CBSE Class 12 Mathematics Matrices MCQs Set 12

Answer: (d) \( \begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix} \)

Explanation:
1. |A| = 2(4) − 1(7) = 1
2. adj A = \( \begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix} \)
3. \( A^{-1} = \begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix} \)

Teacher's Note:
1. When |A| = 1, the inverse equals the adjoint, which is quick to write down.
2. Check: \( \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix} = \begin{bmatrix} 8 - 7 & -2 + 2 \\ 28 - 28 & -7 + 8 \end{bmatrix} = I \).

Answer: (a) \( \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix} \)

Explanation:
1. \( |A| = 0(2 - 3) - 1(1 - 9) + 2(1 - 6) = 8 - 10 = -2 \)
2. Test option (a) as B: row 1 of A gives (0 − 4 + 5, 0 + 3 − 3, 0 − 1 + 1) = (1, 0, 0).
3. Row 2 gives (½ − 8 + 15/2, −½ + 6 − 9/2, ½ − 2 + 3/2) = (0, 1, 0).
4. Row 3 gives (3/2 − 4 + 5/2, −3/2 + 3 − 3/2, 3/2 − 1 + ½) = (0, 0, 1), so AB = I and (a) is the inverse.

Teacher's Note:
1. Since |A| = −2, the inverse must contain halves, which already narrows the options.
2. Checking AB = I row by row is a reliable way to confirm an inverse from given choices.

Answer: (a) \( \frac{1}{10}\begin{bmatrix} 4 & 2 \\ -3 & 1 \end{bmatrix} \)

Explanation:
1. |A| = 1(4) − (−2)(3) = 4 + 6 = 10
2. adj A = \( \begin{bmatrix} 4 & 2 \\ -3 & 1 \end{bmatrix} \)
3. \( A^{-1} = \frac{1}{10}\begin{bmatrix} 4 & 2 \\ -3 & 1 \end{bmatrix} \)

Teacher's Note:
1. Option (d) is adj A without the factor 1/10.
2. Option (c) changes the sign of only one off-diagonal entry.

Answer: (d) \( \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)

Explanation:
1. A interchanges the first and third entries of any column vector.
2. Doing this twice restores the vector, so \( A^2 = I \).
3. Hence \( A^{-1} = A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \).

Teacher's Note:
1. Check: \( A^2 \) has row 1 = (0, 0, 1)A = (1, 0, 0), row 2 = (0, 1, 0), row 3 = (0, 0, 1).
2. Options (a) and (c) have determinant 0 or the wrong structure, so they can be rejected quickly.

Answer: (a) \( A^{-1} = |A|^{-1} \)

Explanation:
1. \( A^{-1} \) is a matrix while \( |A|^{-1} \) is a number, so (a) is not true.
2. \( (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \), so (b) is true.
3. \( (A')^{-1} = (A^{-1})' \) always, so (c) is true.
4. Hence the statement that is not true is (a).

Teacher's Note:
1. The correct link is \( |A^{-1}| = |A|^{-1} \): the determinant of the inverse, not the inverse itself.
2. In 'which is not true' questions, check each option and mark the one that fails.

Answer: (c) \( \begin{bmatrix} -2 & 0 & 1 \\ 0 & 3 & -2 \\ 1 & -2 & 1 \end{bmatrix} \)

Explanation:
1. \( |A| = 1(18 - 16) - 2(12 - 12) + 3(8 - 9) = 2 - 0 - 3 = -1 \)
2. Test option (c) as B: row 1 (1, 2, 3) gives (−2 + 0 + 3, 0 + 6 − 6, 1 − 4 + 3) = (1, 0, 0).
3. Row 2 (2, 3, 4) gives (−4 + 0 + 4, 0 + 9 − 8, 2 − 6 + 4) = (0, 1, 0).
4. Row 3 (3, 4, 6) gives (−6 + 0 + 6, 0 + 12 − 12, 3 − 8 + 6) = (0, 0, 1), so (c) is the inverse.

Teacher's Note:
1. Option (a) is −(c), the result of forgetting that |A| = −1.
2. A symmetric matrix has a symmetric inverse, which is a useful check here.

Answer: (d) \( \frac{1}{7}\begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix} \)

Explanation:
1. |A| = 1(−1) − 2(−4) = −1 + 8 = 7
2. adj A = \( \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix} \)
3. \( A^{-1} = \frac{1}{7}\begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix} \)

Teacher's Note:
1. Option (b) is adj A without dividing by |A| = 7.
2. Check: A(adj A) = [−1 + 8, −2 + 2; 4 − 4, 8 − 1] = 7I.

Answer: (c) \( \begin{bmatrix} 1/3 & -1/3 \\ 1/3 & 1/6 \end{bmatrix} \)

Explanation:
1. \( M = AB = \begin{bmatrix} 0 - 1 + 2 & 0 + 0 + 2 \\ 0 - 2 + 0 & 2 + 0 + 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -2 & 2 \end{bmatrix} \)
2. |M| = 2 + 4 = 6; adj M = \( \begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix} \)
3. \( M^{-1} = \frac{1}{6}\begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1/3 & -1/3 \\ 1/3 & 1/6 \end{bmatrix} \)

Teacher's Note:
1. A (2 × 3) times B (3 × 2) gives a 2 × 2 matrix, which is why M can have an inverse.
2. Options (c) and (d) differ by one sign, so recheck the sign of adj M carefully.

Answer: (d) Invertible matrix

Explanation:
1. \( A^{-1} \) exists, so A is invertible (non-singular).
2. Orthogonal, symmetric or diagonal are special cases not implied by the given condition.

Teacher's Note:
1. Any matrix with |A| ≠ 0 has an inverse, whatever its other properties.
2. An orthogonal matrix would need \( A^{-1} = A' \), which is a much stronger condition.

Answer: (b) a = cos 2θ, b = sin 2θ

Explanation:
1. Let t = tan θ. \( \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}^{-1} = \frac{1}{1 + t^2}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \)
2. \( \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \begin{bmatrix} 1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix} \)
3. So \( a = \frac{1 - t^2}{1 + t^2} = \cos 2\theta \) and \( b = \frac{2t}{1 + t^2} = \sin 2\theta \)

Teacher's Note:
1. The formulas \( \cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \) and \( \sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} \) are the key here.
2. Quick check with θ = 0: both matrices are I, so a = 1, b = 0, matching cos 0 and sin 0.

Answer: (a) det (−A) = −det A

Explanation:
1. For an n × n matrix, det (kA) = \( k^n \) det A.
2. With k = −1 and n = 3: det (−A) = \( (-1)^3 \) det A = −det A.
3. det 2A = 8 det A (not 2 det A), and (b), (c) are not true in general.

Teacher's Note:
1. Multiplying a matrix by k multiplies every row by k, so the determinant gains a factor \( k^n \).
2. For even order, det (−A) = det A; the order of the matrix matters.

Answer: (c) 16

Explanation:
1. A is diagonal: |A| = 2 · 2 · 2 = 8.
2. B is upper triangular: |B| = 1 · 1 · 2 = 2.
3. |AB| = |A||B| = 8 × 2 = 16

Teacher's Note:
1. Use |AB| = |A||B| instead of multiplying the matrices.
2. The determinant of a triangular or diagonal matrix is the product of its diagonal entries.

Answer: (b) −81

Explanation:
1. For order 3, |3AB| = \( 3^3 \)|AB| = 27|A||B|.
2. = 27 × (−1) × 3
3. = −81

Teacher's Note:
1. The scalar 3 is raised to the order of the matrix, giving 27, not 3.
2. Option (a) comes from using 3|A||B|, the most common mistake.

Answer: (c) Determinant is a number associated to a square matrix

Explanation:
1. A determinant is a single number, not a matrix, so (a) is false.
2. It is defined only for square matrices.
3. So a determinant is a number associated to a square matrix.

Teacher's Note:
1. Rectangular matrices such as 2 × 3 have no determinant.
2. Notation: [ ] or ( ) denotes a matrix, while | | denotes its determinant.

Answer: (a) 0

Explanation:
1. A′ = −A, so |A′| = |−A| = \( (-1)^n \)|A|.
2. For odd n, |A| = |A′| = −|A|.
3. So 2|A| = 0, i.e. |A| = 0.

Teacher's Note:
1. Every skew-symmetric matrix of odd order is singular.
2. For even order the determinant can be non-zero, e.g. \( \begin{vmatrix} 0 & 1 \\ -1 & 0 \end{vmatrix} = 1 \).

Answer: (b) 10

Explanation:
1. A(adj A) = |A| I for every square matrix.
2. Here A(adj A) = 10 I.
3. So |A| = 10.

Teacher's Note:
1. Do not confuse with |A(adj A)|, which would be \( 10^2 = 100 \) for a 2 × 2 matrix.
2. The scalar on the diagonal of A(adj A) is always |A|.

Answer: (b) 25

Explanation:
1. \( A^2 - 2A = A(A - 2I) \)
2. |A| = 1 − 6 = −5; \( A - 2I = \begin{bmatrix} -1 & 3 \\ 2 & -1 \end{bmatrix} \), so |A − 2I| = 1 − 6 = −5
3. \( |A^2 - 2A| = |A||A - 2I| = (-5)(-5) = 25 \)

Teacher's Note:
1. Factoring the expression lets you use |AB| = |A||B| instead of squaring the matrix.
2. Direct check: \( A^2 = \begin{bmatrix} 7 & 6 \\ 4 & 7 \end{bmatrix} \), so \( A^2 - 2A = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \) with determinant 25.

Answer: (b) \( -\frac{25}{13} \)

Explanation:
1. Singular means the determinant is 0.
2. Expand along R1: (2 + x)(5 − 2) − 3(−5 − 2x) + 4(1 + x)
3. = 6 + 3x + 15 + 6x + 4 + 4x = 25 + 13x
4. 25 + 13x = 0 gives \( x = -\frac{25}{13} \).

Teacher's Note:
1. Keep 2 + x in brackets while multiplying by its minor to avoid losing a term.
2. The answer has a negative sign, which separates option (b) from (d).

Answer: (c) ±1

Explanation:
1. AA′ = I, so |A||A′| = |I| = 1.
2. |A′| = |A|, so \( |A|^2 = 1 \).
3. Hence |A| = ±1.

Teacher's Note:
1. Such a matrix is called orthogonal; its determinant is always 1 or −1.
2. Both values are possible, e.g. I has determinant 1 and diag(1, −1) has determinant −1.

Answer: (a) 2

Explanation:
1. Expand along R1: |A| = 1(1 − 0) − 0(2 − 0) + 1(4 − 3)
2. = 1 + 1 = 2

Teacher's Note:
1. Choose the row with a zero to reduce work; here R1 has a 0 in the middle.
2. Check by expanding along C3: 1(4 − 3) − 0 + 1(1 − 0) = 2.

Answer: (b) Det (A) = 0 and Det (B) = 0

Explanation:
1. Suppose |A| ≠ 0. Then \( A^{-1} \) exists and \( B = A^{-1}(AB) = A^{-1}O = O \), which contradicts B ≠ O. So |A| = 0.
2. Suppose |B| ≠ 0. Then \( A = (AB)B^{-1} = OB^{-1} = O \), which contradicts A ≠ O. So |B| = 0.
3. Hence Det (A) = 0 and Det (B) = 0.

Teacher's Note:
1. Two non-zero matrices can multiply to give O only when both are singular.
2. Contradiction arguments of this kind are a standard way to prove singularity.

Answer: (a) 0 or 1

Explanation:
1. \( A^2 = A \) gives \( |A|^2 = |A| \).
2. |A|(|A| − 1) = 0
3. So |A| = 0 or |A| = 1.

Teacher's Note:
1. A matrix with \( A^2 = A \) is called idempotent.
2. Both values occur: O has determinant 0 and I has determinant 1.

Answer: (c) \( 2^n \)

Explanation:
1. \( |A^n| = |A \cdot A \cdots A| = |A|^n \) (using |AB| = |A||B| repeatedly)
2. \( = 2^n \)

Teacher's Note:
1. The determinant of a power is the power of the determinant.
2. Do not confuse this with |kA| = kⁿ|A|, which involves a scalar multiple.

Answer: (d) An arbitrary integer

Explanation:
1. With integer entries, |A| is always an integer.
2. All entries 1 gives |A| = 0; \( \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \) gives |A| = 4.
3. \( \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix} \) gives |A| = 1(1) − 2(3) + 1(1) = −4.
4. So |A| can be zero, positive or negative: an arbitrary integer.

Teacher's Note:
1. Positive entries do not force a positive determinant; the minus signs in expansion can win.
2. To reject an option such as 'positive', one counterexample is enough.

Answer: (d) None of these

Explanation:
1. Take A = I and B = −I (2 × 2). Then A + B = O, so det (A + B) = 0.
2. But det A = 1 and det B = (−1)² = 1, so det A + det B = 2 ≠ 0.
3. Neither det A nor det B is 0, so (a), (b) and (c) all fail.
4. Hence none of these follows.

Teacher's Note:
1. The determinant is not additive: det (A + B) ≠ det A + det B in general.
2. A single counterexample is enough to show that a statement does not always hold.

Answer: (c) (2, 3, 4)

Explanation:
1. Equal matrices have equal corresponding entries: x + y + z = 9, x + y = 5, y + z = 7.
2. From the first two: z = 9 − 5 = 4.
3. From y + z = 7: y = 3; then x = 5 − 3 = 2.
4. So (x, y, z) = (2, 3, 4).

Teacher's Note:
1. Subtracting one equation from another often gives a variable directly, as here.
2. Check: 2 + 3 + 4 = 9, 2 + 3 = 5, 3 + 4 = 7.

Answer: (d) (−1, 0, 2)

Explanation:
1. Multiplying out: x + z = 1, −x + y = 1, −y + z = 2.
2. From the first two: z = 1 − x and y = 1 + x.
3. Substitute in the third: −(1 + x) + (1 − x) = 2, so −2x = 2 and x = −1.
4. Then y = 0, z = 2, so (x, y, z) = (−1, 0, 2).

Teacher's Note:
1. Substituting each option into the equations is a quick alternative in MCQs.
2. Check: −1 + 2 = 1, 1 + 0 = 1, 0 + 2 = 2.

Answer: (a) \( \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} \)

Explanation:
1. AX = B gives \( X = A^{-1}B \).
2. Row 1: 3(9) − ½(52) − ½(0) = 27 − 26 = 1
3. Row 2: −4(9) + ¾(52) + 5/4(0) = −36 + 39 = 3
4. Row 3: 2(9) − ¼(52) − ¾(0) = 18 − 13 = 5, so \( X = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} \).

Teacher's Note:
1. X = A⁻¹B, not BA⁻¹; B is a column, so it must be on the right.
2. The zero in B means the third column of A⁻¹ does not affect the answer.

Answer: (c) \( \begin{bmatrix} 14 & -1 \\ 0 & 9 \end{bmatrix} \)

Explanation:
1. \( A^2 = \begin{bmatrix} -2 & 1 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} -2 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 4 & -2 + 3 \\ 0 & 9 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 0 & 9 \end{bmatrix} \)
2. \( 2A^2 = \begin{bmatrix} 8 & 2 \\ 0 & 18 \end{bmatrix} \) and \( 3A = \begin{bmatrix} -6 & 3 \\ 0 & 9 \end{bmatrix} \)
3. \( f(A) = 2A^2 - 3A = \begin{bmatrix} 14 & -1 \\ 0 & 9 \end{bmatrix} \)

Teacher's Note:
1. In f(A), any constant term c must be replaced by cI; here there is none.
2. Subtract entry by entry carefully: 8 − (−6) = 14 and 2 − 3 = −1.

Answer: (c) \( \begin{bmatrix} 2 & 9/2 & 8 & 25/2 \\ 9/2 & 8 & 25/2 & 18 \\ 8 & 25/2 & 18 & 49/2 \end{bmatrix} \)

Explanation:
1. \( a_{ij} = \frac{(i + j)^2}{2} \) with i = 1, 2, 3 and j = 1, 2, 3, 4.
2. Row 1: 4/2, 9/2, 16/2, 25/2 = 2, 9/2, 8, 25/2
3. Row 2: 9/2, 16/2, 25/2, 36/2 = 9/2, 8, 25/2, 18
4. Row 3: 16/2, 25/2, 36/2, 49/2 = 8, 25/2, 18, 49/2, which is option (c).

Teacher's Note:
1. Since \( a_{ij} \) depends only on i + j, each row is the previous one shifted by one place, a quick check.
2. A 3 × 4 matrix has 3 rows and 4 columns, which rules out option (b) at once.

Answer: (c) A(z) = A(x)A(y)

Explanation:
1. \( A(x)A(y) = \frac{1}{(1 - x)(1 - y)}\begin{bmatrix} 1 + xy & -(x + y) \\ -(x + y) & 1 + xy \end{bmatrix} = \frac{1 + xy}{(1 - x)(1 - y)}\begin{bmatrix} 1 & -z \\ -z & 1 \end{bmatrix} \)
2. \( 1 - z = 1 - \frac{x + y}{1 + xy} = \frac{(1 - x)(1 - y)}{1 + xy} \)
3. So \( A(z) = \frac{1}{1 - z}\begin{bmatrix} 1 & -z \\ -z & 1 \end{bmatrix} = \frac{1 + xy}{(1 - x)(1 - y)}\begin{bmatrix} 1 & -z \\ -z & 1 \end{bmatrix} \)
4. Hence A(z) = A(x)A(y).

Teacher's Note:
1. Factor out 1 + xy from the product so that the entries look like those of A(z).
2. The form z = (x + y)/(1 + xy) is the addition rule for tanh, a hint that a product rule is intended.

Answer: (b) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Explanation:
1. \( A^2 = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}\begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} = \begin{bmatrix} 0 - i^2 & 0 \\ 0 & -i^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
2. \( A^{40} = (A^2)^{20} = I^{20} = I \)

Teacher's Note:
1. For high powers, find a small power that gives I (or a scalar multiple of I) and use it to reduce the exponent.
2. Remember \( i^2 = -1 \), so \( -i^2 = 1 \).

Answer: (d) \( A^3 - 3A^2 - A + 9I_3 = O \)

Explanation:
1. \( A^2 = \begin{bmatrix} 4 & 3 & 0 \\ -3 & 2 & -2 \\ 6 & 4 & 5 \end{bmatrix} \)
2. \( A^3 = A^2 A = \begin{bmatrix} 4 & 11 & 1 \\ -9 & -2 & -7 \\ 21 & 11 & 7 \end{bmatrix} \)
3. Check (d): \( A^3 - 3A^2 - A + 9I \) has (1, 1) entry 4 − 12 − 1 + 9 = 0, (1, 2) entry 11 − 9 − 2 = 0, and similarly every entry is 0.
4. So \( A^3 - 3A^2 - A + 9I_3 = O \).

Teacher's Note:
1. Testing one entry, such as (1, 1), eliminates most options quickly: only (d) gives 0 there.
2. This equation comes from the characteristic polynomial of A, a result you will meet in higher studies.

Answer: (a) 4A − 3I

Explanation:
1. \( A^2 = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} \)
2. \( 4A - 3I = \begin{pmatrix} 8 & -4 \\ -4 & 8 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} \)
3. So \( A^2 = 4A - 3I \).

Teacher's Note:
1. To test each option, compare just the (1, 2) entry first: only 4A gives −4.
2. A scalar term in a matrix equation must always be multiplied by I.

Answer: (c) \( \begin{bmatrix} a^n & 0 & 0 \\ 0 & b^n & 0 \\ 0 & 0 & c^n \end{bmatrix} \)

Explanation:
1. \( A^2 = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & b^2 & 0 \\ 0 & 0 & c^2 \end{bmatrix} \), because off-diagonal products are all 0.
2. Each further multiplication by A multiplies each diagonal entry by itself again.
3. So \( A^n = \begin{bmatrix} a^n & 0 & 0 \\ 0 & b^n & 0 \\ 0 & 0 & c^n \end{bmatrix} \).

Teacher's Note:
1. Powers of a diagonal matrix are found by raising each diagonal entry to that power.
2. Option (a) is nA, not Aⁿ; do not confuse a multiple with a power.

Answer: (d) \( A_\alpha.A_\beta = A_{\alpha+\beta} \) and \( (A_\alpha)^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{bmatrix} \)

Explanation:
1. \( A_\alpha A_\beta = \begin{bmatrix} \cos\alpha\cos\beta - \sin\alpha\sin\beta & \cos\alpha\sin\beta + \sin\alpha\cos\beta \\ -(\sin\alpha\cos\beta + \cos\alpha\sin\beta) & \cos\alpha\cos\beta - \sin\alpha\sin\beta \end{bmatrix} \)
2. \( = \begin{bmatrix} \cos(\alpha + \beta) & \sin(\alpha + \beta) \\ -\sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix} = A_{\alpha+\beta} \)
3. Repeating, \( (A_\alpha)^n = A_{n\alpha} = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{bmatrix} \), option (d).

Teacher's Note:
1. Rotation matrices add their angles when multiplied, so powers multiply the angle.
2. Option (c) raises the entries to the power n, which is not how matrix powers work.

Answer: (c) \( DD' = D^n \)

Explanation:
1. \( D' = |\text{adj } A| = |A|^{n-1} = D^{n-1} \)
2. \( DD' = D \cdot D^{n-1} = D^n \)

Teacher's Note:
1. The same result follows from A(adj A) = DI by taking determinants: \( D \cdot D' = D^n \).
2. Keep the order n in mind: for a 3 × 3 matrix, DD′ = D³.

Answer: (a) \( \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} \)

Explanation:
1. \( (A')^{-1} = (A^{-1})' = \frac{1}{|A|}(\text{adj } A)' \), and (adj A)′ is the cofactor matrix of A.
2. \( |A| = 1(-1 - 8) + 2(0 + 8) + 3(0 - 2) = -9 + 16 - 6 = 1 \)
3. Cofactors: row 1: −9, −8, −2; row 2: 8, 7, 2; row 3: −5, −4, −1.
4. So \( (A')^{-1} = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} \), option (a).

Teacher's Note:
1. \( (A')^{-1} = (A^{-1})' \), so the inverse of the transpose is simply the cofactor matrix divided by |A|.
2. Option (c) is \( A^{-1} \) itself, the trap for students who forget the transpose.

Answer: (b) \( \frac{1}{3}\begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{bmatrix} \)

Explanation:
1. Use \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). Let \( B = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{bmatrix} \).
2. Row 1 of A times B: (3, 1 + ω² + ω, 1 + ω + ω²) = (3, 0, 0)
3. Row 2 (1, ω, ω²) times B: (0, 1 + ω³ + ω³, 1 + ω² + ω⁴) = (0, 3, 0); row 3 similarly gives (0, 0, 3).
4. So AB = 3I and \( A^{-1} = \frac{1}{3}B \), option (b).

Teacher's Note:
1. Reduce powers of ω using \( \omega^3 = 1 \) (for example \( \omega^4 = \omega \)) before adding.
2. B is obtained from A by replacing ω with \( \omega^2 \), its complex conjugate.

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

Explanation:
1. \( A^2 = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \)
2. \( A^4 = (A^2)^2 \): row 1 = (9 + 0 − 8, −12 + 4 + 8, 12 + 0 − 12) = (1, 0, 0); row 2 = (0, 1, 0); row 3 = (−6 + 0 + 6, 8 − 2 − 6, −8 + 0 + 9) = (0, 0, 1)
3. So \( A^4 = I \), i.e. \( A \cdot A^3 = I \).
4. Hence \( A^{-1} = A^3 \).

Teacher's Note:
1. If a power of A equals I, the inverse is the next lower power: \( A^k = I \Rightarrow A^{-1} = A^{k-1} \).
2. Squaring \( A^2 \) is faster than multiplying A four times.

Answer: (b) diag \( (d_1^{-1}, d_2^{-1}, \ldots, d_n^{-1}) \)

Explanation:
1. D is diagonal with all \( d_i \neq 0 \).
2. \( D \cdot \text{diag}(d_1^{-1}, \ldots, d_n^{-1}) = \text{diag}(d_1 d_1^{-1}, \ldots, d_n d_n^{-1}) = I \)
3. So \( D^{-1} = \text{diag}(d_1^{-1}, d_2^{-1}, \ldots, d_n^{-1}) \).

Teacher's Note:
1. A diagonal matrix is inverted by taking the reciprocal of each diagonal entry.
2. The condition \( d_i \neq 0 \) is exactly the condition |D| = \( d_1 d_2 \cdots d_n \neq 0 \).

Answer: (b) diag \( (d_1^n, d_2^n, d_3^n, \ldots, d_n^n) \)

Explanation:
1. Product of two diagonal matrices: diag(a₁, …, aₙ) · diag(b₁, …, bₙ) = diag(a₁b₁, …, aₙbₙ).
2. So \( A^2 = \text{diag}(d_1^2, \ldots, d_n^2) \), and repeating n times gives
3. \( A^n = \text{diag}(d_1^n, d_2^n, \ldots, d_n^n) \)

Teacher's Note:
1. Here the power n happens to equal the order of the matrix, but the rule works for any positive power.
2. Option (a) uses the power n − 1, which comes from confusing this with the adjoint formula.

Answer: (d) det A = 0, (adj A)B = O

Explanation:
1. If det A ≠ 0, \( X = A^{-1}B \) is the only solution, so (a) and (b) give a unique solution, not infinitely many.
2. If det A = 0 and (adj A)B ≠ O, the system is inconsistent (no solution).
3. If det A = 0 and (adj A)B = O, the system can have infinitely many solutions.
4. So the required condition is det A = 0, (adj A)B = O.

Teacher's Note:
1. Strictly, det A = 0 with (adj A)B = O means the system is either consistent with infinitely many solutions or inconsistent, so check one solution exists.
2. Remember the three cases: |A| ≠ 0 unique; |A| = 0 and (adj A)B ≠ O none; |A| = 0 and (adj A)B = O infinitely many or none.