CBSE Class 12 Mathematics Matrices MCQs Set 11

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

Explanation:
1. A(adj A) = |A| I.
2. \( |A| = \cos^2 x + \sin^2 x = 1 \)
3. So \( A(\text{adj } A) = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. For this rotation-type matrix, |A| = 1 for every x.
2. There is no need to find adj A; the identity A(adj A) = |A| I does the work.

Answer: (c) \( a^9 \)

Explanation:
1. A is diagonal, so \( |A| = a \cdot a \cdot a = a^3 \).
2. \( |\text{adj } A| = |A|^{3-1} = (a^3)^2 = a^6 \)
3. \( |A|\,|\text{adj } A| = a^3 \cdot a^6 = a^9 \)

Teacher's Note:
1. Alternatively, \( |A|\,|\text{adj } A| = |A|^n \) = \( (a^3)^3 = a^9 \) directly.
2. The determinant of a diagonal matrix is just the product of its diagonal entries.

Answer: (b) \( a^6 \)

Explanation:
1. \( |A| = a^3 \) (product of the diagonal entries).
2. \( |\text{adj } A| = |A|^{n-1} = (a^3)^2 = a^6 \)

Teacher's Note:
1. Directly: adj A = \( a^2 I \), whose determinant is \( (a^2)^3 = a^6 \).
2. Do not confuse this with the previous type of question, which asks for |A| |adj A| = \( a^9 \).

Answer: (d) \( (14)^4 \)

Explanation:
1. \( |A| = 1(1 + 2) - 2(-1 - 4) + (-1)(1 - 2) = 3 + 10 + 1 = 14 \)
2. \( |\text{adj } B| = |B|^{n-1} \), so \( |\text{adj}(\text{adj } A)| = |\text{adj } A|^{2} = (|A|^2)^2 = |A|^4 \) for n = 3
3. So det (adj (adj A)) = \( (14)^4 \).

Teacher's Note:
1. The general result is \( |\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} \).
2. Apply the |adj| rule twice rather than computing any adjoint.

Answer: (c) Null matrix

Explanation:
1. The cofactor of \( a_{ij} \) in A′ equals the cofactor of \( a_{ji} \) in A (transposing does not change minors).
2. So the cofactor matrix of A′ is the transpose of the cofactor matrix of A.
3. Hence adj (A′) = (adj A)′, and adj (A′) − (adj A)′ = O, the null matrix.

Teacher's Note:
1. adj and transpose can be taken in either order: adj (A′) = (adj A)′.
2. Options (a) and (b) are wrong in kind: (a) is a number, not a matrix.

Answer: (a) 13

Explanation:
1. adj A is the transpose of the cofactor matrix, so \( (\text{adj } A)_{23} = A_{32} \), the cofactor of \( a_{32} \).
2. \( A_{32} = (-1)^{3+2}\begin{vmatrix} 1 & 3 \\ 4 & -1 \end{vmatrix} = -(1(-1) - 3(4)) \)
3. = −(−1 − 12) = 13

Teacher's Note:
1. The (i, j) entry of adj A is the cofactor of the (j, i) entry of A; the indices swap.
2. Option (b) comes from forgetting the sign \( (-1)^{i+j} \) of the cofactor.

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

Explanation:
1. A(adj A) = |A| I, a scalar matrix of order 3.
2. \( |A(\text{adj } A)| = |\,|A| I\,| = |A|^3 |I| = |A|^3 \)

Teacher's Note:
1. The rule |kI| = kⁿ (n the order) is the key step here.
2. Equivalently, |A| · |adj A| = |A| · |A|² = |A|³.

Answer: (c) \( \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix} \)

Explanation:
1. \( |A| = 3(1 - 0) + 2(-4 + 2) - 1(0 - 2) = 3 - 4 + 2 = 1 \)
2. Test option (c) as B: row 1 of A times the columns of B gives 3 − 4 + 2 = 1, 6 − 10 + 4 = 0, 9 − 14 + 5 = 0.
3. Row 2: −4 + 2 + 2 = 0, −8 + 5 + 4 = 1, −12 + 7 + 5 = 0; Row 3: 2 + 0 − 2 = 0, 4 + 0 − 4 = 0, 6 + 0 − 5 = 1.
4. So AB = I and the inverse is option (c).

Teacher's Note:
1. When options are given, multiplying A by an option is often faster than finding the inverse from scratch.
2. Since |A| = 1, the inverse here has integer entries and equals adj A.

Answer: (d) \( (AB)^{-1} = B^{-1}A^{-1} \)

Explanation:
1. \( (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I \)
2. So \( (AB)^{-1} = B^{-1}A^{-1} \).
3. AB = BA and (AB)′ = A′B′ are false in general, and \( A^{-1}B^{-1} \) has the wrong order.

Teacher's Note:
1. Remember the 'socks and shoes' rule: the last one put on comes off first.
2. The same reversal applies to transposes: (AB)′ = B′A′.

Answer: (b) \( \begin{bmatrix} -i & 0 \\ 0 & -2i \end{bmatrix} \)

Explanation:
1. The inverse of a diagonal matrix has the reciprocals of the diagonal entries.
2. \( \frac{1}{i} = \frac{i}{i^2} = -i \) and \( \frac{1}{i/2} = \frac{2}{i} = -2i \)
3. So \( A^{-1} = \begin{bmatrix} -i & 0 \\ 0 & -2i \end{bmatrix} \).

Teacher's Note:
1. To simplify 1/i, multiply top and bottom by i and use \( i^2 = -1 \).
2. Check: i × (−i) = 1 and (i/2) × (−2i) = 1, so the product is I.

Answer: (a) \( \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \)

Explanation:
1. \( |A| = \cos^2\alpha + \sin^2\alpha = 1 \)
2. \( \text{adj } A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \) (swap diagonal, negate the others)
3. \( A^{-1} = \frac{1}{|A|}\text{adj } A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \)

Teacher's Note:
1. For a rotation matrix, \( A^{-1} = A' \); rotating back by α undoes the rotation.
2. This is the same as replacing α by −α in A.

Answer: (a) \( \frac{1}{ab - cd}\begin{bmatrix} b & -c \\ -d & a \end{bmatrix} \)

Explanation:
1. \( |A| = ab - cd \) (product of the main diagonal minus the other diagonal).
2. adj A: swap a and b, negate c and d: \( \begin{bmatrix} b & -c \\ -d & a \end{bmatrix} \)
3. \( A^{-1} = \frac{1}{ab - cd}\begin{bmatrix} b & -c \\ -d & a \end{bmatrix} \)

Teacher's Note:
1. Here the diagonal entries are a and b, so |A| = ab − cd, not ad − bc; read the positions carefully.
2. The formula needs ab − cd ≠ 0 for the inverse to exist.

Answer: (b) −1

Explanation:
1. \( |A| = 1(1 - 0) - 2(2 - 0) + 1(0 + 1) = 1 - 4 + 1 = -2 \)
2. \( (A^{-1})_{23} = \frac{A_{32}}{|A|} \), where \( A_{32} \) is the cofactor of \( a_{32} \).
3. \( A_{32} = -\begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = -(0 - 2) = 2 \)
4. \( (A^{-1})_{23} = \frac{2}{-2} = -1 \)

Teacher's Note:
1. To find one entry of the inverse, compute just one cofactor, remembering the index swap.
2. Using \( A_{23} \) instead of \( A_{32} \) is the commonest error in such questions.

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

Explanation:
1. The given matrix is the identity matrix I.
2. I · I = I, so I is its own inverse.
3. So the inverse is \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. The identity matrix is the only matrix that acts like the number 1 in multiplication.
2. Options (a), (c), (d) are permutation matrices, which do not multiply with I to give I.

Answer: (a) \( \frac{-1}{8}\begin{bmatrix} 2 & 3 \\ 4 & 2 \end{bmatrix} \)

Explanation:
1. |A| = 2(2) − (−3)(−4) = 4 − 12 = −8
2. adj A = \( \begin{bmatrix} 2 & 3 \\ 4 & 2 \end{bmatrix} \) (swap 2 and 2, negate −3 and −4)
3. \( A^{-1} = \frac{-1}{8}\begin{bmatrix} 2 & 3 \\ 4 & 2 \end{bmatrix} \)

Teacher's Note:
1. A negative determinant puts a minus sign in front; option (c) misses exactly that.
2. Check: A times adj A should give −8 I.

Answer: (a) \( \begin{bmatrix} \frac{4}{14} & \frac{2}{14} \\ \frac{-1}{14} & \frac{3}{14} \end{bmatrix} \)

Explanation:
1. |A| = 3(4) − (−2)(1) = 12 + 2 = 14
2. adj A = \( \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} \)
3. \( A^{-1} = \frac{1}{14}\begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} \frac{4}{14} & \frac{2}{14} \\ \frac{-1}{14} & \frac{3}{14} \end{bmatrix} \)

Teacher's Note:
1. The fractions are left unsimplified in the options, so match them as they stand.
2. Option (c) forgets to change the signs of the off-diagonal entries.

Answer: (a) \( -(3A^2 + 2A + 5I) \)

Explanation:
1. From \( 3A^3 + 2A^2 + 5A + I = O \): \( 3A^3 + 2A^2 + 5A = -I \)
2. Factor A: \( A(3A^2 + 2A + 5I) = -I \)
3. So \( A[-(3A^2 + 2A + 5I)] = I \), giving \( A^{-1} = -(3A^2 + 2A + 5I) \).

Teacher's Note:
1. Isolate the constant (I) term on one side, then factor out A; whatever multiplies A is the inverse.
2. When factoring, the constant 5 becomes 5I, never just 5.

Answer: (d) \( [G(\beta)]^{-1}[F(\alpha)]^{-1} \)

Explanation:
1. |F(α)| = cos²α + sin²α = 1 and |G(β)| = cos²β + sin²β = 1, so both are invertible.
2. By the reversal law, \( (FG)^{-1} = G^{-1}F^{-1} \).
3. So \( [F(\alpha)G(\beta)]^{-1} = [G(\beta)]^{-1}[F(\alpha)]^{-1} \).

Teacher's Note:
1. Option (c) keeps the original order, the most common trap.
2. Inverses of sums or differences, as in (a) and (b), have no simple rule.

Answer: (b) \( \cos^2\frac{\theta}{2}.A^T \)

Explanation:
1. Write t = tan(θ/2). \( |A| = 1 + t^2 = \sec^2\frac{\theta}{2} \)
2. adj A = \( \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = A^T \)
3. AB = I means \( B = A^{-1} = \frac{1}{\sec^2(\theta/2)}A^T = \cos^2\frac{\theta}{2}.A^T \)

Teacher's Note:
1. Recognise \( 1 + \tan^2 x = \sec^2 x \) so that 1/|A| becomes cos² x.
2. For this pattern of matrix, adj A is just the transpose, which saves work.

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

Explanation:
1. Write the equation as PAQ = I, so \( A = P^{-1}Q^{-1} \).
2. |P| = 4 − 3 = 1, \( P^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \); |Q| = 9 − 10 = −1, \( Q^{-1} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \)
3. \( A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 6 - 5 & 4 - 3 \\ -9 + 10 & -6 + 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \)

Teacher's Note:
1. Pre-multiply by \( P^{-1} \) and post-multiply by \( Q^{-1} \); the sides on which the inverses act must be kept.
2. A faster check: multiply out P(option)Q for each option until you get I.

Answer: (d) |A| ≠ 0

Explanation:
1. A matrix is invertible exactly when |A| ≠ 0.
2. The inverse is unique, so (a) is false; \( A^{-1} \) is itself invertible, so (b) is false.
3. \( (A^{-1})^T = (A^T)^{-1} \) always, so (c) is false.
4. So the correct statement is |A| ≠ 0.

Teacher's Note:
1. The inverse of a matrix, when it exists, is always unique.
2. Transpose and inverse can be taken in either order: \( (A^T)^{-1} = (A^{-1})^T \).

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

Explanation:
1. A is diagonal with |A| = abc (assume a, b, c ≠ 0).
2. AB = I for B = diag(1/a, 1/b, 1/c), since a · (1/a) = 1 and so on.
3. So \( A^{-1} = \begin{bmatrix} 1/a & 0 & 0 \\ 0 & 1/b & 0 \\ 0 & 0 & 1/c \end{bmatrix} \).

Teacher's Note:
1. The inverse of a diagonal matrix is found by taking reciprocals of the diagonal entries.
2. The inverse exists only if no diagonal entry is zero.

Answer: (b) \( \begin{bmatrix} 10 & -3 \\ -3 & 1 \end{bmatrix} \)

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

Teacher's Note:
1. When |A| = 1, the inverse is simply the adjoint.
2. Option (a) forgets to negate the off-diagonal entries.

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

Explanation:
1. |A| = 5(1) − 2(3) = −1
2. adj A = \( \begin{bmatrix} 1 & -2 \\ -3 & 5 \end{bmatrix} \)
3. \( A^{-1} = \frac{1}{-1}\begin{bmatrix} 1 & -2 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 3 & -5 \end{bmatrix} \)

Teacher's Note:
1. Option (a) is adj A; with |A| = −1 every entry must change sign.
2. Check: A times the answer gives [−5 + 6, 10 − 10; −3 + 3, 6 − 5] = I.

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

Explanation:
1. |A| = (−6)(6) − 5(−7) = −36 + 35 = −1
2. adj A = \( \begin{bmatrix} 6 & -5 \\ 7 & -6 \end{bmatrix} \)
3. \( A^{-1} = -\begin{bmatrix} 6 & -5 \\ 7 & -6 \end{bmatrix} = \begin{bmatrix} -6 & 5 \\ -7 & 6 \end{bmatrix} \), which is A itself.

Teacher's Note:
1. A matrix that equals its own inverse satisfies \( A^2 = I \) (an involutory matrix).
2. Check: \( A^2 = \begin{bmatrix} 36 - 35 & -30 + 30 \\ 42 - 42 & -35 + 36 \end{bmatrix} = I \).

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

Explanation:
1. A swaps the first two coordinates, so doing it twice gives back the original: \( A^2 = I \).
2. Check: \( A^2 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
3. So \( A^{-1} = A \).
4. A is symmetric, so \( A^T = A \) is also the inverse; hence both (a) and (b) are correct.

Teacher's Note:
1. A permutation matrix always has \( A^{-1} = A^T \); when it is also symmetric, the inverse is A itself.
2. Option (c) is wrong because A × I = A, not I.

Answer: (d) None of these

Explanation:
1. \( |A| = 3(-6 - 1) - 5(4 - 1) + 7(2 + 3) = -21 - 15 + 35 = -1 \)
2. Cofactor of \( a_{33} \): \( A_{33} = 3(-3) - 5(2) = -19 \)
3. So the (3, 3) entry of \( A^{-1} \) is \( \frac{-19}{-1} = 19 \).
4. No option has 19 in that position, so the answer is none of these.

Teacher's Note:
1. To eliminate options, compute just one entry of the inverse that differs among them.
2. With |A| = −1, \( A^{-1} = -\text{adj } A \), so every cofactor changes sign.

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

Explanation:
1. The matrix A is upper triangular, so |A| = 1 · 1 · 1 = 1 and the inverse exists.
2. Test option (b) as B: row 1 of A times B gives (1, −2 + 2, 1 − 4 + 3) = (1, 0, 0).
3. Row 2 gives (0, 1, −2 + 2) = (0, 1, 0) and row 3 gives (0, 0, 1).
4. So AB = I and \( A^{-1} = \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. The inverse of an upper triangular matrix is also upper triangular, with reciprocals of the diagonal on its diagonal.
2. Option (a) has a zero row, so its determinant is 0 and it cannot be an inverse.

Answer: (b) λ ≠ −17

Explanation:
1. \( |A| = \lambda(0 - 1) - (-1)(-6 + 1) + 4(-3 - 0) \)
2. = −λ − 5 − 12 = −λ − 17
3. A is invertible when |A| ≠ 0, i.e. λ ≠ −17.

Teacher's Note:
1. Invertible and non-singular mean the same thing: |A| ≠ 0.
2. The middle cofactor carries a minus sign, and here the entry itself is −1, so the term becomes −(−1)(−5) = −5.

Answer: (a) \( \frac{1}{27}\begin{pmatrix} 1 & -26 \\ 0 & 27 \end{pmatrix} \)

Explanation:
1. |A| = 3, so \( A^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix} \).
2. \( \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}^2 = \begin{pmatrix} 1 & -8 \\ 0 & 9 \end{pmatrix} \), and \( \begin{pmatrix} 1 & -8 \\ 0 & 9 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -26 \\ 0 & 27 \end{pmatrix} \)
3. \( (A^{-1})^3 = \frac{1}{27}\begin{pmatrix} 1 & -26 \\ 0 & 27 \end{pmatrix} \)

Teacher's Note:
1. \( (A^{-1})^3 = (A^3)^{-1} \), so you could also cube A first and then invert.
2. Keep the scalar 1/3 outside and cube it separately as 1/27.

Answer: (b) 1

Explanation:
1. Not invertible means |A| = 0.
2. \( |A| = 1(2 - 5) - a(1 - 10) + 2(1 - 4) = -3 + 9a - 6 = 9a - 9 \)
3. 9a − 9 = 0 gives a = 1.

Teacher's Note:
1. Check: with a = 1, |A| = 1(−3) − 1(−9) + 2(−3) = −3 + 9 − 6 = 0.
2. Expanding along the row containing the unknown keeps the algebra in one term.

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

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

Teacher's Note:
1. For a 2 × 2 matrix with determinant 1, the inverse is just the adjoint.
2. Option (b) also transposes the entries, which is not part of the 2 × 2 rule.

Answer: (d) 7

Explanation:
1. A is upper triangular, so |A| = 1.
2. \( (A^{-1})_{13} = \frac{A_{31}}{|A|} \), where \( A_{31} \) is the cofactor of \( a_{31} \).
3. \( A_{31} = (+1)\begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = 4 + 3 = 7 \)
4. So the required element is 7.

Teacher's Note:
1. The (1, 3) entry of the inverse uses the cofactor of the (3, 1) entry; the indices swap.
2. Cofactor sign for position (3, 1) is \( (-1)^{3+1} = +1 \).

Answer: (c) \( I_3 \)

Explanation:
1. \( I_3 \cdot I_3 = I_3 \)
2. So \( I_3^{-1} = I_3 \).

Teacher's Note:
1. The identity matrix of any order is its own inverse.
2. The number 0 in option (a) is not even a matrix, so it can never be an inverse.

Answer: (b) \( -(4A^2 + 2A + 7I) \)

Explanation:
1. From \( 4A^3 + 2A^2 + 7A + I = O \): \( 4A^3 + 2A^2 + 7A = -I \)
2. \( A(4A^2 + 2A + 7I) = -I \)
3. \( A[-(4A^2 + 2A + 7I)] = I \), so \( A^{-1} = -(4A^2 + 2A + 7I) \).

Teacher's Note:
1. Move the I term to the other side before factoring out A; the minus sign then appears in the inverse.
2. Option (a) forgets this minus sign, which is the usual error.

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

Explanation:
1. \( (B^{-1}A^{-1})^{-1} = (A^{-1})^{-1}(B^{-1})^{-1} = AB \)
2. \( AB = \begin{bmatrix} 2 & 2 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 + 2 & -2 + 0 \\ 0 + 2 & 3 + 0 \end{bmatrix} \)
3. \( = \begin{bmatrix} 2 & -2 \\ 2 & 3 \end{bmatrix} \)

Teacher's Note:
1. Use the reversal law first; it turns the question into a single multiplication with no inverses at all.
2. \( (A^{-1})^{-1} = A \) for any invertible matrix.

Answer: (c) I − A

Explanation:
1. From \( A^2 - A + I = O \): \( A - A^2 = I \)
2. Factor A: \( A(I - A) = I \)
3. So \( A^{-1} = I - A \).

Teacher's Note:
1. Always rearrange so that I stands alone with a + sign, then factor out A.
2. Check: (I − A)A = A − A² = I as well, so I − A is a two-sided inverse.

Answer: (d) Does not exist

Explanation:
1. |A| = 2(6) − 3(4) = 12 − 12 = 0
2. A is singular, so \( A^{-1} \) does not exist.

Teacher's Note:
1. Always find |A| first; if it is 0, the inverse does not exist and no further work is needed.
2. Here row 2 = 2 × row 1, which shows at once that |A| = 0.

Answer: (a) \( A^2 \)

Explanation:
1. \( A^3 = I \) can be written as \( A \cdot A^2 = I \).
2. So \( A^{-1} = A^2 \).

Teacher's Note:
1. In general, if \( A^n = I \) then \( A^{-1} = A^{n-1} \).
2. Option (b), \( A^3 \), equals I, which is the inverse only when A = I.

Answer: (b) \( B^{-1}A^{-1} \)

Explanation:
1. \( (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I \)
2. So \( (AB)^{-1} = B^{-1}A^{-1} \).

Teacher's Note:
1. \( (BA)^{-1} = A^{-1}B^{-1} \) is a different matrix in general, so option (a) is wrong.
2. The reversal law extends to more factors: \( (ABC)^{-1} = C^{-1}B^{-1}A^{-1} \).

Answer: (b) \( \frac{1}{3}\begin{bmatrix} 5 \\ 7 \end{bmatrix} \)

Explanation:
1. |A| = 1 − 4 = −3; adj A = \( \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} \)
2. \( A^{-1} = \frac{1}{-3}\begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \)
3. \( X = A^{-1}B = \frac{1}{3}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 3 \\ 1 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 5 \\ 7 \end{bmatrix} \)

Teacher's Note:
1. X must be a 2 × 1 column matrix, which rules out the row matrices in (a) and (c) immediately.
2. Check: A × X = (1/3)[−5 + 14; 10 − 7] = (1/3)[9; 3] = [3; 1] = B.

Answer: (b) \( \begin{bmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{3}{11} & -\frac{1}{11} \end{bmatrix} \)

Explanation:
1. |A| = 1(−5) − 2(3) = −11
2. adj A = \( \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix} \)
3. \( A^{-1} = \frac{1}{-11}\begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{3}{11} & -\frac{1}{11} \end{bmatrix} \)

Teacher's Note:
1. Dividing by a negative determinant flips every sign of the adjoint.
2. Option (a) is adj A, which is missing the factor 1/|A|.

Answer: (a) \( \frac{-1}{6} \)

Explanation:
1. \( A^{-1} = \frac{1}{|A|}\text{adj } A \), so λ = 1/|A|.
2. |A| = 0(0) − 3(2) = −6
3. \( \lambda = \frac{-1}{6} \)

Teacher's Note:
1. The scalar multiplying adj A in the inverse is always the reciprocal of the determinant.
2. Watch the sign: |A| here is negative because the diagonal product is 0.