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

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

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$81$

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(D) Given,$A = \begin{bmatrix} 2 & 1 & 0 \ 0 & 2 & 1 \ 1 & 0 & 2 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = 2(4 - 0) - 1(0 - 1) + 0(0 - 2) = 2(4) - 1(-1) + 0 = 8 + 1 = 9$.We know the property of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the square matrix $A$.Here,$n = 3$.Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.Substituting the value of $|A| = 9$,we get:$|\operatorname{adj} A| = 9^2 = 81$.

If $A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$,then $|\operatorname{adj} A|$ is equal to

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