Let A = begin{bmatrix} 2 & 1 & 0 1 & 2 & -1 | vedclass

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(A) Given $A = \begin{bmatrix} 2 & 1 & 0 \ 1 & 2 & -1 \ 0 & -1 & 2 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = 2(4 - 1) - 1(2 - 0) +

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$10$

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(A) Given $A = \begin{bmatrix} 2 & 1 & 0 \ 1 & 2 & -1 \ 0 & -1 & 2 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = 2(4 - 1) - 1(2 - 0) + 0 = 2(3) - 2 = 4$.Since $A$ is a $3 	imes 3$ matrix,for any matrix $M$,$|\operatorname{adj}(M)| = |M|^{3-1} = |M|^2$.Therefore,$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))| = |2A|^{(3-1)^3} = |2A|^{2^3} = |2A|^8$.Since $|kA| = k^n|A|$ for an $n 	imes n$ matrix,$|2A| = 2^3|A| = 8 	imes 4 = 32 = 2^5$.Substituting this into the expression:$|2A|^8 = (2^5)^8 = 2^{40}$.We are given that this equals $(16)^n = (2^4)^n = 2^{4n}$.Equating the exponents: $4n = 40 \Rightarrow n = 10$.

Let $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$. If $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))| = (16)^n$,then $n$ is equal to

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