CBSE Class 12 Mathematics Matrices MCQs Set 08

Answer: (a) \( \begin{pmatrix} 1 & -4i \\ 0 & 1 \end{pmatrix} \)

Explanation:
1. Write A = iI + N with \( N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \); N² = O and iI commutes with N.
2. By the binomial theorem, A⁴ = (iI)⁴ + 4(iI)³N = i⁴I + 4i³N = I − 4iN.
3. \( A^4 = \begin{pmatrix} 1 & -4i \\ 0 & 1 \end{pmatrix} \).

Teacher's Note:
1. Splitting a triangular matrix into scalar part plus nilpotent part makes high powers easy.
2. Remember i³ = −i and i⁴ = 1 when simplifying powers of i.

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

Explanation:
1. A 3 × 1 column times a 1 × 3 row gives a 3 × 3 matrix.
2. Each row of the product is the row [2 1 −1] multiplied by 1, −1 and 2 in turn.
3. \( \begin{bmatrix} 2 & 1 & -1 \\ -2 & -1 & 1 \\ 4 & 2 & -2 \end{bmatrix} \).

Teacher's Note:
1. Column times row gives a full matrix; row times column gives a single number in brackets.
2. The order check (3 × 1)(1 × 3) = 3 × 3 rules out (a), (b) and (d) straight away.

Answer: (c) \( \begin{bmatrix} 2 & 2 \\ 7/2 & 1 \end{bmatrix} \)

Explanation:
1. \( 2X = \begin{bmatrix} 3 & 2 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 7 & 2 \end{bmatrix} \).
2. \( X = \begin{bmatrix} 2 & 2 \\ 7/2 & 1 \end{bmatrix} \).

Teacher's Note:
1. Add the subtracted matrix to both sides before dividing by 2.
2. Option (b) comes from subtracting instead of adding; keep track of the sign when moving a term across.

Answer: (c) −6, −4, −9

Explanation:
1. \( kA = \begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix} = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix} \).
2. −4k = 24 gives k = −6.
3. 2k = 3a gives a = −4, and 3k = 2b gives b = −9. So (k, a, b) = (−6, −4, −9).

Teacher's Note:
1. Find the scalar k from the entry that involves only k, then use it in the others.
2. Option (b) has the right size but the wrong signs; substitute back to check.

Answer: (a) \( \begin{bmatrix} 1 & 2n \\ 0 & 1 \end{bmatrix} \)

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} \).
2. Each multiplication adds 2 to the top-right entry, so \( A^n = \begin{bmatrix} 1 & 2n \\ 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. [[1,a],[0,1]]ⁿ = [[1,na],[0,1]] is a standard result worth remembering.
2. Check any option with n = 1; only (a) gives back A.

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

Explanation:
1. \( A^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I \).
2. A¹⁶ = (A²)⁸ = (−I)⁸ = I.
3. \( A^{16} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. This A is a rotation by 90°, so A⁴ = I and every power that is a multiple of 4 is I.
2. Reduce the exponent: 16 = 4 × 4, so A¹⁶ = (A⁴)⁴ = I.

Answer: (b) 14I

Explanation:
1. \( A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \).
2. \( 5A = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} \).
3. \( A^2 - 5A = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = 14I \).

Teacher's Note:
1. Cayley–Hamilton shortcut: A² − (trace)A + |A|I = O with trace 5 and |A| = 6 − 20 = −14, so A² − 5A = 14I.
2. The trace-determinant check is a fast way to verify any 2 × 2 squaring.

Answer: (b) \( 2^{99} A \)

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2A \).
2. Then A³ = 2A² = 2²A, and in general Aⁿ = 2ⁿ⁻¹A.
3. So A¹⁰⁰ = 2⁹⁹A.

Teacher's Note:
1. Whenever A² = kA, all powers follow: Aⁿ = kⁿ⁻¹A.
2. Check with n = 2: 2¹A = 2A agrees, so the exponent is n − 1, not n.

Answer: (b) It is associative

Explanation:
1. (AB)C = A(BC) whenever the products are defined, so multiplication is associative.
2. Commutativity fails: for \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \), AB = O but BA ≠ O.
3. So only (b) is true.

Teacher's Note:
1. Associativity lets us write ABC without brackets, but the order of the factors must stay the same.
2. Because (a) is false, the option 'Both (a) and (b)' is also false.

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

Explanation:
1. P is 3 × 3 and Q is 3 × 2, so PQ is 3 × 2.
2. Row 1 of P (i, 0, −i): (i(−i) + 0 − i(i), i(i) + 0 − i(−i)) = (1 + 1, −1 − 1) = (2, −2).
3. Row 2 (0, −i, i): (0 + 0 + i·i, 0 + 0 + i(−i)) = (−1, 1); row 3 (−i, i, 0): ((−i)(−i), (−i)(i)) = (−1, 1).
4. \( PQ = \begin{pmatrix} 2 & -2 \\ -1 & 1 \\ -1 & 1 \end{pmatrix} \).

Teacher's Note:
1. Use i² = −1 and (−i)(i) = 1 carefully; option (a) is the same matrix with every sign flipped.
2. The order 3 × 2 alone removes options (c) and (d).

Answer: (a) \( \begin{bmatrix} 43 \\ 44 \end{bmatrix} \)

Explanation:
1. \( \begin{bmatrix} 7 & 1 & 2 \\ 9 & 2 & 1 \end{bmatrix}\begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 21+4+10 \\ 27+8+5 \end{bmatrix} = \begin{bmatrix} 35 \\ 40 \end{bmatrix} \).
2. \( 2\begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix} \).
3. Sum: \( \begin{bmatrix} 43 \\ 44 \end{bmatrix} \).

Teacher's Note:
1. Do the product first, then the scalar multiple, then add; the order of operations is the same as for numbers.
2. The result is 2 × 1, so it must be compared with column matrices only.

Answer: (a) \( \begin{bmatrix} 5 & 1 & -3 \\ 3 & 2 & 6 \\ 14 & 5 & 0 \end{bmatrix} \)

Explanation:
1. Row 1 of A (1, 2, −1) times B: (1 + 4 + 0, 0 + 2 − 1, 0 + 0 − 3) = (5, 1, −3).
2. Row 2 (3, 0, 2): (3, 2, 6); row 3 (4, 5, 0): (4 + 10, 5, 0) = (14, 5, 0).
3. \( AB = \begin{bmatrix} 5 & 1 & -3 \\ 3 & 2 & 6 \\ 14 & 5 & 0 \end{bmatrix} \).

Teacher's Note:
1. Computing just the (1, 1) entry, 5, already eliminates options (c) and (d), and the (1, 2) entry removes (b).
2. In MCQs, compute one distinguishing entry first instead of the whole product.

Answer: (c) \( \begin{bmatrix} \frac{3}{4} & 0 \\ 0 & \frac{3}{4} \end{bmatrix} \)

Explanation:
1. From 2A − 2B = I, A − B = \( \frac{1}{2}I \).
2. Adding to A + B = I: 2A = \( \frac{3}{2}I \), so A = \( \frac{3}{4}I \).
3. \( A = \begin{bmatrix} \frac{3}{4} & 0 \\ 0 & \frac{3}{4} \end{bmatrix} \).

Teacher's Note:
1. Divide the second equation by 2 first so that the two equations can be added directly.
2. Then B = I − A = (1/4)I, which is option (a); do not mix up A and B.

Answer: (c) 3A

Explanation:
1. A is the unit matrix I, so A² = I = A.
2. A² + 2A = A + 2A = 3A.

Teacher's Note:
1. The identity satisfies Iⁿ = I for every n, which makes such expressions linear.
2. Write the answer in terms of A as the options demand.

Answer: (b) \( \begin{bmatrix} -8 \\ 3 \end{bmatrix} \)

Explanation:
1. A is 2 × 3 and B is 3 × 1, so AB is 2 × 1.
2. Row 1: 1(2) − 6(2) + 2(1) = −8; row 2: 0(2) − 1(2) + 5(1) = 3.
3. \( AB = \begin{bmatrix} -8 \\ 3 \end{bmatrix} \).

Teacher's Note:
1. The product of a 2 × 3 and a 3 × 1 matrix is a column, so the row [−8 3] in (a) has the wrong order.
2. Getting the right numbers in the wrong shape is still a wrong answer in matrices.

Answer: (b) AB = O, BA ≠ O

Explanation:
1. \( AB = \begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 1 & 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
2. \( BA = \begin{bmatrix} 0 & 0 \\ 1 & 12 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1+24 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 25 & 0 \end{bmatrix} \neq O \).
3. So AB = O but BA ≠ O.

Teacher's Note:
1. The second column of A is zero and the first row of B is zero, which forces AB = O.
2. Always compute both orders when a question compares AB and BA.

Answer: (c) \( A^2 - AB - BA + B^2 \)

Explanation:
1. (A − B)² = (A − B)(A − B).
2. Expanding with the distributive law and keeping the order: A·A − A·B − B·A + B·B.
3. So (A − B)² = A² − AB − BA + B².

Teacher's Note:
1. −2AB is correct only when AB = BA; in general the two middle terms stay separate.
2. Never swap the order inside a product when expanding matrix brackets.

Answer: (b) \( \begin{bmatrix} 3 & 19 \\ -10 & 29 \end{bmatrix} \)

Explanation:
1. \( 2A = \begin{bmatrix} 6 & 10 \\ -4 & 8 \end{bmatrix} \) and \( 3B = \begin{bmatrix} 3 & -9 \\ 6 & -21 \end{bmatrix} \).
2. \( 2A - 3B = \begin{bmatrix} 6-3 & 10+9 \\ -4-6 & 8+21 \end{bmatrix} = \begin{bmatrix} 3 & 19 \\ -10 & 29 \end{bmatrix} \).

Teacher's Note:
1. Subtracting a negative entry adds it: 10 − (−9) = 19 and 8 − (−21) = 29.
2. The options differ only in signs, so check the sign of each entry.

Answer: (c) \( \begin{bmatrix} 5 & 10 \\ 0 & -3 \end{bmatrix} \)

Explanation:
1. \( 4A = \begin{bmatrix} 8 & 16 \\ 0 & 12 \end{bmatrix} \) and \( 3B = \begin{bmatrix} 3 & 6 \\ 0 & 15 \end{bmatrix} \).
2. \( 4A - 3B = \begin{bmatrix} 5 & 10 \\ 0 & -3 \end{bmatrix} \).

Teacher's Note:
1. Multiply every entry by the scalar, including the zeros.
2. The (2, 2) entry 12 − 15 = −3 is the one that separates the options.

Answer: (b) \( \begin{pmatrix} 1 & -1 \\ 10 & 20 \end{pmatrix} \)

Explanation:
1. \( 5A = \begin{pmatrix} 5 & -10 \\ 15 & 20 \end{pmatrix} \) and \( 3B = \begin{pmatrix} 6 & 3 \\ -12 & 9 \end{pmatrix} \).
2. \( 5A - 3B = \begin{pmatrix} -1 & -13 \\ 27 & 11 \end{pmatrix} \).
3. \( 5A - 3B + C = \begin{pmatrix} -1+2 & -13+12 \\ 27-17 & 11+9 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 10 & 20 \end{pmatrix} \).

Teacher's Note:
1. Do the combination in two stages to keep the arithmetic clean.
2. Options (a) and (b) are transposes of each other, so check the position of each entry, not just the values.

Answer: (a) \( \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \)

Explanation:
1. \( A + B = \begin{bmatrix} 4 & 2 \\ 5 & 4 \end{bmatrix} \).
2. \( A + B - C = \begin{bmatrix} 4-1 & 2 \\ 5 & 4-1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \).

Teacher's Note:
1. C is the identity, so subtracting it only lowers the diagonal entries by 1.
2. Recognising I in a question saves arithmetic.

Answer: (b) 1, 2

Explanation:
1. \( LHS = \begin{bmatrix} 2x & y + 1 \\ 2 & 2 \end{bmatrix} \).
2. Equating with \( \begin{bmatrix} 2 & 3 \\ y & 2 \end{bmatrix} \): 2x = 2, y + 1 = 3 and 2 = y.
3. So x = 1 and y = 2.

Teacher's Note:
1. When a variable appears in two entries, both equations must give the same value; here both give y = 2.
2. That consistency is a good check that the addition was done correctly.

Answer: (c) 2

Explanation:
1. \( 3X - \begin{bmatrix} 2 & 3 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3-2 & 3a-3 \\ 0 & 3-2 \end{bmatrix} = \begin{bmatrix} 1 & 3a - 3 \\ 0 & 1 \end{bmatrix} \).
2. Equating with \( \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \): 3a − 3 = 3.
3. So a = 2.

Teacher's Note:
1. Only the entry containing a gives an equation; the rest are checks.
2. Remember to multiply a by 3 as well when forming 3X.

Answer: (c) AB = −BA

Explanation:
1. \( AB = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \).
2. \( BA = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \).
3. So BA = −AB, i.e. AB = −BA.

Teacher's Note:
1. Matrices with AB = −BA are said to anticommute; then (A + B)² = A² + B².
2. Also B² = I, so option (b) would need AB = I, which is false.

Answer: (a) \( \begin{bmatrix} 8 & 15 & 16 \\ 5 & 9 & 10 \end{bmatrix} \)

Explanation:
1. A is 2 × 2 and B is 2 × 3, so AB is 2 × 3.
2. Row 1 (2, 3): (2 + 6, 6 + 9, 4 + 12) = (8, 15, 16); row 2 (1, 2): (1 + 4, 3 + 6, 2 + 8) = (5, 9, 10).
3. \( AB = \begin{bmatrix} 8 & 15 & 16 \\ 5 & 9 & 10 \end{bmatrix} \).

Teacher's Note:
1. The order 2 × 3 immediately rules out options (b) and (c).
2. Option (b) is the transpose of the correct product, a common mix-up.

Answer: (c) \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)

Explanation:
1. \( AB = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix} \).
2. Entries: (5 − 4, −10 + 10; 2 − 2, −4 + 5).
3. \( AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) = I.

Teacher's Note:
1. AB = I means B is the inverse of A: for A = [[5,2],[2,1]], |A| = 1 and A⁻¹ = [[1,−2],[−2,5]].
2. Recognising the inverse pattern (swap diagonal, negate off-diagonal, divide by |A|) predicts the answer.

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

Explanation:
1. Let \( M = \begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix} \); then A = (given product)M⁻¹, with |M| = −1 − 6 = −7.
2. Test option (c): \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} -1 & 2 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} -1-3 & 2-1 \\ -2+9 & 4+3 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 7 & 7 \end{bmatrix} \), which is the given matrix.
3. Since M is invertible, A is unique, so \( A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \).

Teacher's Note:
1. In an MCQ it is quicker to multiply each option by M than to find M⁻¹.
2. AM = C gives A = CM⁻¹, with M⁻¹ on the right; the side matters.

Answer: (d) \( A^2 + 2B - 2A \)

Explanation:
1. Orders: A is 3 × 3, B is 3 × 2, C is 3 × 1, so B′ is 2 × 3 and C′ is 1 × 3.
2. AB is (3 × 3)(3 × 2) = 3 × 2, defined; B′C is (2 × 3)(3 × 1) = 2 × 1, defined; CC′ is 3 × 3, defined.
3. A² is 3 × 3 but 2B is 3 × 2, and matrices of different orders cannot be added. So A² + 2B − 2A is not defined.

Teacher's Note:
1. Addition needs identical orders, while multiplication needs only matching inner dimensions.
2. List the order of every matrix (and its transpose) first; the answer then follows quickly.

Answer: (b) 3A = B

Explanation:
1. Multiplying each element of A by 3 is the definition of the scalar multiple 3A.
2. So B = 3A, i.e. 3A = B.

Teacher's Note:
1. Do not confuse kA with the determinant rule: |3A| = 3²|A| for a 2 × 2 matrix, but 3A itself is just 3 times each entry.
2. The 9 in options (c) and (d) is the trap that comes from that determinant rule.

Answer: (d) None of these

Explanation:
1. If A = O or B = O, then certainly AB = O, so the condition 'A = O or B = O' is sufficient.
2. It is not necessary: \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \) are both non-zero, yet AB = O.
3. Options (a), (b) and (c) are therefore not 'if and only if' conditions, so the answer is None of these.

Teacher's Note:
1. Unlike real numbers, matrices can have zero divisors: two non-zero matrices whose product is O.
2. For an 'if and only if' statement, test both directions; one counter-example breaks it.

Answer: (c) 1 × 2

Explanation:
1. \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) is 2 × 2 and [1  2] is 1 × 2.
2. If A is m × n, then for A × (2 × 2) to be defined n = 2, and the product is m × 2.
3. The product is 1 × 2, so m = 1 and A is of order 1 × 2.

Teacher's Note:
1. The product of an m × n and an n × p matrix is m × p: rows from the first, columns from the second.
2. Since [[1,2],[3,4]] is invertible, A = [1 2] × [[1,2],[3,4]]⁻¹, a 1 × 2 times 2 × 2 product, so A is 1 × 2.

Answer: (d) \( A_{3 \times 2}, B_{2 \times 3}, C_{3 \times 3} \)

Explanation:
1. For AB to be defined and equal C, A is m × n, B is n × p and C is m × p.
2. (a): (2 × 3)(3 × 2) gives 2 × 2, not 2 × 3. (b): (3 × 2)(2 × 3) gives 3 × 3, not 3 × 2. (c): A has 3 columns but B has 2 rows, so AB is not defined.
3. (d): (3 × 2)(2 × 3) = 3 × 3, matching C. So (d) is correct.

Teacher's Note:
1. Check two things: the inner numbers must match and the outer numbers give the order of C.
2. Writing the orders side by side, like (3 × 2)(2 × 3), makes the check instant.

Answer: (c) m = n

Explanation:
1. A square matrix has the same number of rows and columns.
2. For an m × n matrix this means m = n.

Teacher's Note:
1. Only square matrices have determinants, inverses and powers like A², so this definition is used everywhere.
2. m < n gives a horizontal matrix and m > n a vertical matrix.

Answer: (b) 4

Explanation:
1. The (3, 3) entry of AB is row 3 of A times column 3 of B.
2. Row 3 of A is (−2, 2, 0) and column 3 of B is (3, 5, 0)ᵀ.
3. (−2)(3) + 2(5) + 0(0) = −6 + 10 = 4.

Teacher's Note:
1. To find a single entry of a product, there is no need to compute the whole product.
2. Entry (i, j) always uses row i of the first matrix and column j of the second.

Answer: (b) Skew-symmetric matrix

Explanation:
1. (AB − BA)ᵀ = (AB)ᵀ − (BA)ᵀ = BᵀAᵀ − AᵀBᵀ.
2. Since Aᵀ = A and Bᵀ = B, this equals BA − AB = −(AB − BA).
3. So AB − BA is skew-symmetric.

Teacher's Note:
1. Similarly AB + BA is symmetric for symmetric A and B; both facts use (AB)ᵀ = BᵀAᵀ.
2. It is not always the null matrix: it is O only when A and B commute.

Answer: (b) \( A^2 + AB + BA + B^2 \)

Explanation:
1. (A + B)² = (A + B)(A + B).
2. Expanding while keeping the order: A² + AB + BA + B².
3. AB and BA need not be equal, so they cannot be combined into 2AB.

Teacher's Note:
1. (A + B)² = A² + 2AB + B² holds only when AB = BA.
2. This is one of the most common errors in matrix algebra; always write both middle terms.

Answer: (d) None of these

Explanation:
1. Two matrices can be added only if they have the same order.
2. A is 2 × 3 and B is 3 × 2, which are different orders.
3. So A + B is not defined, and the answer is None of these.

Teacher's Note:
1. Addition needs identical orders; multiplication only needs matching inner numbers.
2. Here AB (2 × 2) and BA (3 × 3) are defined, but A + B is not.

Answer: (c) \( k^n |B| \)

Explanation:
1. A = kB means every entry of A is k times the matching entry of B.
2. Taking k out of each of the n rows of the determinant gives a factor k each time.
3. So |A| = |kB| = kⁿ|B|.

Teacher's Note:
1. A common slip is writing |kB| = k|B|; that would be true only for n = 1.
2. Example: |2I₃| = 2³ = 8, not 2.

Answer: (c) A + B′

Explanation:
1. A is 3 × 3, B is 3 × 2, C is 1 × 3, so B′ is 2 × 3.
2. B′B is (2 × 3)(3 × 2) = 2 × 2, defined. CAB is (1 × 3)(3 × 3)(3 × 2) = 1 × 2, defined. A² + A is 3 × 3, defined.
3. A + B′ adds a 3 × 3 matrix to a 2 × 3 matrix, which is not defined.

Teacher's Note:
1. For a chain like CAB, check each adjacent pair: the inner numbers 3 and 3, then 3 and 3, must match.
2. Addition fails as soon as the orders differ, however the matrices were formed.

Answer: (c) AC = BC

Explanation:
1. \( AC = [a\ b]\begin{bmatrix} a \\ -a \end{bmatrix} = [a^2 - ab] \).
2. \( BC = [-b\ -a]\begin{bmatrix} a \\ -a \end{bmatrix} = [-ab + a^2] \).
3. So AC = BC.

Teacher's Note:
1. −B = [b a] is A with the entries swapped, so A = −B is false in general.
2. CA and CB are 2 × 2 matrices with different rows, so (d) also fails; compute before assuming.

Answer: (a) AB = BA

Explanation:
1. (A + B)(A − B) = A² − AB + BA − B².
2. This equals A² − B² exactly when −AB + BA = O.
3. So the condition is AB = BA.

Teacher's Note:
1. Every algebraic identity for numbers that needs commutativity holds for matrices only when they commute.
2. Keep the order of each product while expanding brackets.

Answer: (a) A + B = B + A

Explanation:
1. Addition is done entry by entry, and addition of numbers is commutative.
2. So A + B = B + A for all matrices of the same order.
3. The other statements fail in general (for example AB ≠ BA).

Teacher's Note:
1. Matrix addition is commutative and associative; matrix multiplication is only associative.
2. A − B = B − A would force A = B, so (c) is not true in general.

Answer: (a) \( A^2 - B^2 = (A + B)(A - B) \)

Explanation:
1. (b) is true: (Aᵀ)ᵀ = A. (c) also holds: if AB = BA then (AB)ⁿ = AⁿBⁿ.
2. (d) is true: (A − I)(I + A) = A + A² − I − A = A² − I, which is O exactly when A² = I.
3. (a) is incorrect: (A + B)(A − B) = A² − AB + BA − B², which equals A² − B² only if AB = BA.

Teacher's Note:
1. (A + B)(A − B) = A² − B² is not an identity for matrices unless they commute.
2. I commutes with every matrix, which is why (d) can be expanded like ordinary algebra.

Answer: (d) None of these

Explanation:
1. (i) If Aᵀ = A, then the cofactor matrix is also symmetric, so adj A is symmetric: true.
2. (ii) adj I = I because every cofactor of I is 0 off the diagonal and 1 on it: true. (iii) A(adj A) = (adj A)A = |A|I is a standard result: true.
3. (iv) For a diagonal matrix the off-diagonal cofactors are all zero, so adj A is diagonal: true.
4. All four statements are correct, so none is incorrect: answer (d).

Teacher's Note:
1. For diag(a, b, c), adj A = diag(bc, ca, ab), which shows (iv) directly.
2. In 'is/are incorrect' questions, check each statement independently before choosing.