Discussion
If \(A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -2 \end{bmatrix}\) then \(|adj(adjA)|\) is equal to
Answer: B
\(|adj(adj A)|\) equals 256.
A is upper triangular, so its determinant is the product of the diagonal: \(|A| = 1 \times 2 \times (-2) = -4\). For an n-by-n matrix, \(|adj A| = |A|^{n-1}\), and applying this twice:
\[ \begin{aligned} |adj(adj A)| &= |A|^{(n-1)^{2}} = (-4)^{4} \\ &= 256 \end{aligned} \]
With n = 3 the exponent is 4, an even power, so the sign disappears; -256 keeps a sign that should not be there. 16 is \(|adj A| = (-4)^{2}\), one step only.
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