If A = begin{bmatrix} 1 & 0 & 0 0 & 2 & 3 1 | vedclass

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(B) Given the matrix $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 3 \ 1 & 0 & 1 \end{bmatrix}$.First,we calculate the determinant of $A$,denoted as $|A|

Vedclass · Accepted Answer

$4$

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(B) Given the matrix $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 3 \ 1 & 0 & 1 \end{bmatrix}$.First,we calculate the determinant of $A$,denoted as $|A|$.Expanding along the first row: $|A| = 1(2 	imes 1 - 3 	imes 0) - 0 + 0 = 1(2) = 2$.The order of the matrix $A$ is $n = 3$.We use the property of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$.Substituting the values: $|\operatorname{adj} A| = 2^{3-1} = 2^2 = 4$.Therefore,the correct option is $B$.

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 1 & 0 & 1 \end{bmatrix}$,then $|\operatorname{adj} A| = $ . . . . . . .

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