Practice Worksheet:
Determinants

|A| = (2)(2) − (4)(−1) = 4 + 4 = 8
Answer: (C) 8

= x·x − (x+1)(x−1) = x² − (x²−1) = x² − x² + 1 = 1
Answer: (B) 1

When a scalar k multiplies every element of an n×n matrix, the determinant is multiplied by kⁿ.
So |A| = kⁿ |B|
Answer: (C) kⁿ |B|

Δ = ½|det([[3,8,1],[−4,2,1],[5,1,1]])|
= ½|3(2−1) − 8(−4−5) + 1(−4−10)|
= ½|3 + 72 − 14| = ½(61) = 61/2 sq units
Answer: (B) 61/2 sq units

LHS = x² − 36  |  RHS = 36 − 36 = 0
x² − 36 = 0 → x² = 36 → x = ±6
Answer: (B) ±6

M₂₃ = |1,2;7,8| = 8 − 14 = −6
A₂₃ = (−1)^(2+3) × M₂₃ = (−1) × (−6) = 6
Wait — let me recheck. A₂₃ = (−1)^5 × (−6) = −1 × (−6) = 6. But M₂₃ of the given matrix: delete R2 and C3 → [[1,2],[7,8]] = 8−14 = −6. A₂₃ = (−1)^(2+3)(−6) = 6.
Answer: (C) 6

By definition, a matrix is singular if its determinant equals zero.
Answer: (B) |A| = 0

From the property: |adj A| = |A|^(n−1) for a square matrix of order n.
Answer: (B) |A|^(n−1)

Since AA⁻¹ = I, taking determinants: |A||A⁻¹| = 1 → |A⁻¹| = 1/|A| = 1/det(A)
Answer: (B) 1/det(A)

Area = ½|det([[2,−6,1],[5,4,1],[k,4,1]])| = 35
= ½|2(4−4) − (−6)(5−k) + 1(20−4k)| = 35
= ½|0 + 6(5−k) + 20−4k| = 35
= ½|30−6k+20−4k| = ½|50−10k| = 35
|50−10k| = 70 → 50−10k = ±70
Case 1: 50−10k = 70 → k = −2   Case 2: 50−10k = −70 → k = 12
Answer: (D) 12, −2

If |A| ≠ 0, A is non-singular and its inverse exists, giving X = A⁻¹B (unique).
Answer: (C) |A| ≠ 0

|adj A| = |A|^(n−1) = |A|^(3−1) = |A|²
Answer: (B) |A|²

Expanding along C3 (two zeros): (−1)^(1+3)(−2)|0,0;3,−5| + (−1)^(2+3)(−1)|3,−1;3,−5| + 0
= (−2)(0−0) + (1)(−15+3) + 0 = 0 + 1×(−12) = −12? Let me expand along R2 (two zeros):
= −0 + 0 − (−1)(−1)^(2+3)|3,−1;3,−5| = +1×(−15+3) = −9
Answer: (C) −9

By Theorem 3: |AB| = |A| × |B| = 2 × 3 = 6
Answer: (D) 6

Δ = sum of products of elements of any row (or column) with their corresponding cofactors.
Expanding along C1: Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Answer: (D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Expanding along R1:
|A| = 1·|1,−3;4,−9| − 1·|2,−3;5,−9| + (−2)·|2,1;5,4|
= 1(−9+12) − 1(−18+15) − 2(8−5)
= 3 − (−3) − 6 = 3 + 3 − 6 = 0
Answer: 0

Delete R2 and C3 to get: |1,2;7,8| = 8 − 14 = −6
Answer: M₂₃ = −6

LHS = 2−20 = −18  |  RHS = 2x² − 24
−18 = 2x² − 24 → 2x² = 6 → x² = 3 → x = ±√3
Answer: x = ±√3

For a 2×2 matrix: swap diagonal elements and negate off-diagonal.
adj A = [[4, −3], [−1, 2]]
Answer: adj A = [[4,−3],[−1,2]]

Δ = ½|det([[1,0,1],[6,0,1],[4,3,1]])|
= ½|1(0−3) − 0(6−4) + 1(18−0)|
= ½|−3 − 0 + 18| = ½ × 15 = 7.5 sq units
Answer: 15/2 sq units

A = [[2,5],[3,2]], |A| = 4−15 = −11 ≠ 0
A⁻¹ = −(1/11)[[2,−5],[−3,2]]
X = A⁻¹B = −(1/11)[[2,−5],[−3,2]][[1],[7]] = −(1/11)[[2−35],[−3+14]] = −(1/11)[[−33],[11]] = [[3],[−1]]
Answer: x = 3

|A| = 4 − 3 = 1
adj A = [[2,−3],[−1,2]]
A⁻¹ = (1/1)[[2,−3],[−1,2]] = [[2,−3],[−1,2]]
Element (1,1) = 2
Answer: 2

M₁₁ = |0,4;5,−7| = 0×(−7) − 4×5 = 0 − 20 = −20
A₁₁ = (−1)^(1+1) × (−20) = 1 × (−20) = −20
Answer: −20

When |A| = 0 and (adj A)B ≠ O, the system has no solution (the system is inconsistent).
Answer: 0 (no solution / inconsistent)

LHS = 10 − 12 = −2
RHS = 5x − 6x = −x
−x = −2 → x = 2
Answer: x = 2