side is built from the first two, then test the right sides.
2. Equivalently, compute \( (\text{adj } A)B \): if it is non-zero while |A| = 0, there is no solution.

Answer: (d) Infinitely many solutions

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 2 & 3 \\ 1 & -1 & 4 \\ 2 & 1 & 7 \end{vmatrix} \) = 1(−7 − 4) − 2(7 − 8) + 3(1 + 2) = −11 + 2 + 9 = 0
2. The third equation is the sum of the first two: (x + 2y + 3z) + (x − y + 4z) = 2x + y + 7z and 1 + 0 = 1
3. So only two independent equations remain for three unknowns
4. The system has infinitely many solutions

Teacher's Note:
1. A linear system can never have exactly two solutions; it has none, one, or infinitely many.
2. Adding or subtracting the given equations is often the fastest way to see a dependency.

Answer: (b) 1

Explanation:
1. Solve the first two: from x − 2y = 2, x = 2 + 2y; then 2(2 + 2y) + y = 4 ⇒ 5y = 0 ⇒ y = 0, x = 2
2. Check in the third: 3(2) + 5(0) = 6 ✓
3. All three lines pass through (2, 0), so there is exactly 1 solution

Teacher's Note:
1. Three equations in two unknowns are consistent only if the solution of any two satisfies the third.
2. The determinant test for this is \( \begin{vmatrix} 2 & 1 & -4 \\ 1 & -2 & -2 \\ 3 & 5 & -6 \end{vmatrix} = 0 \), which is another way to confirm concurrency.