If A=0-4121-312-1, then A−1 exists (i.e., A is invertible) if

0-4121-312-1≠00+4(−2+3)+1(4−λ)≠04+4−λ≠0⇒λ≠8

If A=0-4121-312-1, then A−1 exists (i.e., A is invertible) if

0-4121-312-1≠00+4(−2+3)+1(4−λ)≠04+4−λ≠0⇒λ≠8

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If $A = [\begin{matrix} 0 & - 4 & 1 \\ 2 & 1 & - 3 \\ 1 & 2 & - 1 \end{matrix}]$ , then $A^{- 1}$ exists (i.e., A is invertible) if

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$λ \neq 8$

$λ = 4$

$λ = 8$

$λ \neq 4$

answer is A.

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$|\begin{matrix} 0 & - 4 & 1 \\ 2 & 1 & - 3 \\ 1 & 2 & - 1 \end{matrix}| \neq 0$

$0 + 4 (- 2 + 3) + 1 (4 - λ) \neq 0$

$4 + 4 - λ \neq 0 \Rightarrow λ \neq 8$

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If A=0-4121-312-1, then A−1 exists (i.e., A is invertible) if

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If $A = [\begin{matrix} 0 & - 4 & 1 \\ 2 & 1 & - 3 \\ 1 & 2 & - 1 \end{matrix}]$ , then $A^{- 1}$ exists (i.e., A is invertible) if