CBSE Class 12 Mathematics Matrices MCQs Set 07

Answer: (b) 0

Explanation:
1. In a diagonal matrix every element off the leading diagonal must be zero.
2. Here b is the element in row 3, column 2, which is off the diagonal.
3. So b = 0.

Teacher's Note:
1. A diagonal matrix may have zeros on the diagonal too; only the off-diagonal entries are forced to be zero.
2. Locate the position (i, j) of the unknown first: if i ≠ j it must vanish.

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

Explanation:
1. A diagonal matrix is a square matrix in which every element with i ≠ j is zero.
2. Option (a) has a₁₃ = 2 ≠ 0 and option (c) has a₁₂ = a₂₁ = 2 ≠ 0, so neither is diagonal.
3. Option (b) has non-zero entries only on the leading diagonal, so it is diagonal.

Teacher's Note:
1. Scan only the off-diagonal entries; the diagonal values themselves can be anything.
2. Matrix (b) is in fact 2I, a scalar matrix, and every scalar matrix is diagonal.

Answer: (c) A scalar matrix

Explanation:
1. \( 3I = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \) (for order 3), a diagonal matrix with all diagonal entries equal to 3.
2. A diagonal matrix with all diagonal entries equal is a scalar matrix.
3. Its diagonal entries are 3, not 1, so it is not a unit matrix. Hence 3I is a scalar matrix.

Teacher's Note:
1. Every scalar matrix has the form kI; the unit matrix is the special case k = 1.
2. A scalar matrix is also triangular, but the most precise description asked for here is scalar.

Answer: (d) None of these

Explanation:
1. A + X = I gives X = I − A.
2. \( X = \begin{bmatrix} 1-1 & 0-2 & 0+1 \\ 0-3 & 1-4 & 0-7 \\ 0-5 & 0-1 & 1-6 \end{bmatrix} = \begin{bmatrix} 0 & -2 & 1 \\ -3 & -3 & -7 \\ -5 & -1 & -5 \end{bmatrix} \).
3. Option (a) has −6 in the last place instead of −5, and (b), (c) differ more, so none of (a), (b), (c) matches. Answer: None of these.

Teacher's Note:
1. Subtract A from I entry by entry; only the diagonal entries get the extra 1.
2. Check the last diagonal entry carefully: 1 − 6 = −5, which is exactly where option (a) goes wrong.

Answer: (d) A is a scalar matrix

Explanation:
1. Let \( A = \begin{bmatrix} p & 0 \\ 0 & q \end{bmatrix} \). Any two diagonal matrices of the same order commute, and their product is diagonal, so (a) and (b) are true.
2. A diagonal matrix is symmetric, so \( A^T = A \) and (c) is true.
3. A is scalar only when p = q; for example \( \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \) is diagonal but not scalar. So (d) is the wrong statement.

Teacher's Note:
1. Every scalar matrix is diagonal, but a diagonal matrix is scalar only when all its diagonal entries are equal.
2. When a question asks for the wrong statement, a single counter-example like diag(1, 2) settles it.

Answer: (d) 4

Explanation:
1. \( M^2 = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} \).
2. \( M^2 - \lambda M - I = \begin{bmatrix} 4-\lambda & 8-2\lambda \\ 8-2\lambda & 12-3\lambda \end{bmatrix} \).
3. Setting this equal to O gives 4 − λ = 0, 8 − 2λ = 0 and 12 − 3λ = 0, all satisfied by λ = 4.

Teacher's Note:
1. Equate one simple entry first (here 4 − λ = 0), then check that the same λ makes every other entry zero.
2. By Cayley–Hamilton, M² − (trace M)M + |M|I = O; here |M| = −1, so λ equals the trace, 1 + 3 = 4.

Answer: (c) AB = BA

Explanation:
1. A and B are rotation matrices: A rotates by α and B by β.
2. \( AB = \begin{bmatrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix} \) using the cos and sin addition formulae.
3. Computing BA gives the same matrix (β + α = α + β), so AB = BA.

Teacher's Note:
1. Rotations about the same origin always commute because the angles simply add.
2. Remember the pattern: the product of rotations by α and β is the rotation by α + β.

Answer: (c) AB = O, BA ≠ O

Explanation:
1. \( AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
2. \( BA = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \neq O \).
3. Hence AB = O but BA ≠ O.

Teacher's Note:
1. This example shows matrix multiplication is not commutative and that AB = O does not force A or B to be O.
2. Multiply row by column carefully: entry (i, j) of AB is row i of A times column j of B.

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

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \) and \( A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \).
2. The top-right entry increases by 1 at each multiplication, so by induction \( A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \).
3. Hence the answer is \( \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. Compute A² and A³, spot the pattern, then confirm it with one more multiplication.
2. Test any option with n = 1: it must give back A, which quickly removes (b), (c) and (d).

Answer: (b) 2A

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix} \).
2. \( A^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 2\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \).
3. So A² = 2A.

Teacher's Note:
1. After squaring, try to factor out a common scalar to compare with A.
2. In general, for A = [[a,−a],[−a,a]], A² = 2aA, so here with a = 1 we get 2A.

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

Explanation:
1. \( 2A = \begin{bmatrix} -3 & 7 \\ 4 & 3 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} -5 & 8 \\ 2 & 3 \end{bmatrix} \).
2. Dividing every entry by 2: \( A = \begin{bmatrix} -5/2 & 4 \\ 1 & 3/2 \end{bmatrix} \).

Teacher's Note:
1. Move the known matrix to the right side first, then divide by the scalar.
2. Option (a) is 2A, not A; forgetting to divide by 2 is the usual slip.

Answer: (b) (3, 4)

Explanation:
1. \( [m\ n]\begin{bmatrix} m \\ n \end{bmatrix} = [m^2 + n^2] \), so m² + n² = 25.
2. Among the options, (3, 4) gives 9 + 16 = 25 with m < n.
3. (2, 3) gives 13 and (4, 3) has m > n, so (m, n) = (3, 4).

Teacher's Note:
1. A 1 × 2 row times a 2 × 1 column is a 1 × 1 matrix: the sum of products.
2. Read the extra condition m < n; it rules out (4, 3) even though 16 + 9 = 25.

Answer: (b) O

Explanation:
1. Write A = uuᵀ with u = (cos θ, sin θ)ᵀ and B = vvᵀ with v = (cos φ, sin φ)ᵀ.
2. Then AB = u(uᵀv)vᵀ, and uᵀv = cos θ cos φ + sin θ sin φ = cos(θ − φ).
3. Since θ − φ is an odd multiple of \( \frac{\pi}{2} \), cos(θ − φ) = 0, so AB = O.

Teacher's Note:
1. Each entry of AB contains the factor cos(θ − φ); spotting this saves four long multiplications.
2. Matrices of the form uuᵀ multiply as scalars uᵀv, a very handy shortcut.

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

Explanation:
1. Row 1 of A times column 1 of B: 0·p² + r·pq − q·pr = 0.
2. Row 1 of A times column 2 of B: 0·pq + r·q² − q·qr = 0; similarly every other entry cancels in pairs.
3. Hence AB is the 3 × 3 zero matrix.

Teacher's Note:
1. Shortcut: B = uuᵀ with u = (p, q, r)ᵀ, and Au = 0, so AB = (Au)uᵀ = O.
2. Compute one or two entries first; if they vanish by cancellation, the symmetric pattern usually continues.

Answer: (c) −5I

Explanation:
1. \( A^2 = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix} \).
2. \( 6A = \begin{bmatrix} 24 & 6 \\ 18 & 12 \end{bmatrix} \).
3. \( A^2 - 6A = \begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix} = -5I \).

Teacher's Note:
1. By Cayley–Hamilton, A² − (trace)A + |A|I = O; trace = 6 and |A| = 5, so A² − 6A = −5I at once.
2. The trace-and-determinant shortcut works for every 2 × 2 matrix.

Answer: (b) ±1

Explanation:
1. \( A^2 = \begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix}\begin{bmatrix} \lambda & 1 \\ -1 & -\lambda \end{bmatrix} = \begin{bmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{bmatrix} \).
2. A² = O requires λ² − 1 = 0.
3. So λ = ±1.

Teacher's Note:
1. The off-diagonal entries cancel automatically here (λ − λ = 0), so only the diagonal gives a condition.
2. A 2 × 2 matrix with A² = O must have trace 0 and determinant 0; here |A| = 1 − λ² = 0 gives the same result.

Answer: (a) 2

Explanation:
1. \( A^2 = \begin{bmatrix} a^2b^2 - a^2b^2 & ab^3 - ab^3 \\ -a^3b + a^3b & -a^2b^2 + a^2b^2 \end{bmatrix} \).
2. Every entry is zero, so A² = O.
3. A itself is not O in general, so the least n with Aⁿ = O is 2.

Teacher's Note:
1. A has trace ab − ab = 0 and determinant −a²b² + a²b² = 0, which forces A² = O for a 2 × 2 matrix.
2. Such a matrix is called nilpotent of index 2.

Answer: (b) 1

Explanation:
1. \( AB = \begin{bmatrix} 1/3 & 2 \\ 0 & 2x - 3 \end{bmatrix}\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 - 2 \\ 0 & -(2x - 3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 3 - 2x \end{bmatrix} \).
2. AB = I gives 3 − 2x = 1.
3. So x = 1.

Teacher's Note:
1. Only the entry that contains x needs to be matched; the others are a quick check that the data is consistent.
2. Watch the sign: (2x − 3)(−1) = 3 − 2x.

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

Explanation:
1. Adding the two equations: (2A + B) + (A − B) = 3A.
2. \( 3A = \begin{bmatrix} 6 & 4 \\ 6 & -11 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ 6 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 12 & -9 \end{bmatrix} \).
3. \( A = \begin{bmatrix} 2 & 2 \\ 4 & -3 \end{bmatrix} \).

Teacher's Note:
1. Treat matrix equations like simultaneous linear equations and eliminate B by adding.
2. Check: B = 2A + B − 2A should then give a sensible matrix; here B = [[2,0],[−2,−5]].

Answer: (c) [−2  −1  4]

Explanation:
1. A is 1 × 3 and B is 3 × 3, so AB is 1 × 3.
2. First entry: 1(−5) + 2(0) + 3(1) = −2; second: 1(4) + 2(2) + 3(−3) = −1; third: 1(0) + 2(−1) + 3(2) = 4.
3. So AB = [−2  −1  4].

Teacher's Note:
1. Fix the order of the product first: a 1 × 3 matrix times a 3 × 3 matrix must be a 1 × 3 row.
2. That order check alone rules out options (a), (b) and (d).

Answer: (b) Null matrix

Explanation:
1. Row 1 of A times the columns of B: (−1 + 6 − 5, −2 + 12 − 10, −1 + 6 − 5) = (0, 0, 0).
2. Row 2: (−2 − 18 + 20, −4 − 36 + 40, −2 − 18 + 20) = (0, 0, 0).
3. Row 3: (−3 − 12 + 15, −6 − 24 + 30, −3 − 12 + 15) = (0, 0, 0).
4. So AB is the null matrix.

Teacher's Note:
1. The columns of B are multiples of (−1, 6, 5)ᵀ, so it is enough to check that A sends this one column to zero.
2. A product of two non-zero matrices can be the zero matrix; this is a standard property to remember.

Answer: (c) 16A

Explanation:
1. A = 2I, so A⁵ = (2I)⁵ = 2⁵I⁵ = 32I.
2. Since A = 2I, 32I = 16(2I) = 16A.
3. So A⁵ = 16A.

Teacher's Note:
1. For a scalar matrix kI, (kI)ⁿ = kⁿI, because I commutes with everything and Iⁿ = I.
2. Do not choose 32A: rewrite 32I in terms of A = 2I before comparing with options.

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

Explanation:
1. Let \( B = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \). Then \( AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} p & q \\ r & s \end{bmatrix} = \begin{bmatrix} r & s \\ 0 & 0 \end{bmatrix} \).
2. AB = O requires r = 0 and s = 0, i.e. the second row of B is zero.
3. Only \( \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} \) has a zero second row.

Teacher's Note:
1. Left-multiplying by [[0,1],[0,0]] moves the second row of B up to the first row.
2. Find the general condition on B first; then scanning the options takes seconds.

Answer: (c) R(s + t)

Explanation:
1. \( R(s)R(t) = \begin{bmatrix} \cos s\cos t - \sin s\sin t & \cos s\sin t + \sin s\cos t \\ -\sin s\cos t - \cos s\sin t & -\sin s\sin t + \cos s\cos t \end{bmatrix} \).
2. Using the addition formulae, this is \( \begin{bmatrix} \cos(s + t) & \sin(s + t) \\ -\sin(s + t) & \cos(s + t) \end{bmatrix} \).
3. So R(s)R(t) = R(s + t).

Teacher's Note:
1. Products of rotation matrices add their angles, just as products of powers add exponents.
2. A quick check with t = 0: R(0) = I, so R(s)R(0) must equal R(s), which only (c) gives.

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

Explanation:
1. 4B = 3A − (3A − 4B).
2. \( 3A = \begin{bmatrix} 0 & 3 & 6 \\ 6 & 9 & 12 \\ 12 & 15 & 18 \end{bmatrix} \), so \( 4B = \begin{bmatrix} 0+4 & 3-3 & 6-6 \\ 6-6 & 9-5 & 12-12 \\ 12-12 & 15-15 & 18-14 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \).
3. Hence B = I, the unit matrix.

Teacher's Note:
1. Isolate the unknown matrix exactly as in ordinary algebra, then divide by the scalar.
2. Keep the sign right: 4B = 3A − (given matrix), not the given matrix − 3A.

Answer: (d) \( \begin{bmatrix} 1 & 4a \\ 0 & 1 \end{bmatrix} \)

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} \).
2. \( A^4 = (A^2)^2 = \begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2a \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4a \\ 0 & 1 \end{bmatrix} \).

Teacher's Note:
1. In general [[1,a],[0,1]]ⁿ = [[1,na],[0,1]]; the top-right entry adds, it does not multiply.
2. Squaring twice (A² then A⁴) is faster than multiplying four times.

Answer: (a) \( \begin{bmatrix} -3 & 4 \\ 14 & -13 \end{bmatrix} \)

Explanation:
1. Let \( M = \begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} \); |M| = 3 − 4 = −1, so \( M^{-1} = \frac{1}{-1}\begin{bmatrix} 1 & -1 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 4 & -3 \end{bmatrix} \).
2. \( X = M^{-1}\begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} -5+2 & 1+3 \\ 20-6 & -4-9 \end{bmatrix} \).
3. \( X = \begin{bmatrix} -3 & 4 \\ 14 & -13 \end{bmatrix} \).

Teacher's Note:
1. MX = C gives X = M⁻¹C with M⁻¹ on the left; writing CM⁻¹ instead is a common error.
2. Check: [[3,1],[4,1]][[−3,4],[14,−13]] = [[5,−1],[2,3]], which confirms the answer.

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

Explanation:
1. Subtract entry by entry: \( A - B = \begin{bmatrix} 5-6 & -3+4 \\ 2-3 & 4-6 \end{bmatrix} \).
2. \( A - B = \begin{bmatrix} -1 & 1 \\ -1 & -2 \end{bmatrix} \).

Teacher's Note:
1. Subtraction is done element by element, and the matrices must have the same order.
2. Option (a) is A + B; read the sign in the question carefully.

Answer: (c) \( X = \frac{1}{7}I, Y = \frac{2}{7}I \)

Explanation:
1. From 2X − Y = O we get Y = 2X.
2. Substituting in 3X + 2Y = I: 3X + 4X = I, so 7X = I.
3. \( X = \frac{1}{7}I \) and \( Y = \frac{2}{7}I \).

Teacher's Note:
1. X and Y are matrices, so they must be written as multiples of I, not as plain numbers; that removes (a) and (b).
2. Solve matrix equations by substitution or elimination exactly as with ordinary simultaneous equations.

Answer: (b) O

Explanation:
1. \( A - 2I = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \) and \( A - 3I = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \).
2. \( (A - 2I)(A - 3I) = \begin{bmatrix} 2-2 & 4-4 \\ -1+1 & -2+2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).
3. So the product is the null matrix O.

Teacher's Note:
1. The characteristic equation of A is λ² − 5λ + 6 = 0, i.e. (λ − 2)(λ − 3) = 0, so by Cayley–Hamilton (A − 2I)(A − 3I) = O.
2. Trace 5 and determinant 6 give this equation at once for any 2 × 2 matrix.

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

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. So A⁴ = (A²)² = I² = I.

Teacher's Note:
1. This A swaps the two rows; swapping twice gives back the identity.
2. Once A² = I, every even power of A is I and every odd power is A.

Answer: (d) \( \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \)

Explanation:
1. \( A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \).
2. Entries: (9 − 1, 3 + 2; −3 − 2, −1 + 4).
3. \( A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \).

Teacher's Note:
1. Signs in the off-diagonal entries are where most errors happen; compute each one separately.
2. Check with Cayley–Hamilton: A² = 5A − 7I = [[8,5],[−5,3]].

Answer: (d) None of these

Explanation:
1. \( X^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix} \).
2. Pattern: \( X^n = \begin{bmatrix} 1 + 2n & -4n \\ n & 1 - 2n \end{bmatrix} \) (true for n = 1, 2 and by induction).
3. (b) fails already at n = 1, and at n = 2 option (a) gives \( \begin{bmatrix} 6 & -8 \\ 2 & -2 \end{bmatrix} \) and (c) gives \( \begin{bmatrix} 9 & 16 \\ 1 & 1 \end{bmatrix} \), neither equal to X². So the answer is None of these.

Teacher's Note:
1. Always test a proposed formula for Xⁿ with n = 1 and n = 2; one failure is enough to reject it.
2. Powers do not act entry-wise, so an option like [[3ⁿ, (−4)ⁿ],[1ⁿ, (−1)ⁿ]] is never right in general.

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

Explanation:
1. A = iI, so A² = i²I = −I.
2. \( A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \).

Teacher's Note:
1. A scalar matrix kI squares to k²I; here k = i and i² = −1.
2. Do not square only one diagonal entry; both become −1.

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

Explanation:
1. Subtracting the equations: (A + B) − (A − 2B) = 3B, so \( 3B = \begin{bmatrix} 1+1 & 0-1 \\ 1-0 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix} \).
2. \( B = \begin{bmatrix} 2/3 & -1/3 \\ 1/3 & 2/3 \end{bmatrix} \).
3. \( A = (A + B) - B = \begin{bmatrix} 1-2/3 & 0+1/3 \\ 1-1/3 & 1-2/3 \end{bmatrix} = \begin{bmatrix} 1/3 & 1/3 \\ 2/3 & 1/3 \end{bmatrix} \).

Teacher's Note:
1. Alternatively 2(A + B) + (A − 2B) = 3A eliminates B in one step: 3A = [[1,1],[2,1]].
2. Option (b) is B, not A; keep track of which matrix you have found.

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

Explanation:
1. \( AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \).
2. Entries: (0 + 1, 1 + 0; 0 + 1, 0 + 0).
3. \( AB = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \).

Teacher's Note:
1. Right-multiplying by [[0,1],[1,0]] swaps the two columns of A, which gives the answer at a glance.
2. Left-multiplying by the same matrix would swap rows instead, so BA ≠ AB here.

Answer: (b) \( \begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix} \)

Explanation:
1. Row 1 of A (1, 3, 0) times B: (2 + 3, 3 + 6, 4 + 9) = (5, 9, 13).
2. Row 2 (−1, 2, 1): (−2 + 2 − 1, −3 + 4 + 1, −4 + 6 + 2) = (−1, 2, 4).
3. Row 3 (0, 0, 2): (−2, 2, 4).
4. So \( AB = \begin{bmatrix} 5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4 \end{bmatrix} \).

Teacher's Note:
1. Row 3 of A is (0, 0, 2), so row 3 of AB is just twice row 3 of B; such shortcuts separate close options.
2. Options (a) and (b) differ in one entry only, so compute that entry with extra care.

Answer: (b) x = 3, y = −2

Explanation:
1. \( LHS = \begin{bmatrix} x - 2 & 1 \\ 4 & y + 4 \end{bmatrix} \) and \( RHS = \begin{bmatrix} 3-2 & 5-4 \\ 6-2 & 3-1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix} \).
2. Equating entries: x − 2 = 1 and y + 4 = 2.
3. So x = 3 and y = −2.

Teacher's Note:
1. Equal matrices have equal corresponding entries; the entries without x or y act as a check.
2. Simplify each side fully before equating to avoid sign slips.

Answer: (b) \( \begin{bmatrix} 8 & -20 \\ 7 & -9 \end{bmatrix} \)

Explanation:
1. \( 5A = \begin{bmatrix} 5 & -10 \\ 15 & 0 \end{bmatrix} \), \( 3B = \begin{bmatrix} -3 & 12 \\ 6 & 9 \end{bmatrix} \), \( 2C = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \).
2. \( 5A - 3B - 2C = \begin{bmatrix} 5+3-0 & -10-12+2 \\ 15-6-2 & 0-9-0 \end{bmatrix} \).
3. \( = \begin{bmatrix} 8 & -20 \\ 7 & -9 \end{bmatrix} \).

Teacher's Note:
1. Write the three scalar multiples separately before combining; it keeps track of signs.
2. Subtracting 2C means adding +2 where C has −1, which is the step most often missed.

Answer: (b) \( a^2 I + 2abA \)

Explanation:
1. \( A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = O \).
2. aI and bA commute, so (aI + bA)² = a²I + 2abA + b²A².
3. With A² = O this is a²I + 2abA.

Teacher's Note:
1. The binomial expansion is valid here only because I commutes with every matrix.
2. Spotting A² = O early removes the b² term immediately.

Answer: (d) [20]

Explanation:
1. UV = [2(3) − 3(2) + 4(1)] = [4].
2. XY = [0(2) + 2(2) + 3(4)] = [16].
3. UV + XY = [20].

Teacher's Note:
1. A row matrix times a column matrix is a 1 × 1 matrix, so the answer is written [20], not the number 20.
2. Each product is the dot product of the row with the column.

Answer: (c) Matrix multiplication is commutative

Explanation:
1. Matrix addition is commutative and associative, and matrix multiplication is associative.
2. Multiplication is not commutative: 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 the statement that is not true is (c).

Teacher's Note:
1. Keep one ready counter-example for AB ≠ BA; the pair above is the simplest.
2. Some pairs, such as A and I, do commute, but commutativity fails in general.

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

Explanation:
1. Orders: A is 1 × 3, B is 3 × 1 and C is 2 × 2.
2. AB: columns of A (3) = rows of B (3), so AB is defined (1 × 1). BA: columns of B (1) = rows of A (1), so BA is defined (3 × 3).
3. AB is 1 × 1 and C is 2 × 2, so (AB)C is not defined; AC is not defined since A has 3 columns and C has 2 rows, so (AC)B is not defined either.
4. Hence both AB and BA are defined: option (d).

Teacher's Note:
1. A product PQ is defined only when the number of columns of P equals the number of rows of Q.
2. Write the order under each matrix and check the inner numbers; the outer numbers give the order of the product.

Answer: (d) None of these

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & 2 & 3 \\ -2 & 3 & -1 \\ 3 & 1 & 2 \end{bmatrix}^2 = \begin{bmatrix} 6 & 11 & 7 \\ -11 & 4 & -11 \\ 7 & 11 & 12 \end{bmatrix} \).
2. \( A^2 + 9I = \begin{bmatrix} 15 & 11 & 7 \\ -11 & 13 & -11 \\ 7 & 11 & 21 \end{bmatrix} \).
3. Comparing with kA: the (1, 1) entry needs k = 15 but the (1, 2) entry needs k = 11/2, so A² + 9I is not 2A, 4A or 6A. Answer: None of these.

Teacher's Note:
1. To test whether a matrix equals kA, find k from one entry and check it against another entry.
2. Two entries giving different values of k is enough to reject every option of the form kA.