Matrices & Determinants — Practice

Question 1

The value of the determinant |5 3; 4 2| is:
(a) -2  (b) 2  (c) 22  (d) 14

Accepted Answer

|a b; c d| = ad − bc = 5(2) − 3(4) = 10 − 12 = −2.

Answer: (a) -2. Trap (b) 2 comes from computing bc − ad (sign flip); (c) 22 comes from adding instead of subtracting (ad + bc); (d) 14 is the sum of all four entries, not a determinant operation at all.

Question 2

The value of |1 2 3; 0 1 4; 5 6 0| is:
(a) 1  (b) -79  (c) 24  (d) -1

Accepted Answer

Expand along row 1: 1(1×0 − 4×6) − 2(0×0 − 4×5) + 3(0×6 − 1×5)
= 1(0−24) − 2(0−20) + 3(0−5)
= 1(−24) − 2(−20) + 3(−5)
= −24 + 40 − 15 = 1.

Answer: (a) 1. Trap (b) -79 comes from forgetting the alternating sign pattern and adding all three cofactor terms as positive: −24 + (−40) + (−15) = −79. Always write the +,−,+ checkerboard before expanding.

Question 3

If A is a matrix of order 2×3 and B is a matrix of order 3×4, then the order of AB is:
(a) 2×4  (b) 3×3  (c) 4×2  (d) Not defined

Accepted Answer

For A(m×n)·B(n×p), the inner dimensions (3 and 3) must match — they do — and the product takes the outer dimensions m×p = 2×4.

Answer: (a) 2×4. Trap (b) 3×3 confuses the matching inner dimension for the answer's order; (c) 4×2 reverses the correct order; (d) wrongly assumes a mismatch.

Question 4

If A and B are matrices such that the product AB is defined, then (AB)ᵀ equals:
(a) AᵀBᵀ  (b) BᵀAᵀ  (c) BA  (d) AB

Accepted Answer

The transpose of a product reverses the order of multiplication: (AB)ᵀ = BᵀAᵀ. This is provable directly from the definition of matrix multiplication and holds even when A and B are not square.

Answer: (b) BᵀAᵀ. Trap (a) AᵀBᵀ is the single most common transpose error — order does NOT stay the same.

Question 5

A is a 3×3 matrix with |A| = 5. The value of |2A| is:
(a) 10  (b) 40  (c) 15  (d) 5

Accepted Answer

|kA| = kⁿ|A| for an n×n matrix. Here n = 3, k = 2: |2A| = 2³ × 5 = 8 × 5 = 40.

Answer: (b) 40. Trap (a) 10 uses |kA| = k|A| (forgetting the exponent entirely); (c) 15 uses k = 3 instead of kⁿ; (d) 5 wrongly assumes scaling the matrix doesn't change the determinant.

Question 6

If A = [4 3; 2 1], then A⁻¹ equals:
(a) [-1/2 3/2; 1 -2]  (b) [1 -3; -2 4]  (c) [1/2 -3/2; -1 2]  (d) [-1/2 -3/2; -1 -2]

Accepted Answer

|A| = 4(1) − 3(2) = 4 − 6 = −2.

adj(A) = [1 -3; -2 4] (swap diagonal 4↔1, negate off-diagonal 3,2).

A⁻¹ = (1/−2) × [1 -3; -2 4] = [-1/2 3/2; 1 -2].

Answer: (a). Trap (b) is exactly adj(A) with the 1/|A| division forgotten — the classic inverse error. Trap (c) uses |A| = +2 instead of the correct −2 (sign error in the determinant).

Question 7

A is a 3×3 matrix with |A| = 3. The value of |adj A| is:
(a) 3  (b) 9  (c) 27  (d) 6

Accepted Answer

|adj A| = |A|ⁿ⁻¹ for an n×n matrix. Here n = 3, so |adj A| = |A|² = 3² = 9.

Answer: (b) 9. Trap (a) 3 forgets the exponent entirely; (c) 27 uses |A|ⁿ (= |A|³) instead of |A|ⁿ⁻¹; (d) 6 uses a linear (n×|A|) relationship that doesn't exist.

Question 8

The system x + 2y = 3 and 2x + 4y = 9:
(a) has a unique solution  (b) has no solution  (c) has infinitely many solutions  (d) has only the trivial solution

Accepted Answer

Δ = |1 2; 2 4| = 1(4) − 2(2) = 4 − 4 = 0.

Δx = |3 2; 9 4| = 3(4) − 2(9) = 12 − 18 = −6 ≠ 0.

Since Δ = 0 but Δx ≠ 0, the system is inconsistent — the two lines are parallel and distinct.

Answer: (b) has no solution. Trap (c) is the classic error of assuming Δ = 0 always means infinite solutions without checking Δx.

Question 9

For the system x+y+z=6, 2x−y+z=3, x+2y−z=2, the value of y is:
(a) 1  (b) 2  (c) 3  (d) 4

Accepted Answer

Δ = |1 1 1; 2 -1 1; 1 2 -1| = 1(1−2) − 1(−2−1) + 1(4+1) = −1 + 3 + 5 = 7.

Δy (replace column 2 with constants 6,3,2) = |1 6 1; 2 3 1; 1 2 -1| = 1(−3−2) − 6(−2−1) + 1(4−3) = −5 + 18 + 1 = 14.

y = Δy/Δ = 14/7 = 2. (Full solution: x=1, y=2, z=3 — check row 2: 2(1)−2+3=3 ✓; row 3: 1+4−3=2 ✓.)

Answer: (b) 2.

Question 10

If A is a skew-symmetric matrix of order 3×3, then |A| equals:
(a) 0  (b) 1  (c) It depends on the entries of A  (d) -|A|

Accepted Answer

For a skew-symmetric matrix, Aᵀ = −A. So |A| = |Aᵀ| = |−A| = (−1)³|A| = −|A|, which gives 2|A| = 0, i.e. |A| = 0.

Answer: (a) 0. This holds for any ODD order — for even order (e.g. 2×2), a skew-symmetric matrix can have a non-zero determinant, so option (c) would be the answer to that variant.

Question 11

A is a matrix of order 3×4 and B is a matrix of order 4×3. Which statement is correct?
(a) Both AB and BA are defined; AB is 3×3 and BA is 4×4  (b) Only AB is defined  (c) Only BA is defined  (d) Neither product is defined

Accepted Answer

AB: inner dimensions 4 = 4 match, so AB is defined and has order 3×3 (outer dimensions).

BA: inner dimensions 3 = 3 match, so BA is also defined, with order 4×4.

Both products exist but have different orders — a direct illustration of why AB ≠ BA even makes sense as a question (they're not even the same size here).

Answer: (a). This is a favourite NDA trap for testing whether students check dimensions instead of assuming non-square matrices can't be multiplied both ways.

Practice — Matrices & Determinants

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