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(C) Given the diagonal matrix $A = \begin{bmatrix} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{bmatrix}$.For a diagonal matrix,$A^n = \begin{bmatrix} 3^n

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$\begin{bmatrix} \frac{1}{27} & 0 & 0 \ 0 & \frac{1}{125} & 0 \ 0 & 0 & \frac{1}{64} \end{bmatrix}$

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(C) Given the diagonal matrix $A = \begin{bmatrix} 3 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 4 \end{bmatrix}$.For a diagonal matrix,$A^n = \begin{bmatrix} 3^n & 0 & 0 \ 0 & 5^n & 0 \ 0 & 0 & 4^n \end{bmatrix}$.Therefore,$B = A^3 = \begin{bmatrix} 3^3 & 0 & 0 \ 0 & 5^3 & 0 \ 0 & 0 & 4^3 \end{bmatrix} = \begin{bmatrix} 27 & 0 & 0 \ 0 & 125 & 0 \ 0 & 0 & 64 \end{bmatrix}$.The inverse of a diagonal matrix $D = 	ext{diag}(d_1, d_2, d_3)$ is $D^{-1} = 	ext{diag}(\frac{1}{d_1}, \frac{1}{d_2}, \frac{1}{d_3})$.Thus,$B^{-1} = \begin{bmatrix} \frac{1}{27} & 0 & 0 \ 0 & \frac{1}{125} & 0 \ 0 & 0 & \frac{1}{64} \end{bmatrix}$.Hence,option $C$ is correct.

If $A=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ and $B=A^3$,then $B^{-1}=$

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