CBSE Class 12 Mathematics Determinants MCQs Set 04

Answer: (d) None of these

Explanation:
1. Expand along \( R_1 \): 1(−k + 6) − k(−3 + 4) + 3(9 − 2k)
2. = 6 − k − k + 27 − 6k = 33 − 8k
3. 33 − 8k = 0 ⇒ \( k = \frac{33}{8} \)
4. This value is not among −1, 0, 1, so the answer is None of these

Teacher's Note:
1. When k appears in more than one place, collect all the k-terms carefully before equating to zero.
2. A quick check is to substitute each listed value: none of −1, 0, 1 makes the determinant zero.

Answer: (d) xy

Explanation:
1. Expand along \( R_1 \): 1[(1 + x)(1 + y) − 1] − 1[(1 + y) − 1] + 1[1 − (1 + x)]
2. = (xy + x + y) − y − x
3. = xy

Teacher's Note:
1. Expand along the row of 1s so that each term is just a 2 × 2 minor with no extra multiplier.
2. Check with x = y = 0: every entry becomes 1 and the value 0 matches xy = 0.

Answer: (d) −25

Explanation:
1. Expand along \( R_1 \): 4[(−1)(−1) − (1)(11)] − (−6)[(−1)(−1) − (1)(−4)] + 1[(−1)(11) − (−1)(−4)]
2. = 4(1 − 11) + 6(1 + 4) + (−11 − 4)
3. = −40 + 30 − 15 = −25

Teacher's Note:
1. Keep the alternating signs + − + for the first-row expansion; the middle term flips the sign of −6 to +6.
2. Write each 2 × 2 minor in brackets before multiplying, since most errors here come from sign slips inside the minors.

Answer: (b) 0

Explanation:
1. \( \begin{vmatrix} 1 & \log_b a \\ \log_a b & 1 \end{vmatrix} = 1 - \log_b a \cdot \log_a b \)
2. By the change of base rule, \( \log_b a \cdot \log_a b = \frac{\log a}{\log b} \cdot \frac{\log b}{\log a} = 1 \)
3. Value = 1 − 1 = 0

Teacher's Note:
1. Remember that \( \log_a b \) and \( \log_b a \) are reciprocals of each other.
2. This reciprocal pair appears often in determinant questions, so spot it before expanding anything bigger.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): 0(0 + c^2) − a(0 − bc) + (−b)(ac − 0)
2. = abc − abc
3. = 0

Teacher's Note:
1. This determinant is skew-symmetric (\( a_{ij} = -a_{ji} \) with zero diagonal) and of odd order, so its value is always 0.
2. Recognising the skew-symmetric pattern saves the whole expansion in an exam.

Answer: (b) 0

Explanation:
1. Expand along \( R_1 \): 31(58 × 24 − 71 × 105) − 37(31 × 24 − 71 × 31) + 92(31 × 105 − 58 × 31)
2. = 31(1392 − 7455) − 37(31 × (−47)) + 92(31 × 47)
3. = 31(−6063) + 37(1457) + 92(1457) = −187953 + 53909 + 134044
4. Value = 0

Teacher's Note:
1. Write the second and third minors as 31 × (−47) and 31 × 47 before multiplying; it keeps the arithmetic short.
2. Large-number determinants in MCQs are often 0, but confirm it by a careful expansion rather than guessing.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): 1(5 × 20 − 7 × 14) − 2(3 × 20 − 7 × 8) + 3(3 × 14 − 5 × 8)
2. = 1(100 − 98) − 2(60 − 56) + 3(42 − 40)
3. = 2 − 8 + 6 = 0

Teacher's Note:
1. The minors here are small (2, 4, 2), so a direct expansion along the first row is quick.
2. Keep the sign pattern + − + for the first row; the middle term is subtracted.

Answer: (b) 0

Explanation:
1. Expand along \( R_1 \): 7(1 × 55 − 41 × 5) − 9(4 × 55 − 41 × 5) + 79(4 × 5 − 1 × 5)
2. = 7(55 − 205) − 9(220 − 205) + 79(20 − 5)
3. = 7(−150) − 9(15) + 79(15) = −1050 − 135 + 1185
4. Value = 0

Teacher's Note:
1. Notice the last two minors are both 15, so the last two terms combine as (79 − 9) × 15 = 1050.
2. That exactly cancels 7 × (−150), a neat check on the arithmetic.

Answer: (a) 0

Explanation:
1. Expand along \( R_3 \) = (1, 1, 1): 1(17 × 7 − 15 × 8) − 1(19 × 7 − 15 × 9) + 1(19 × 8 − 17 × 9)
2. = (119 − 120) − (133 − 135) + (152 − 153)
3. = −1 + 2 − 1 = 0

Teacher's Note:
1. Expanding along the row of 1s avoids multiplying minors by large numbers.
2. The sign pattern for the third row is + − +, the same as for the first row.

Answer: (a) \( 1 + a^2 + b^2 + c^2 \)

Explanation:
1. Expand along \( R_1 \): 1(1 + c^2) − a(−a + bc) + b(ac + b)
2. = 1 + c^2 + a^2 − abc + abc + b^2
3. = \( 1 + a^2 + b^2 + c^2 \)

Teacher's Note:
1. The off-diagonal part is skew-symmetric, so the cross terms ±abc cancel and only squares survive.
2. A quick check: put a = b = c = 0 and the determinant becomes 1, matching the formula.

Answer: (b) 0

Explanation:
1. Expand along \( R_1 \): 2[6 × 2 − (−10)(7)] − 8[(−5)(2) − (−10)(1)] + 4[(−5)(7) − 6 × 1]
2. = 2(12 + 70) − 8(−10 + 10) + 4(−35 − 6)
3. = 164 − 0 − 164 = 0

Teacher's Note:
1. Bracket every minor before multiplying; the double negatives such as −(−10)(7) are where most slips happen.
2. A zero minor, like the middle one here, removes a whole term from the expansion.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): 19(3 × 6 − 15 × 11) − 6(21 × 6 − 15 × 28) + 7(21 × 11 − 3 × 28)
2. = 19(18 − 165) − 6(126 − 420) + 7(231 − 84)
3. = 19(−147) − 6(−294) + 7(147) = −2793 + 1764 + 1029
4. Value = 0

Teacher's Note:
1. All three minors are multiples of 147 (−147, −294, 147), so the value is 147(−19 + 12 + 7) = 0.
2. Spotting a common factor in the minors makes the final addition easy to check.

Answer: (a) 0

Explanation:
1. Expand along \( R_1 \): \( 1(2\pi - 2\sqrt{2}) - 1(2e - 2\sqrt{2}) + 1(2e - 2\pi) \)
2. = \( 2\pi - 2\sqrt{2} - 2e + 2\sqrt{2} + 2e - 2\pi \)
3. Value = 0

Teacher's Note:
1. Irrational entries like e, π, √2 are a distraction; treat them as letters and expand.
2. Every term cancels in pairs, so the middle row has no effect on the answer.

Answer: (b) 64Δ

Explanation:
1. Δ = 1(−2 − 4) − 3(4 − 0) + 1(8 − 0) = −6 − 12 + 8 = −10
2. New determinant: 4(−32 − 64) − 12(64 − 0) + 4(128 − 0) = −384 − 768 + 512 = −640
3. −640 = 64 × (−10) = 64Δ

Teacher's Note:
1. Every element is multiplied by 4, so the new matrix is 4A and \( |4A| = 4^3|A| = 64|A| \) for a 3 × 3 matrix.
2. Do not confuse this with \( 4^2 \); the power equals the order of the matrix.

Answer: (b) −2

Explanation:
1. Expand along \( R_1 \): 1(1 × 4 − 2 × 1) − 1(3 × 4 − 2 × 1) + 3(3 × 1 − 1 × 1)
2. = 1(2) − 1(10) + 3(2)
3. = 2 − 10 + 6 = −2

Teacher's Note:
1. Write each 2 × 2 minor in brackets first, then attach the signs + − +.
2. The middle minor is the largest here, so its minus sign decides the negative answer.

Answer: (a) 0

Explanation:
1. Expand along \( C_1 \), which has two zeros: value = −5 × (3 · 8 − 4 · 6)
2. = −5 × (24 − 24)
3. = 0

Teacher's Note:
1. Choose the row or column with the most zeros for expansion.
2. Here the first column has two zeros, so only one 2 × 2 minor needs to be calculated.

Answer: (b) −8

Explanation:
1. Expand along \( R_1 \): 1[1 × 4 − (−2)(2)] − 2[(−3)(4) − (−2)(2)] + 4[(−3)(2) − 1 × 2]
2. = 1(8) − 2(−8) + 4(−8)
3. = 8 + 16 − 32 = −8

Teacher's Note:
1. Bracket each minor before multiplying so that signs such as −(−2)(2) are handled correctly.
2. The middle term becomes positive because both its sign and its minor are negative.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): 43(7 × 2 − 4 × 3) − 1(35 × 2 − 4 × 17) + 6(35 × 3 − 7 × 17)
2. = 43(14 − 12) − 1(70 − 68) + 6(105 − 119)
3. = 43(2) − 2 + 6(−14) = 86 − 2 − 84
4. Value = 0

Teacher's Note:
1. The large entry 43 is multiplied by a very small minor (2), so the arithmetic stays light.
2. Always check the last minor carefully: 105 − 119 = −14, not +14.

Answer: (b) 0

Explanation:
1. Minors of \( R_1 \): \( M_{11} \) = 225 × 181 − 198² = 40725 − 39204 = 1521
2. \( M_{12} \) = 240 × 181 − 198 × 219 = 43440 − 43362 = 78; \( M_{13} \) = 240 × 198 − 225 × 219 = 47520 − 49275 = −1755
3. Value = 265(1521) − 240(78) + 219(−1755) = 403065 − 18720 − 384345
4. Value = 0

Teacher's Note:
1. With large entries, compute and write down each minor separately before combining them.
2. Symmetric determinants with large numbers are often designed to be zero, but always confirm by calculation.

Answer: (a) 4

Explanation:
1. Expand along \( R_1 \): 2(2 × 2 − 1 × 1) − 1(1 × 2 − 1 × 1) + 1(1 × 1 − 2 × 1)
2. = 2(3) − 1(1) + 1(−1)
3. = 6 − 1 − 1 = 4

Teacher's Note:
1. For small entries a straight expansion along the first row is the fastest method.
2. For the n × n matrix with 2 on the diagonal and 1 elsewhere the value is n + 1, which gives 4 for n = 3.

Answer: (a) 0

Explanation:
1. Expand along \( R_1 \): 1(7 × 15 − 8 × 14) − 2(6 × 15 − 8 × 13) + 3(6 × 14 − 7 × 13)
2. = 1(105 − 112) − 2(90 − 104) + 3(84 − 91)
3. = −7 + 28 − 21 = 0

Teacher's Note:
1. The minors −7, −14, −7 share the factor 7, so the value is 7(−1 + 4 − 3) = 0.
2. Each row here rises in equal steps, which is a hint that the answer may be 0; the expansion confirms it.

Answer: (d) 4xyz

Explanation:
1. Expand along \( R_1 \): x[yz − z(−y)] − (−y)[(−x)z − z(−x)] + z[(−x)(−y) − y(−x)]
2. = x(2yz) + y(0) + z(2xy)
3. = 2xyz + 2xyz = 4xyz

Teacher's Note:
1. The middle minor is 0, so only two terms survive the expansion.
2. Check by putting x = y = z = 1: the numeric determinant equals 4, matching 4xyz.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): 13(17 × 21 − 20 × 18) − 16(14 × 21 − 20 × 15) + 19(14 × 18 − 17 × 15)
2. = 13(357 − 360) − 16(294 − 300) + 19(252 − 255)
3. = −39 + 96 − 57 = 0

Teacher's Note:
1. The minors −3, −6, −3 are small even though the entries are large, so the expansion is quick.
2. Entries rising in equal steps along each row are a hint that the value may be 0; confirm by expanding.

Answer: (b) 4

Explanation:
1. Expand along \( R_1 \): a[bc − (−c)(−b)] − (−b)[(−a)c − (−c)(−a)] + (−c)[(−a)(−b) − b(−a)]
2. = a(0) + b(−2ac) − c(2ab) = −4abc
3. −4abc + λabc = 0 ⇒ λ = 4

Teacher's Note:
1. The first minor is 0, which removes one term straight away.
2. Watch the sign of the determinant: it is −4abc, so λ must be +4 to cancel it.

Answer: (d) \( \sin(\theta - \phi) \)

Explanation:
1. Expand along \( R_1 \): \( \sin\theta(\cos\phi - 0) - 1(0 + \cos\theta\sin\phi) + 0 \)
2. = \( \sin\theta\cos\phi - \cos\theta\sin\phi \)
3. = \( \sin(\theta - \phi) \)

Teacher's Note:
1. Expand along a row with a zero to save one minor.
2. Recognise the compound-angle pattern sin A cos B − cos A sin B = sin(A − B) at the end.

Answer: (d) 0

Explanation:
1. Expand along \( R_1 \): \( 1(1 - \log_y z\log_z y) - \log_x y(\log_y x - \log_y z\log_z x) + \log_x z(\log_y x\log_z y - \log_z x) \)
2. Change of base: \( \log_y z\log_z y = 1 \), \( \log_y z\log_z x = \log_y x \), \( \log_y x\log_z y = \log_z x \)
3. So the value = \( 1(1 - 1) - \log_x y(\log_y x - \log_y x) + \log_x z(\log_z x - \log_z x) \)
4. Value = 0

Teacher's Note:
1. The chain rule \( \log_b a \cdot \log_c b = \log_c a \) collapses every product in the minors.
2. Note the diagonal entries are \( \log_x x = 1 \), which is why each minor contains a 1.

Answer: (d) [2, 4]

Explanation:
1. Expand along \( R_1 \): \( 1(1 + \sin^2\alpha) - \sin\alpha(-\sin\alpha + \sin\alpha) + 1(\sin^2\alpha + 1) \)
2. = \( 2 + 2\sin^2\alpha \)
3. Since \( 0 \leq \sin^2\alpha \leq 1 \), Δ ranges from 2 to 4
4. Δ ∈ [2, 4]

Teacher's Note:
1. After expanding, use the known range of \( \sin^2\alpha \) to bound the expression.
2. Check the extremes: α = 0 gives 2 and α = 90° gives 4.

Answer: (b) 0

Explanation:
1. Expand along \( R_1 \): 11(13 × 15 − 14 × 14) − 12(12 × 15 − 14 × 13) + 13(12 × 14 − 13 × 13)
2. = 11(195 − 196) − 12(180 − 182) + 13(168 − 169)
3. = −11 + 24 − 13 = 0

Teacher's Note:
1. Consecutive integers give minors of just −1, −2, −1, so the expansion is very short.
2. Rows of consecutive integers are a hint that the value is 0; the expansion confirms it.

Answer: (d) 4

Explanation:
1. Expand along \( R_1 \): −1[(−1)(−1) − 1 × 1] − 1[1 × (−1) − 1 × 1] + 1[1 × 1 − (−1)(1)]
2. = −1(0) − 1(−2) + 1(2)
3. = 0 + 2 + 2 = 4

Teacher's Note:
1. The first minor is 0, which removes one term at once.
2. Keep the signs straight: −1 × (−2) = +2, and forgetting it gives 0 instead of 4.

Answer: (a) 0

Explanation:
1. Let \( s = \sin^2 x \), \( c = \cos^2 x \). Expand along \( R_1 \): s(2s − 12) − c(2c + 10) + 1(12c + 10s)
2. = 2s² − 12s − 2c² − 10c + 12c + 10s = 2(s² − c²) − 2(s − c)
3. = 2(s − c)(s + c − 1)
4. Since s + c = sin²x + cos²x = 1, the value is 0

Teacher's Note:
1. Replacing \( \sin^2 x \) and \( \cos^2 x \) by single letters keeps the expansion tidy.
2. Factorise before substituting; the factor (s + c − 1) is exactly what the identity sin²x + cos²x = 1 kills.

Answer: (b) −8

Explanation:
1. The determinant is \( \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \)
2. Minors of the first row: 9 × 25 − 16 × 16 = −31, 4 × 25 − 16 × 9 = −44, 4 × 16 − 9 × 9 = −17
3. Value = 1(−31) − 4(−44) + 9(−17) = −31 + 176 − 153 = −8

Teacher's Note:
1. Write the squares as numbers first, then compute the three minors separately.
2. Unlike rows of consecutive integers, rows of consecutive squares do not give 0; here the value is −8.

Answer: (c) 0

Explanation:
1. Expand along \( R_1 \): 41(45 × 49 − 46 × 48) − 42(44 × 49 − 46 × 47) + 43(44 × 48 − 45 × 47)
2. = 41(2205 − 2208) − 42(2156 − 2162) + 43(2112 − 2115)
3. = −123 + 252 − 129 = 0

Teacher's Note:
1. Consecutive integers give tiny minors (−3, −6, −3), so the expansion is quick despite large entries.
2. Each row is in A.P. here, a hint that the value is 0; the expansion confirms it.

Answer: (d) None of these

Explanation:
1. Expand along \( R_1 \): \( C(C^2 - 1) - 1(C - 6) + 0 = C^3 - 2C + 6 \)
2. With \( C = 2\cos\theta \): \( 8\cos^3\theta - 4\cos\theta + 6 \)
3. But \( \frac{\sin 4\theta}{\sin\theta} = 4\cos\theta\cos 2\theta = 8\cos^3\theta - 4\cos\theta \), which misses the +6; (b) and (c) do not match either
4. So the answer is None of these

Teacher's Note:
1. If the bottom-left entry were 0 instead of 6, the value would be exactly \( \frac{\sin 4\theta}{\sin\theta} \); a single entry changes the answer.
2. Test an easy angle: at θ = 90°, C = 0 and Δ = 6, while (a), (b), (c) give 0, 0, 0.