If A is a square matrix such that A(operatorn | vedclass

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(B) Given $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4 \end{bmatrix}$.Taking the determinant on both sides,we get $|

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$16$

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(B) Given $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4 \end{bmatrix}$.Taking the determinant on both sides,we get $|A(\operatorname{adj} A)| = \begin{vmatrix} 4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4 \end{vmatrix}$.Using the property $|AB| = |A||B|$,we have $|A| |\operatorname{adj} A| = 4^3 = 64$.We know that $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the matrix. Here $n = 3$,so $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.Substituting this into the equation,we get $|A| \cdot |A|^2 = 64$,which implies $|A|^3 = 64$.Thus,$|A| = 4$.Finally,$|\operatorname{adj} A| = |A|^2 = 4^2 = 16$.

If $A$ is a square matrix such that $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,then $\operatorname{det}(\operatorname{adj} A)$ is equal to

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