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

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(A) Given $A = \begin{bmatrix} 2 & -1 \ 3 & -2 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (2)(-2) - (-1)(3) = -4 + 3 = -1$.Since

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(A) Given $A = \begin{bmatrix} 2 & -1 \ 3 & -2 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (2)(-2) - (-1)(3) = -4 + 3 = -1$.Since $|A| 
eq 0$,$A$ is invertible.Now,calculate $A^2$:$A^2 = \begin{bmatrix} 2 & -1 \ 3 & -2 \end{bmatrix} \begin{bmatrix} 2 & -1 \ 3 & -2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \ 6-6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $A^2 = I$,we can find $A^3$:$A^3 = A^2 \cdot A = I \cdot A = A$.We need to find the inverse of $A^3$,which is $(A^3)^{-1}$.Since $A^3 = A$,then $(A^3)^{-1} = A^{-1}$.From $A^2 = I$,we know that $A \cdot A = I$,which implies $A^{-1} = A$.Therefore,$(A^3)^{-1} = A$.

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$,then the inverse of the matrix $A^3$ is

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