If A=12-1-1122-11, then det. [adj (adj A)] is

We know that adj (adj A) = |A|n–2 A if |A| ≠ 0, provided order of A is n.∴ adj (adj A) = |A| A (as n = 3)∴ det [adj (adj A)] = |A|3 det A = |A|4.But A=12-1-1122-11=14∴ det [adj (adj A)] = (14)4.

If A=12-1-1122-11, then det. [adj (adj A)] is

We know that adj (adj A) = |A|n–2 A if |A| ≠ 0, provided order of A is n.∴ adj (adj A) = |A| A (as n = 3)∴ det [adj (adj A)] = |A|3 det A = |A|4.But A=12-1-1122-11=14∴ det [adj (adj A)] = (14)4.

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If $A = [\begin{matrix} 1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix}]$ , then det. [adj (adj A)] is

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(14)¹

(14)⁴

(14)²

(14)³

answer is A.

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Detailed Solution

We know that adj (adj A) = |A|^n–2 A if |A| ≠ 0, provided order of A is n.
$∴$ adj (adj A) = |A| A (as n = 3)
$∴$ det [adj (adj A)] = |A|³ det A = |A|⁴.
But $|A| = |\begin{matrix} 1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix}| = 14$
$∴$ det [adj (adj A)] = (14)⁴.