CBSE Class 12 Mathematics Determinants MCQs Set 06

Answer: (b) a only

Explanation:
1. A unique solution exists exactly when \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & -1 \\ 5 & -1 & a \end{vmatrix} \neq 0 \)
2. |A| = 1(3a − 1) − 1(2a + 5) + 1(−2 − 15) = a − 23
3. So the condition is a ≠ 23, which involves only a
4. The right-hand constant b plays no part in uniqueness

Teacher's Note:
1. Uniqueness depends only on the coefficient matrix; the constants decide consistency only when |A| = 0.
2. So a letter that appears only on the right-hand side can never affect whether the solution is unique.

Answer: (b) 1, 2, 3

Explanation:
1. Subtract the second equation from the first: 2y = 4 ⇒ y = 2
2. Then x + z = 4 and 2x + 2 − z = 1, i.e. 2x − z = −1
3. Adding: 3x = 3 ⇒ x = 1, and z = 4 − 1 = 3
4. (x, y, z) = (1, 2, 3)

Teacher's Note:
1. Always substitute the answer back into all three equations: 1 + 2 + 3 = 6, 1 − 2 + 3 = 2, 2 + 2 − 3 = 1.
2. Elimination is quicker than finding \( A^{-1} \) when two equations differ in only one variable’s sign.

Answer: (a) A unique solution

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 3 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} \)
2. = 1(−1 − 2) − 1(3 − 6) + 1(3 + 3) = −3 + 3 + 6 = 6
3. |A| ≠ 0, so the system has a unique solution (it is x = −10, y = −4, z = 16)

Teacher's Note:
1. A non-zero coefficient determinant guarantees exactly one solution, whatever the constants are.
2. Zero is the only solution only for a homogeneous system, and this system is not homogeneous.

Answer: (c) No solution

Explanation:
1. Multiply the first equation by 2: 2x + 2y = 4
2. The second says 2x + 2y = 3
3. 4 ≠ 3, so the equations contradict each other; the lines are parallel
4. No solution

Teacher's Note:
1. Equal ratios of coefficients but a different ratio of constants mean parallel lines.
2. In matrix terms |A| = 0 and \( (\text{adj } A)B \neq O \), which is the test for inconsistency.

Answer: (c) Infinite solutions

Explanation:
1. \( |A| = \begin{vmatrix} 1 & -1 & 1 \\ 3 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} \) = 1(−1 − 2) + 1(3 − 6) + 1(3 + 3) = −3 − 3 + 6 = 0
2. Find the relation: −3(eq1) + 2(eq2) gives \( (-3 + 6)x_1 + (3 - 2)x_2 + (-3 + 4)x_3 = 3x_1 + x_2 + x_3 \)
3. Right sides: −3(2) + 2(−6) = −18, which matches the third equation
4. The third equation is dependent, so there are infinitely many solutions

Teacher's Note:
1. After finding |A| = 0, always check the constants with the same combination to separate infinite from no solutions.
2. Compare with the previous question: a single change in a constant turns infinitely many solutions into none.

Answer: (a) μ only

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 5 & -1 & \mu \\ 2 & 3 & -1 \end{vmatrix} \)
2. = 1(1 − 3μ) − 1(−5 − 2μ) + 1(15 + 2) = 23 − μ
3. Unique solution ⇔ μ ≠ 23, which involves only μ
4. λ is a constant on the right side, so it does not matter

Teacher's Note:
1. The coefficient matrix alone decides whether the solution is unique.
2. Note that the order of equations here differs from the earlier similar question, but the logic is the same.

Answer: (a) Zero

Explanation:
1. \( |A| = \begin{vmatrix} 0 & 1 & -1 \\ -1 & 0 & 2 \\ 1 & -2 & 0 \end{vmatrix} \) = 0 − 1(0 − 2) − 1(2 − 0) = 0
2. Add the second and third equations: \( -2x_2 + 2x_3 = 1 \), i.e. \( x_2 - x_3 = -\frac{1}{2} \)
3. But the first equation says \( x_2 - x_3 = 1 \)
4. Contradiction, so there are zero solutions

Teacher's Note:
1. Read the variables carefully: the coefficient matrix columns are for \( x_1, x_2, x_3 \) in that order.
2. Combining equations to reproduce another equation's left side is the quickest consistency check.

Answer: (b) 1

Explanation:
1. \( |A| = \begin{vmatrix} k + 1 & 8 \\ k & k + 3 \end{vmatrix} = (k + 1)(k + 3) - 8k = k^2 - 4k + 3 = (k - 1)(k - 3) \)
2. k = 1: 2x + 8y = 4 and x + 4y = 2 are the same line, giving infinitely many solutions
3. k = 3: 4x + 8y = 12 (x + 2y = 3) and 3x + 6y = 8 (x + 2y = 8/3) are parallel, giving no solution
4. So only one value, k = 1, works

Teacher's Note:
1. |A| = 0 gives the candidates; each must then be tested, because some candidates lead to no solution.
2. For two lines, infinitely many solutions means all three ratios \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) are equal.

Answer: (d) λ = 3

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \) = 1(2λ − 6) − 1(λ − 3) + 1(2 − 2) = λ − 3
2. |A| = 0 ⇒ λ = 3
3. With λ = 3 the last two equations read x + 2y + 3z = 10 and x + 2y + 3z = 12
4. These contradict each other, so the system is inconsistent for λ = 3

Teacher's Note:
1. Inconsistency needs |A| = 0 together with conflicting constants.
2. Compare with the solved example where the right side was n: n = 10 would have given infinitely many solutions instead.

Answer: (d) 1, 2, −1

Explanation:
1. Here |A| = 8 ≠ 0, so the solution is unique; test the options
2. (1, 2, −1): 3(1) + 2 + 2(−1) = 3 ✓; 2(1) − 3(2) − (−1) = −3 ✓; 1 + 2(2) + (−1) = 4 ✓
3. So x = 1, y = 2, z = −1

Teacher's Note:
1. In an MCQ with a unique solution, substituting the options is faster than computing \( X = A^{-1}B \).
2. Check all three equations, because a wrong option can sometimes satisfy one or two of them.

Answer: (d) 0

Explanation:
1. \( |A| = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{vmatrix} \) = 2(9 − 8) − 1(−3 − 2) − 1(4 + 3) = 2 + 5 − 7 = 0
2. Left sides: (2x + y − z) − (x − 3y + 2z) = x + 4y − 3z, the left side of the third equation
3. Right sides: 7 − 1 = 6, but the third equation has 5
4. Inconsistent, so the number of solutions is 0

Teacher's Note:
1. Once |A| = 0, express the third left side through the first two and compare constants.
2. A linear system cannot have exactly 2 or 3 solutions, so those options can be rejected at once.

Answer: (b) λ = 3, μ ≠ 10

Explanation:
1. Expand along the first row: |A| = 1(2λ − 6) − 1(λ − 3) + 1(2 − 2) = λ − 3, so there is no unique solution only when λ = 3
2. With λ = 3 the last two equations are x + 2y + 3z = 10 and x + 2y + 3z = μ
3. They contradict each other when μ ≠ 10
4. No solution for λ = 3, μ ≠ 10

Teacher's Note:
1. For λ ≠ 3 the system always has a unique solution, whatever μ is.
2. λ = 3 with μ = 10 gives infinitely many solutions, so both conditions matter.

Answer: (b) 8

Explanation:
1. Solve the first two: 2x + 3y = 8 and 7x − 5y = −3
2. Multiply the first by 5 and the second by 3 and add: 10x + 21x = 40 − 9 ⇒ x = 1, then y = 2
3. Put (1, 2) into the third: 4(1) − 6(2) + λ = 0 ⇒ λ = 8

Teacher's Note:
1. For three equations in two unknowns, the solution of any two must satisfy the third.
2. Remember to rewrite 7x − 5y + 3 = 0 as 7x − 5y = −3 before eliminating.

Answer: (a) Infinitely many solutions

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} \) = 1(15 − 16) − 2(10 − 12) + 3(8 − 9) = −1 + 4 − 3 = 0
2. Check dependency: 2(eq2) − eq1 = 3x + 4y + 5z and 2(5) − 4 = 6, exactly the third equation
3. Only two independent equations remain, so there are infinitely many solutions

Teacher's Note:
1. When |A| = 0, combine the first two equations to rebuild the third and compare the constants.
2. If the third constant had been anything other than 6, the system would have had no solution.

Answer: (a) \( abc + 2fgh - af^2 - bg^2 - ch^2 \)

Explanation:
1. Expand along \( R_1 \): \( a(bc - f^2) - h(hc - fg) + g(hf - bg) \)
2. = \( abc - af^2 - ch^2 + fgh + fgh - bg^2 \)
3. = \( abc + 2fgh - af^2 - bg^2 - ch^2 \)

Teacher's Note:
1. This symmetric determinant appears in the condition for a pair of straight lines, so the result is worth remembering.
2. The term 2fgh arises because fgh comes once from the second minor and once from the third.

Answer: (a) 0

Explanation:
1. Expand along \( R_1 \): 0 − (a − b)[0 − (b − c)(c − a)] + (a − c)[(b − a)(c − b) − 0]
2. = (a − b)(b − c)(c − a) + (a − c)(b − a)(c − b)
3. (a − c)(b − a)(c − b) = [−(c − a)][−(a − b)][−(b − c)] = −(a − b)(b − c)(c − a)
4. So the two terms cancel and the value is 0

Teacher's Note:
1. Rewrite each factor so that all of them read (a − b), (b − c), (c − a); each reversal contributes a minus sign.
2. The pattern \( a_{ji} = -a_{ij} \) with a zero diagonal (skew-symmetric, order 3) is a hint that the terms will cancel.

Answer: (c) 0

Explanation:
1. The determinant is skew-symmetric: \( a_{12} = -h \) and \( a_{21} = h \), and so on, with zero diagonal
2. Direct expansion: 0 + h(0 + fg) − g(hf − 0) = fgh − fgh
3. = 0

Teacher's Note:
1. Odd-order skew-symmetric determinants are always zero, so the expansion just confirms it.
2. Even-order skew-symmetric determinants need not be zero; they turn out to be perfect squares.

Answer: (b) Zero

Explanation:
1. In a skew-symmetric matrix, \( a_{ij} = -a_{ji} \) for all i, j
2. Put j = i: \( a_{ii} = -a_{ii} \) ⇒ \( 2a_{ii} = 0 \)
3. So every diagonal element is zero

Teacher's Note:
1. This one-line argument is a standard 1-mark board question, so practise writing it.
2. In contrast, a symmetric matrix can have any values on its diagonal.

Answer: (c) 4!

Explanation:
1. With factorial values: \( \begin{vmatrix} 1 & 2 & 6 \\ 2 & 6 & 24 \\ 6 & 24 & 120 \end{vmatrix} \)
2. Expand along \( R_1 \): 1(720 − 576) − 2(240 − 144) + 6(48 − 36)
3. = 144 − 192 + 72 = 24
4. 24 = 4!

Teacher's Note:
1. Convert factorials to numbers first: 2! = 2, 3! = 6, 4! = 24, 5! = 120.
2. Compute each 2 × 2 minor separately (144, 96, 12) before attaching the signs + − +.

Answer: (d) None of these

Explanation:
1. Using \( {}^5C_0 = 1, {}^5C_1 = 5, {}^5C_2 = 10, {}^5C_3 = 10, {}^5C_4 = 5, {}^5C_5 = 1 \): \( \begin{vmatrix} 1 & 10 & 14 \\ 5 & 5 & 1 \\ 10 & 1 & 1 \end{vmatrix} \)
2. Expand along \( R_1 \): 1(5 − 1) − 10(5 − 10) + 14(5 − 50)
3. = 4 + 50 − 630 = −576
4. −576 is not 0, −720 or 80, so the answer is None of these

Teacher's Note:
1. Use \( {}^nC_r = {}^nC_{n-r} \) to write the binomial coefficients quickly.
2. Note that −(6!) = −720, which is close but not equal to −576, so compute rather than guess.

Answer: (a) 0

Explanation:
1. Expand along \( R_1 \): \( \cos C(0 + \sin B\tan A) - \tan A(\sin B\cos C - 0) + 0 \)
2. = \( \sin B\tan A\cos C - \tan A\sin B\cos C \)
3. = 0

Teacher's Note:
1. No trigonometric identity is needed; the two terms are identical products and cancel.
2. Expanding along the row with a zero leaves only two minors to compute.

Answer: (c) (1, −3)

Explanation:
1. Expand along \( R_1 \): \( 2(3 - 7m^2) - 3(6 - 6m^2) + m(14 - 6) \)
2. = \( 6 - 14m^2 - 18 + 18m^2 + 8m = 4m^2 + 8m - 12 \)
3. \( 4(m^2 + 2m - 3) = 4(m - 1)(m + 3) = 0 \)
4. m = 1 or m = −3, i.e. the solution set is (1, −3)

Teacher's Note:
1. Check m = 1: then \( R_1 \) = (2, 3, 1) and \( R_2 \) = (2, 1, 1); a quick expansion confirms zero.
2. Divide out the common factor 4 before factorising to keep the quadratic simple.

Answer: (b) All real x

Explanation:
1. \( \Delta_1 \) along \( R_1 \): 7[(x + 1)7 − 3x] − x[(−5)(7) − 3 × 4] + 2[−5x − 4(x + 1)] = 7(4x + 7) + 47x + 2(−9x − 4)
2. = 28x + 49 + 47x − 18x − 8, so \( \Delta_1 = 57x + 41 \)
3. \( \Delta_2 \) along \( R_1 \): x[3 × 4 − (−5)(7)] − 2[4(x + 1) + 5x] + 7[7(x + 1) − 3x] = 47x − 2(9x + 4) + 7(4x + 7)
4. = 47x − 18x − 8 + 28x + 49, so \( \Delta_2 = 57x + 41 \)
5. Hence \( \Delta_1 - \Delta_2 = 0 \) for all real x

Teacher's Note:
1. Both determinants are linear in x because x appears in only one column, so each expansion is short.
2. Getting the same expression 57x + 41 for both shows the difference is zero identically, not just for one x.

Answer: (a) x = 5

Explanation:
1. \( \Delta_1 \) = 10(49 + 20) − 4(119 − 16) + 3(−85 − 28) = 690 − 412 − 339 = −61
2. \( \Delta_2 \) along \( R_1 \): 4[7(x + 12) − 4(x − 1)] − (x + 5)[49 + 20] + 3[7(x − 1) + 5(x + 12)]
3. = 4(3x + 88) − 69(x + 5) + 3(12x + 53) = 12x + 352 − 69x − 345 + 36x + 159 = 166 − 21x
4. \( \Delta_1 + \Delta_2 \) = 105 − 21x = 0 ⇒ x = 5

Teacher's Note:
1. Δ2 is linear in x because x appears in only one column, so expect an answer of the form px + q.
2. Check: at x = 5, Δ2 = 166 − 105 = 61, which is exactly −Δ1.

Answer: (d) 6561

Explanation:
1. For a determinant of order n, the determinant of cofactors is \( \Delta' = \Delta^{n-1} \)
2. Here n = 3: \( \Delta' = 9^2 = 81 \)
3. \( \Delta'^2 = 81^2 = 6561 \)

Teacher's Note:
1. Read the question carefully: it asks for the square of \( \Delta' \), not \( \Delta' \) itself.
2. The same result in matrix form is \( |\text{adj } A| = |A|^{n-1} \).

Answer: (c) λ = 1 or 2

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix} \) = 1(20 − 16) − 1(10 − 4) + 1(4 − 2) = 0
2. Compute \( (\text{adj } A)B \) with \( B = \begin{bmatrix} 1 \\ \lambda \\ \lambda^2 \end{bmatrix} \): it equals \( (\lambda^2 - 3\lambda + 2)\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} \)
3. For a solution we need \( (\text{adj } A)B = O \): \( \lambda^2 - 3\lambda + 2 = 0 \)
4. (λ − 1)(λ − 2) = 0 ⇒ λ = 1 or 2

Teacher's Note:
1. When |A| = 0, the system is consistent (with infinitely many solutions) only if \( (\text{adj } A)B = O \).
2. Equivalently, eq3 = 3(eq2) − 2(eq1) on the left, so the constants must satisfy λ² = 3λ − 2.

Answer: (d) Finitely many solutions

Explanation:
1. Let \( X = \frac{x^2}{a^2}, Y = \frac{y^2}{b^2}, Z = \frac{z^2}{c^2} \): X + Y − Z = 1, X − Y + Z = 1, −X + Y + Z = 1
2. Adding the first two: 2X = 2 ⇒ X = 1; similarly Y = 1 and Z = 1
3. So x = ±a, y = ±b, z = ±c
4. That gives 2 × 2 × 2 = 8 solutions: finitely many

Teacher's Note:
1. Substitution turns a non-linear system into a linear one with a unique solution in X, Y, Z.
2. Each square equation like \( x^2 = a^2 \) gives two values, so count the sign choices at the end.

Answer: (a) k ≠ 0

Explanation:
1. \( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix} \)
2. = 1(k + 2) − 1(2k + 3) + 1(4 − 3) = −k
3. Unique solution ⇔ |A| ≠ 0 ⇔ k ≠ 0

Teacher's Note:
1. Notice that eq1 + eq2 gives 3x + 2y + 0z = 5, so at k = 0 the left sides are dependent.
2. Interval options like −1 < k < 1 include k = 0, so they cannot be correct.

Answer: (c) 0

Explanation:
1. Given \( a_{ij} = -a_{ji} \) for every i and j
2. Put j = i: \( a_{ii} = -a_{ii} \)
3. ⇒ \( 2a_{ii} = 0 \) ⇒ every principal diagonal element is 0

Teacher's Note:
1. This condition defines a skew-symmetric determinant, whose diagonal must be zero.
2. Since the order here is 3 (odd), the whole determinant also equals zero.

Answer: (d) −108

Explanation:
1. g is the constant term, so put x = 0 on both sides: g = f(0)
2. \( f(0) = \begin{vmatrix} 3 & 2 & 4 \\ 7 & 2 & 0 \\ 3 & -1 & 7 \end{vmatrix} \)
3. = 3(14 − 0) − 2(49 − 0) + 4(−7 − 6) = 42 − 98 − 52
4. = −108

Teacher's Note:
1. The constant term of a polynomial written as a determinant is the determinant evaluated at x = 0.
2. Similarly, the sum of all coefficients is obtained by putting x = 1.

Answer: (b) ax(2a + 3x)

Explanation:
1. Expand along \( R_1 \): \( f(x) = a(a^2 + ax) + 1(a^2x + ax^2) = a^3 + 2a^2x + ax^2 \)
2. \( f(2x) = a^3 + 4a^2x + 4ax^2 \)
3. \( f(2x) - f(x) = 2a^2x + 3ax^2 = ax(2a + 3x) \)

Teacher's Note:
1. Find f(x) as a polynomial first; then substituting 2x is easy.
2. Check with a = 1, x = 1: f(2) = 9 and f(1) = 4, and the difference 5 equals 1 × 1 × (2 + 3).

Answer: (a) 0

Explanation:
1. Since x + y + z = π, \( \sin(x + y + z) = 0 \); since A + B + C = π, \( \cos(A + B) = -\cos C \)
2. The determinant becomes \( \begin{vmatrix} 0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0 \end{vmatrix} \)
3. Expand along \( R_1 \): \( 0 - \sin B(0 + \tan A\cos C) + \cos C(\sin B\tan A - 0) \)
4. = \( -\sin B\tan A\cos C + \sin B\tan A\cos C = 0 \)

Teacher's Note:
1. Simplify the entries with the given angle conditions first; the hidden structure then appears.
2. With a zero in the first row only two minors are needed, and they cancel exactly.

Answer: (c) [z]

Explanation:
1. From the given ranges: [x] = −1, [y] = 0, [z] = 1
2. The determinant becomes \( \begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{vmatrix} \)
3. Expand along \( R_1 \): \( 1 \times \begin{vmatrix} -1 & 1 \\ -1 & 0 \end{vmatrix} = 1(0 + 1) = 1 \)
4. 1 = [z]

Teacher's Note:
1. Replace each greatest integer value first; the determinant then becomes purely numerical.
2. Remember that [x] = −1 for −1 ≤ x < 0, not 0.