Class 12 Maths Chapter 4: Determinants worksheet with answer key

Answer: (c) 78

Expand along R₁ (it has a zero): |A| = 3(4·5 − 0·1) − 0 + 2(1·1 − 4·(−2)) = 3(20) + 2(9) = 60 + 18 = 78.

Answer: (a) 3√2

x² − 12 = 20 − 14 = 6, so x² = 18 and, as x > 0, x = √18 = 3√2.

Answer: (a) inconsistent, as it has no solution

Doubling the first equation gives 2x + 4y = 6, which cannot also equal 7. No (x, y) satisfies both, so the system is inconsistent.

Answer: (d) 39

M₂₃ = |[[2, 7], [5, −2]]| = −4 − 35 = −39, and A₂₃ = (−1)²⁺³ M₂₃ = −(−39) = 39.

Answer: (c) 9

Singular means the determinant is 0: 2k − 18 = 0, so k = 9.

For P(x, y) on the line, the area of the triangle with vertices (x, y), (2, 3), (−1, −3) is zero: |[[x, y, 1], [2, 3, 1], [−1, −3, 1]]| = 0. Expanding along R₁: x(3 + 3) − y(2 + 1) + 1(−6 + 3) = 6x − 3y − 3 = 0, i.e. 2x − y − 1 = 0. (Both points satisfy it.)

• sets the determinant with (x, y) and the two points equal to zero (1)
• expands to 2x − y − 1 = 0 (1)

A₁₁ = |[[3, 1], [2, −2]]| = −6 − 2 = −8; A₁₂ = −|[[0, 1], [5, −2]]| = −(0 − 5) = 5; A₁₃ = |[[0, 3], [5, 2]]| = 0 − 15 = −15. Δ = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = 2(−8) + (−1)(5) + 4(−15) = −16 − 5 − 60 = −81.

• the three cofactors −8, 5, −15 (1)
• Δ = −81 (1)

Expand along C₃, which has two zeros: the determinant is 3 |[[λ, 2], [2, λ]]| = 3(λ² − 4). It is singular when 3(λ² − 4) = 0, i.e. λ = 2 or λ = −2.

• determinant 3(λ² − 4), expanded along C₃ (1)
• λ = ±2 (1)

(a) A₁₂ = −|[[−1, 1], [5, −2]]| = −(2 − 5) = 3; A₂₂ = |[[2, 3], [5, −2]]| = −4 − 15 = −19; A₃₂ = −|[[2, 3], [−1, 1]]| = −(2 + 3) = −5. Δ = 0(3) + 4(−19) + 0(−5) = −76. (b) a₁₁A₁₂ + a₂₁A₂₂ + a₃₁A₃₂ = 2(3) + (−1)(−19) + 5(−5) = 6 + 19 − 25 = 0: the elements of C₁ times the cofactors of C₂.

• the cofactors 3, −19, −5 (1)
• Δ = −76 (1)
• (b) sum 6 + 19 − 25 = 0 (1)

A² = [[2, 1, 1], [1, 2, 1], [1, 1, 2]], so A² − A − 2I = [[2 − 0 − 2, 1 − 1, 1 − 1], [1 − 1, 2 − 0 − 2, 1 − 1], [1 − 1, 1 − 1, 2 − 0 − 2]] = O. |A| = 0 − 1(0 − 1) + 1(1 − 0) = 2 ≠ 0, so A⁻¹ exists. Post-multiplying A² − A = 2I by A⁻¹: A − I = 2A⁻¹, so A⁻¹ = 12(A − I) = 12[[−1, 1, 1], [1, −1, 1], [1, 1, −1]].

• A² computed and the identity shown (1)
• |A| ≠ 0 and A − I = 2A⁻¹ (1)
• A⁻¹ = 12[[−1, 1, 1], [1, −1, 1], [1, 1, −1]] (1)

Let x, y be the prices (₹) of a football and a bat: 3x + 4y = 5600, 5x + 2y = 4900, i.e. AX = B with A = [[3, 4], [5, 2]], B = [[5600], [4900]]. |A| = 6 − 20 = −14 ≠ 0. adj A = [[2, −4], [−5, 3]]. X = A⁻¹B = −114[[2(5600) − 4(4900)], [−5(5600) + 3(4900)]] = −114[[−8400], [−13300]] = [[600], [950]]. A football costs ₹600 and a bat ₹950.

• equations as AX = B and |A| = −14 (1)
• adj A and A⁻¹ (1)
• x = 600, y = 950 (1)