DeterminantHard

Question

Let A = , then A-1 exists if

Options

A.x ≠ 0
B.λ ≠ 0
C.3x + λ ≠ 0, λ ≠ 0
D.x ≠ 0, λ ≠ 0

Solution

A =
|A| = (3x + λ)
by R2 → R2 - R1, R3 → R3 - R1
= (3x + λ)  = (3x + λ) (λ2)
|A| ≠ 0 for non-singular matrix
∴  3x + λ ≠ 0, λ ≠ 0

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