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(C) Given $C = \operatorname{adj} A = \begin{bmatrix} 2 & -1 & 1 \ -1 & 0 & 2 \ 1 & -2 & -1 \end{bmatrix}$.Calculating the determinant $|C| = |\oper

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$(3, 1/81)$

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(C) Given $C = \operatorname{adj} A = \begin{bmatrix} 2 & -1 & 1 \ -1 & 0 & 2 \ 1 & -2 & -1 \end{bmatrix}$.Calculating the determinant $|C| = |\operatorname{adj} A| = 2(0 - (-4)) - (-1)(1 - 2) + 1(2 - 0) = 2(4) + 1(-1) + 1(2) = 8 - 1 + 2 = 9$.We know that $|\operatorname{adj} A| = |A|^{n-1}$,where $n=3$. So,$|A|^2 = 9$,which implies $|A| = \pm 3$. Thus,$\lambda = \pm 3$ and $|\lambda| = 3$.Given $B = \operatorname{adj} C = \operatorname{adj}(\operatorname{adj} A)$.Using the property $|\operatorname{adj} M| = |M|^{n-1}$,we have $|B| = |\operatorname{adj} C| = |C|^{n-1} = |C|^2 = 9^2 = 81$.We need to find $\mu = |(B^{-1})^T|$. Since $|(B^{-1})^T| = |B^{-1}| = \frac{1}{|B|}$,we get $\mu = \frac{1}{81}$.Therefore,the ordered pair $(|\lambda|, \mu) = (3, 1/81)$.

Let $A$ be a $3 \times 3$ matrix such that $\operatorname{adj} A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{bmatrix}$ and $B = \operatorname{adj}(\operatorname{adj} A)$. If $|A| = \lambda$ and $|(B^{-1})^T| = \mu$,then the ordered pair $(|\lambda|, \mu)$ is equal to

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