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

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

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$\frac{1}{27} \begin{bmatrix} 1 & -26 \ 0 & 27 \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} 3 & 2 \ 0 & 1 \end{bmatrix}$.First,we calculate $A^2 = \begin{bmatrix} 3 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 \ 0 & 1 \end{bmatrix}$.Then,$A^3 = A^2 \cdot A = \begin{bmatrix} 9 & 8 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 27 & 26 \ 0 & 1 \end{bmatrix}$.We know that $(A^{-1})^3 = (A^3)^{-1}$.For a matrix $M = \begin{bmatrix} a & b \ 0 & d \end{bmatrix}$,$M^{-1} = \frac{1}{ad} \begin{bmatrix} d & -b \ 0 & a \end{bmatrix}$.Here,$A^3 = \begin{bmatrix} 27 & 26 \ 0 & 1 \end{bmatrix}$,so $a=27, b=26, d=1$.$(A^3)^{-1} = \frac{1}{27 	imes 1} \begin{bmatrix} 1 & -26 \ 0 & 27 \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 1 & -26 \ 0 & 27 \end{bmatrix}$.

If $A = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix}$,then $(A^{-1})^3 = $

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