CBSE Class 12 Mathematics Matrices MCQs Set 10

Answer: (b) Symmetric matrix

Explanation:
1. (AAᵀ)ᵀ = (Aᵀ)ᵀAᵀ = AAᵀ.
2. So AAᵀ is symmetric.

Teacher's Note:
1. The reversal law (XY)ᵀ = YᵀXᵀ is the whole proof; the same holds for AᵀA.
2. AAᵀ is symmetric even when A is not square.

Answer: (d) Skew-symmetric matrix

Explanation:
1. aⱼᵢ = j² − i² = −(i² − j²) = −aᵢⱼ for all i, j.
2. In particular aᵢᵢ = i² − i² = 0.
3. So Aᵀ = −A and A is skew-symmetric.

Teacher's Note:
1. Swap i and j in the formula: if the formula changes sign, the matrix is skew-symmetric; if it is unchanged, symmetric.
2. A is not the zero matrix, since a₁₂ = 1 − 4 = −3 ≠ 0.

Answer: (c) Skew-symmetric matrix

Explanation:
1. (A − Aᵀ)ᵀ = Aᵀ − (Aᵀ)ᵀ = Aᵀ − A = −(A − Aᵀ).
2. So A − Aᵀ is skew-symmetric.

Teacher's Note:
1. This holds for every square matrix A; the given fact about A + Aᵀ is always true and adds nothing.
2. A − Aᵀ is zero only when A is symmetric.

Answer: (b) a = 1

Explanation:
1. A is singular when |A| = 0.
2. |A| = 4a − 2 × 2 = 4a − 4.
3. 4a − 4 = 0 gives a = 1.

Teacher's Note:
1. Singular means determinant zero; non-singular means determinant non-zero.
2. Option (a) a ≠ 1 is exactly the non-singular condition, the opposite of what is asked.

Answer: (d) y − u = 0

Explanation:
1. A matrix is symmetric when its transpose equals itself.
2. \( \begin{bmatrix} x & y \\ u & v \end{bmatrix}^T = \begin{bmatrix} x & u \\ y & v \end{bmatrix} \), so we need u = y.
3. So y − u = 0.

Teacher's Note:
1. The diagonal entries x and v are unrestricted, so (a) and (b) are not required.
2. y + u = 0 would be the condition for the off-diagonal part of a skew-symmetric matrix.

Answer: (d) None of these

Explanation:
1. (a) If A is invertible, (adj A)⁻¹ = A/|A| and adj(A⁻¹) = |A⁻¹|(A⁻¹)⁻¹ = A/|A|, so they are equal.
2. (b) (Aᵀ)⁻¹ = (A⁻¹)ᵀ, from transposing AA⁻¹ = I. (c) (A³)⁻¹ = (A⁻¹)³, since A³(A⁻¹)³ = I.
3. All three statements are true (for invertible A), so the wrong statement is None of these.

Teacher's Note:
1. Inverse commutes with transpose, powers and adjoint, which is what all three statements say.
2. These results need A to be non-singular; always assume that when inverses appear.

Answer: (d) |A| ≠ 0

Explanation:
1. A square matrix A is non-singular if its determinant is non-zero.
2. So the condition is |A| ≠ 0, which covers both |A| > 0 and |A| < 0.

Teacher's Note:
1. Options (b) and (c) are each only part of the full condition, so (d) is the best answer.
2. Non-singular, invertible and |A| ≠ 0 all mean the same thing for square matrices.

Answer: (a) Symmetric

Explanation:
1. Compare symmetric positions: a₁₂ = a₂₁ = h, a₁₃ = a₃₁ = g, a₂₃ = a₃₂ = f.
2. So Aᵀ = A and the matrix is symmetric.

Teacher's Note:
1. This matrix appears in the general second-degree equation ax² + 2hxy + by² + 2gx + 2fy + c = 0.
2. It is not scalar unless a = b = c and f = g = h = 0.

Answer: (b) Unit matrix

Explanation:
1. A = −I.
2. A² = (−I)² = I.
3. So A² is the unit matrix.

Teacher's Note:
1. (−I)ⁿ is I for even n and −I for odd n.
2. Recognising A as a scalar matrix saves a full 3 × 3 multiplication.

Answer: (a) Symmetric

Explanation:
1. (M′AM)′ = M′A′(M′)′ by the reversal law.
2. Since A′ = A and (M′)′ = M, this is M′AM.
3. So M′AM is symmetric.

Teacher's Note:
1. The reversal law for three factors is (XYZ)′ = Z′Y′X′.
2. This result is used in coordinate changes for quadratic forms.

Answer: (d) A − A′

Explanation:
1. (A + A′)′ = A′ + A, so A + A′ is symmetric.
2. (AA′)′ = AA′ and (A′A)′ = A′A, so both are symmetric.
3. (A − A′)′ = A′ − A = −(A − A′), so A − A′ is skew-symmetric, not symmetric (unless it is O).

Teacher's Note:
1. A − A′ is the skew-symmetric part of A; it is symmetric only if A is itself symmetric.
2. The four expressions here are the standard symmetric and skew-symmetric constructions to remember.

Answer: (d) 2

Explanation:
1. A singular matrix has determinant 0.
2. Expanding along R1: \( |A| = 0(0 + 9) - 1(0 - 3\lambda) + (-2)(3 - 0) = 3\lambda - 6 \)
3. 3λ − 6 = 0 gives λ = 2.

Teacher's Note:
1. Expanding along the row or column with a zero saves a whole minor, so R1 is the natural choice here.
2. Watch the alternating signs + − + of the cofactors; most wrong answers come from missing the minus on the middle term.

Answer: (d) 4

Explanation:
1. Non-singular means |A| ≠ 0, so find the λ that makes |A| = 0.
2. \( |A| = 1(25 - 6\lambda) - 2(20 - 18) + 3(4\lambda - 15) \)
3. = 25 − 6λ − 4 + 12λ − 45 = 6λ − 24
4. |A| = 0 when λ = 4, so λ should not be equal to 4.

Teacher's Note:
1. For 'non-singular' questions, find where the determinant vanishes and exclude that value.
2. Collect the λ terms and the constants separately while expanding to avoid sign slips.

Answer: (c) Non-singular

Explanation:
1. Expand along R3: \( |A| = 1(0 - 1) - 0 + 0 = -1 \)
2. Since |A| = −1 ≠ 0, A is non-singular.
3. A is not symmetric (\( a_{23} = 1 \) but \( a_{32} = 0 \)) and not skew-symmetric (its diagonal is not zero).

Teacher's Note:
1. A row with two zeros, like R3 here, makes expansion almost instant.
2. A skew-symmetric matrix must have every diagonal entry 0, so a quick look at the diagonal rules that option out.

Answer: (a) Unit matrix

Explanation:
1. Row 1 and row 2 of \( A^2 \) are (1, 0, 0) and (0, 1, 0), since the first two rows of A are those of I.
2. Row 3 of \( A^2 \) = (a·1 + b·0 + (−1)a, a·0 + b·1 + (−1)b, (−1)(−1)) = (0, 0, 1)
3. So \( A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \), the unit matrix.

Teacher's Note:
1. A matrix with \( A^2 = I \) is called involutory and is its own inverse.
2. Compute one row at a time; here only the third row needs real work.

Answer: (b) 4

Explanation:
1. Singular means |A| = 0.
2. \( |A| = 1(40 - 40) - 3(20 - 24) + (\lambda + 2)(10 - 12) \)
3. = 0 + 12 − 2λ − 4 = 8 − 2λ
4. 8 − 2λ = 0 gives λ = 4.

Teacher's Note:
1. Keep the expression λ + 2 in brackets while multiplying by its minor so the sign of the constant is not lost.
2. Check: with λ = 4 the first row is (1, 3, 6), and the determinant indeed becomes 0.

Answer: (a) \( \begin{bmatrix} 0 & 4 & 5 \\ -4 & 0 & -6 \\ -5 & 6 & 0 \end{bmatrix} \)

Explanation:
1. A skew-symmetric matrix satisfies \( a_{ji} = -a_{ij} \), so every diagonal entry must be 0.
2. Options (b), (c), (d) have non-zero diagonal entries, so they are not skew-symmetric.
3. In (a): \( a_{12} = 4, a_{21} = -4 \); \( a_{13} = 5, a_{31} = -5 \); \( a_{23} = -6, a_{32} = 6 \), and the diagonal is 0.
4. So (a) is skew-symmetric.

Teacher's Note:
1. Check the diagonal first: any non-zero diagonal entry immediately rules out skew-symmetry.
2. Then pair each entry above the diagonal with its mirror below; they must be negatives of each other.

Answer: (a) Singular

Explanation:
1. Compare the columns: C1 = (1, 2, 3) and C3 = (2, 4, 6) = 2 × C1.
2. A determinant with two proportional columns is 0, so |A| = 0.
3. Hence A is singular.

Teacher's Note:
1. Before expanding, look for proportional rows or columns; it often gives the determinant as 0 at once.
2. A is also not symmetric because \( a_{12} = 3 \) but \( a_{21} = 2 \).

Answer: (b) A non-singular matrix

Explanation:
1. If A is non-singular, \( A^{-1} \) exists.
2. Pre-multiply AB = AC by \( A^{-1} \): \( A^{-1}AB = A^{-1}AC \), so IB = IC, i.e. B = C.
3. If A were singular, cancellation can fail (for example AB = AC = O with B ≠ C).
4. So A must be non-singular.

Teacher's Note:
1. Cancellation in matrix equations is allowed only through multiplication by an inverse.
2. A common trap is to cancel A as with numbers; this is wrong whenever |A| = 0.

Answer: (b) A is a null matrix

Explanation:
1. Symmetric: A′ = A. Skew-symmetric: A′ = −A.
2. So A = −A, which gives 2A = O.
3. Hence A = O, the null matrix.

Teacher's Note:
1. The zero matrix is the only matrix that is both symmetric and skew-symmetric.
2. Write both conditions as equations and combine them, rather than testing examples.

Answer: (c) \( \frac{A' - A}{2} \)

Explanation:
1. Let \( P = \frac{A' - A}{2} \).
2. \( P' = \frac{(A')' - A'}{2} = \frac{A - A'}{2} = -P \)
3. So P is skew-symmetric, while \( \frac{A + A'}{2} \) is symmetric and \( \frac{A + B}{2} \) need not be either.

Teacher's Note:
1. Use (A′)′ = A and (A ± B)′ = A′ ± B′ to test any such expression quickly.
2. A − A′ and A′ − A are always skew-symmetric; A + A′ is always symmetric.

Answer: (a) Sum of a symmetric matrix and a skew-symmetric matrix

Explanation:
1. Write \( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') \).
2. \( \frac{1}{2}(A + A') \) is symmetric because its transpose is itself.
3. \( \frac{1}{2}(A - A') \) is skew-symmetric because its transpose is its negative.
4. So every square matrix is the sum of a symmetric and a skew-symmetric matrix.

Teacher's Note:
1. This decomposition is unique and is a favourite CBSE question, so learn the two formulas P = ½(A + A′) and Q = ½(A − A′).
2. Check your answer by verifying P + Q = A, P′ = P and Q′ = −Q.

Answer: (b) A skew-symmetric matrix

Explanation:
1. A is skew-symmetric, so A′ = −A.
2. \( (A^n)' = (A')^n = (-A)^n = (-1)^n A^n \)
3. For odd n, \( (-1)^n = -1 \), so \( (A^n)' = -A^n \).
4. Hence \( A^n \) is skew-symmetric.

Teacher's Note:
1. For even n the same working gives \( (A^n)' = A^n \), so even powers of a skew-symmetric matrix are symmetric.
2. The rule \( (A^n)' = (A')^n \) comes from applying (AB)′ = B′A′ repeatedly.

Answer: (b) \( k^2 I \)

Explanation:
1. kI is a diagonal matrix with every diagonal entry k.
2. Each diagonal cofactor is a 2 × 2 minor \( \begin{vmatrix} k & 0 \\ 0 & k \end{vmatrix} = k^2 \); every off-diagonal cofactor is 0.
3. So \( \text{adj}(kI) = k^2 I \).

Teacher's Note:
1. In general \( \text{adj}(kA) = k^{n-1}\,\text{adj } A \) for an n × n matrix, here with n = 3.
2. Check: (kI)(k²I) = k³I = |kI| I, as A(adj A) = |A| I requires.

Answer: (a) A′

Explanation:
1. \( |A| = \cos^2\theta + \sin^2\theta = 1 \) (expanding along R3).
2. Cofactors: \( A_{11} = \cos\theta, A_{12} = -\sin\theta, A_{21} = \sin\theta, A_{22} = \cos\theta, A_{33} = 1 \), others 0.
3. \( \text{adj } A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \), which is A′.

Teacher's Note:
1. For a rotation matrix, |A| = 1 and adj A = A⁻¹ = A′, which is a quick check.
2. Remember that adj A is the transpose of the cofactor matrix, not the cofactor matrix itself.

Answer: (b) \( |A|^{n-2} A \)

Explanation:
1. Use B(adj B) = |B| I with B = adj A: \( (\text{adj } A)\,\text{adj}(\text{adj } A) = |\text{adj } A|\, I = |A|^{n-1} I \)
2. Pre-multiply by A: \( |A|\,\text{adj}(\text{adj } A) = |A|^{n-1} A \) (since A(adj A) = |A| I)
3. Dividing by |A| (A non-singular): \( \text{adj}(\text{adj } A) = |A|^{n-2} A \)

Teacher's Note:
1. Pair this with \( |\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} \) for determinant questions.
2. For n = 2, the formula gives adj(adj A) = A, which you can check on any 2 × 2 matrix.

Answer: (a) N

Explanation:
1. Cofactors of row 1: \( A_{11} = 0 - 4 = -4 \), \( A_{12} = -(3 - 4) = 1 \), \( A_{13} = 4 - 0 = 4 \)
2. Row 2: \( A_{21} = -(-9 + 12) = -3 \), \( A_{22} = -12 + 12 = 0 \), \( A_{23} = -(-16 + 12) = 4 \)
3. Row 3: \( A_{31} = -3 - 0 = -3 \), \( A_{32} = -(-4 + 3) = 1 \), \( A_{33} = 0 + 3 = 3 \)
4. \( \text{adj } N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} = N \)

Teacher's Note:
1. Write the cofactors of row i as column i of adj A; this transpose step is where most errors occur.
2. Check: |N| = 1 here, so N(adj N) = N² should equal I, which you can verify quickly.

Answer: (c) |A| I

Explanation:
1. For any square matrix, A(adj A) = (adj A)A = |A| I.
2. This follows because a row of A times its own cofactors gives |A|, and times another row's cofactors gives 0.
3. So A(adj A) = |A| I.

Teacher's Note:
1. This identity is the basis of the formula \( A^{-1} = \frac{1}{|A|}\,\text{adj } A \).
2. Note it is |A| I, a scalar matrix, and not the number |A| itself.

Answer: (b) 1

Explanation:
1. A(adj A) = |A| I, so k = |A|.
2. \( |A| = \cos\alpha\cos\alpha - \sin\alpha(-\sin\alpha) = \cos^2\alpha + \sin^2\alpha = 1 \)
3. So k = 1.

Teacher's Note:
1. Whenever A(adj A) is written as a scalar matrix, the scalar is simply |A|.
2. Rotation-type matrices always have determinant 1, a fact worth remembering.

Answer: (d) None of these

Explanation:
1. Cofactors of row 1: \( A_{11} = 2 \), \( A_{12} = -(5 + 0) = -5 \), \( A_{13} = 30 + 2 = 32 \)
2. Row 2: \( A_{21} = 0 \), \( A_{22} = 1 \), \( A_{23} = -(6 - 0) = -6 \); Row 3: \( A_{31} = 0, A_{32} = 0, A_{33} = 2 \)
3. Transposing: \( \text{adj } A = \begin{bmatrix} 2 & 0 & 0 \\ -5 & 1 & 0 \\ 32 & -6 & 2 \end{bmatrix} \)
4. Option (a) is the untransposed cofactor matrix, and (b), (c) are different, so the answer is none of these.

Teacher's Note:
1. Option (a) is a deliberate trap: it is the cofactor matrix, which must be transposed to get adj A.
2. For a lower triangular matrix, adj A is also lower triangular, a quick check on your answer.

Answer: (a) \( \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} \)

Explanation:
1. A(adj A) = |A| I.
2. |A| = 3(4) − 2(1) = 10
3. So \( A(\text{adj } A) = 10 I = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix} \).

Teacher's Note:
1. Never multiply A by adj A directly in such questions; |A| I gives the answer instantly.
2. For a 2 × 2 matrix, adj A is found by swapping the diagonal entries and changing the signs of the other two.

Answer: (a) Singular

Explanation:
1. \( |\text{adj } A| = |A|^{n-1} \) for an n × n matrix.
2. A is singular, so |A| = 0, giving |adj A| = 0 (n ≥ 2).
3. So adj A is also singular.

Teacher's Note:
1. adj A is defined for every square matrix, singular or not; only the inverse fails to exist.
2. The formula \( |\text{adj } A| = |A|^{n-1} \) settles most singular/non-singular questions about the adjoint.

Answer: (b) \( \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} \)

Explanation:
1. Row 1 cofactors: \( A_{11} = 6 - 3 = 3 \), \( A_{12} = -(3 + 6) = -9 \), \( A_{13} = -1 - 4 = -5 \)
2. Row 2: \( A_{21} = -(3 + 1) = -4 \), \( A_{22} = 3 - 2 = 1 \), \( A_{23} = -(-1 - 2) = 3 \)
3. Row 3: \( A_{31} = -3 - 2 = -5 \), \( A_{32} = -(-3 - 1) = 4 \), \( A_{33} = 2 - 1 = 1 \)
4. Transposing: \( \text{adj} = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} \), option (b).

Teacher's Note:
1. Option (a) is the cofactor matrix before transposing, the classic distractor.
2. Check one entry: R1 of A times column 1 of adj must give |A| = 3 − 9 − 5 = −11, and it does: 3 − 9 − 5 = −11.

Answer: (c) O

Explanation:
1. For square matrices of the same order, adj (AB) = (adj B)(adj A).
2. So adj (AB) − (adj B)(adj A) = O, the null matrix.

Teacher's Note:
1. The order reverses, just like \( (AB)^{-1} = B^{-1}A^{-1} \) and (AB)′ = B′A′.
2. Writing adj (AB) = (adj A)(adj B) is the common mistake to avoid.

Answer: (b) \( \begin{pmatrix} 1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1 \end{pmatrix} \)

Explanation:
1. Row 1 cofactors: \( A_{11} = 1 - 0 = 1 \), \( A_{12} = -(0 - 4) = 4 \), \( A_{13} = 0 - 2 = -2 \)
2. Row 2: \( A_{21} = -(2 - 0) = -2 \), \( A_{22} = 1 - 0 = 1 \), \( A_{23} = -(0 - 4) = 4 \)
3. Row 3: \( A_{31} = 4 - 0 = 4 \), \( A_{32} = -(2 - 0) = -2 \), \( A_{33} = 1 - 0 = 1 \)
4. Transposing the cofactor matrix: \( \text{adj } A = \begin{pmatrix} 1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1 \end{pmatrix} \), option (b).

Teacher's Note:
1. Option (a) is the cofactor matrix itself; adj A needs one more step, the transpose.
2. Check: R1 of A times column 1 of adj A is 1(1) + 2(4) + 0 = 9 = |A|, as it must be.

Answer: (c) 144

Explanation:
1. \( |A| = 1(108 - 72) - 2(27 - 9) + 3(8 - 4) = 36 - 36 + 12 = 12 \)
2. For a 3 × 3 matrix, \( |\text{adj } A| = |A|^{3-1} = |A|^2 \)
3. So |adj A| = 12² = 144.

Teacher's Note:
1. Never compute adj A just to find its determinant; use \( |\text{adj } A| = |A|^{n-1} \).
2. Taking out common factors (here the matrix is a Vandermonde type) is another quick way to get |A|.

Answer: (d) \( 2^6 \)

Explanation:
1. \( |\text{adj } A| = |A|^{n-1} \) with n = 3.
2. |adj A| = 8² = 64
3. 64 = \( 2^6 \)

Teacher's Note:
1. Write the answer in the form the options use; here 64 has to be recognised as \( 2^6 \).
2. A common slip is using \( |A|^n \) instead of \( |A|^{n-1} \), which would give 512.

Answer: (b) (adj B)(adj A)

Explanation:
1. For non-singular A, B: \( \text{adj } A = |A| A^{-1} \) and \( \text{adj } B = |B| B^{-1} \).
2. \( \text{adj}(AB) = |AB|(AB)^{-1} = |A||B|\,B^{-1}A^{-1} \)
3. \( = (|B|B^{-1})(|A|A^{-1}) = (\text{adj } B)(\text{adj } A) \)

Teacher's Note:
1. The reversal of order is the same pattern as in \( (AB)^{-1} = B^{-1}A^{-1} \).
2. Options (c) and (d) use adjoints of inverses, which bring in extra factors of 1/|A| and 1/|B|.

Answer: (b) \( d^{n-1} \)

Explanation:
1. A(adj A) = |A| I = d I
2. Take determinants: \( |A|\,|\text{adj } A| = d^n \) (an n × n scalar matrix dI has determinant \( d^n \))
3. So \( d\,|\text{adj } A| = d^n \), giving \( |\text{adj } A| = d^{n-1} \).

Teacher's Note:
1. The derivation uses |kI| = kⁿ, not k; this is the step students most often get wrong.
2. The result also holds when d = 0, since adj A is then singular.

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

Explanation:
1. For \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), adj = \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
2. Here a = 1, b = −1, c = 2, d = 3.
3. \( \text{adj } A = \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix} \)

Teacher's Note:
1. For 2 × 2 matrices: swap the diagonal entries and change the signs of the off-diagonal entries.
2. Option (d) changes the signs of only one off-diagonal entry, a common slip.

Answer: (d) Both (a) and (c)

Explanation:
1. A(adj A) = |A| I for every square matrix.
2. |A| = 3(7) − 4(5) = 21 − 20 = 1
3. So A(adj A) = |A| I = 1·I = I.
4. Both (a) and (c) describe this matrix, so the answer is (d).

Teacher's Note:
1. Option (b) is a number, not a matrix, so it can never equal A(adj A).
2. When |A| = 1, A(adj A) = I and adj A is exactly \( A^{-1} \).

Answer: (a) \( \begin{bmatrix} 7 & -6 \\ 5 & -2 \end{bmatrix} \)

Explanation:
1. For \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), adj = \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
2. Here a = −2, b = 6, c = −5, d = 7.
3. \( \text{adj } A = \begin{bmatrix} 7 & -6 \\ 5 & -2 \end{bmatrix} \)

Teacher's Note:
1. Check: A(adj A) should be |A| I = (−14 + 30) I = 16 I.
2. Option (c) also transposes the entries, which is wrong for the 2 × 2 shortcut.

Answer: (b) \( \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \)

Explanation:
1. Row 1 cofactors: \( A_{11} = -3 + 4 = 1 \), \( A_{12} = -(2 - 0) = -2 \), \( A_{13} = -2 - 0 = -2 \)
2. Row 2: \( A_{21} = -(-3 + 4) = -1 \), \( A_{22} = 3 - 0 = 3 \), \( A_{23} = -(-3 - 0) = 3 \)
3. Row 3: \( A_{31} = -12 + 12 = 0 \), \( A_{32} = -(12 - 8) = -4 \), \( A_{33} = -9 + 6 = -3 \)
4. Transposing: \( \text{adj} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \), option (b).

Teacher's Note:
1. Verify with one product: R1 of A times column 1 of adj gives 3 + 6 − 8 = 1 = |A|.
2. Here |A| = 1, so this adjoint is also the inverse of the matrix.