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

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(C) Given $A = \begin{bmatrix} 2 & 3 \ 0 & -1 \end{bmatrix}$. The determinant $|A| = (2)(-1) - (3)(0) = -2$.First,we calculate $\det(A^4) = |A|^4 = (

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$16$

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(C) Given $A = \begin{bmatrix} 2 & 3 \ 0 & -1 \end{bmatrix}$. The determinant $|A| = (2)(-1) - (3)(0) = -2$.First,we calculate $\det(A^4) = |A|^4 = (-2)^4 = 16$.Next,consider $\operatorname{adj}(2A)$. Since $2A = \begin{bmatrix} 4 & 6 \ 0 & -2 \end{bmatrix}$,its determinant is $|2A| = 2^2 |A| = 4(-2) = -8$.The adjoint of a $2 	imes 2$ matrix $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$\operatorname{adj}(2A) = \begin{bmatrix} -2 & -6 \ 0 & 4 \end{bmatrix}$.Note that $\operatorname{adj}(2A) = -2 \begin{bmatrix} 1 & 3 \ 0 & -2 \end{bmatrix}$.Alternatively,using the property $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,for $n=2$,$\operatorname{adj}(2A) = 2 \operatorname{adj}(A) = 2 \begin{bmatrix} -1 & -3 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & -6 \ 0 & 4 \end{bmatrix}$.The eigenvalues of $A$ are $\lambda_1 = 2$ and $\lambda_2 = -1$. The eigenvalues of $A^{10}$ are $2^{10}$ and $(-1)^{10} = 1$.The eigenvalues of $\operatorname{adj}(2A)$ are $2 	imes (-1) = -2$ and $2 	imes 2 = 4$. Thus,the eigenvalues of $(\operatorname{adj}(2A))^{10}$ are $(-2)^{10} = 2^{10}$ and $4^{10} = 2^{20}$.Since $A^{10}$ and $(\operatorname{adj}(2A))^{10}$ share the eigenvalue $2^{10}$,the matrix $A^{10} - (\operatorname{adj}(2A))^{10}$ has at least one eigenvalue equal to $2^{10} - 2^{10} = 0$.Therefore,$\det(A^{10} - (\operatorname{adj}(2A))^{10}) = 0$.Finally,the required value is $16 + 0 = 16$.

If $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$,then the value of $\det(A^4) + \det(A^{10} - (\operatorname{adj}(2A))^{10})$ is equal to ........

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