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Question:

Let M = ⟦0 1 a; 1 2 3; 3 b 1⟧ and adj M = ⟦-1 1 -1; 8 -6 2; -5 3 -1⟧ where a and b are real numbers. Which of the following option is/are correct?

det(adj M²) = 81

If M⟦α β γ⟧ = ⟦1 2 3⟧, then α - β + γ = 3

(adj M)⁻¹ + adj M⁻¹ = -M

a + b = 3

Solution:

Correct option is D (adj M)⁻¹ + adj M⁻¹ = -M

M = ⟦0 1 a; 1 2 3; 3 b 1⟧ and adj M = ⟦-1 1 -1; 8 -6 2; -5 3 -1⟧
(adj M)11 = 2 - 3b, (adj M)22 = -3 ⇒ 2 - 3b = -3 ⇒ b = 1 and -3a = -6; ⇒ a = 2
|adj M| = -1(6 - 6) - 1(-8 + 10) - 1(24 - 30) = 2 - 6 = -4
det(adj(M²)) = |det(adj M)|² = (-4)² = 16
Now ⟦0 1 2; 1 2 3; 3 1 1⟧ ⟦α β γ⟧ = ⟦1 2 3⟧ ⇒ β + 2γ = 1, α + 2β + 3γ = 2, 3α + β + γ = 3
On solving α = 1, β = -1, γ = 1 so α - β + γ = 3
Now (adj M)⁻¹ + (adj M)⁻¹ = 2(adj M)⁻¹ = 2 adj(adj M) |adj M| = 12 |M|³ -2