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(B) First,we calculate $A^{2} = A \cdot A = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} = \begin{bmatrix

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$\begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix}$

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(B) First,we calculate $A^{2} = A \cdot A = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6 \ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \ 4 & 7 \end{bmatrix}$.Now,substitute into the equation $A^{2} - 4A + I$:$\begin{bmatrix} 7 & 12 \ 4 & 7 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 12 \ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7-8+1 & 12-12+0 \ 4-4+0 & 7-8+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = O$.To find $A^{-1}$,start with $A^{2} - 4A + I = O$.Subtract $I$ from both sides: $A^{2} - 4A = -I$.Multiply by $A^{-1}$ on both sides: $A^{-1}(A^{2} - 4A) = A^{-1}(-I)$.$(A^{-1}A)A - 4(A^{-1}A) = -A^{-1}$.$IA - 4I = -A^{-1}$.$A - 4I = -A^{-1}$,which implies $A^{-1} = 4I - A$.$A^{-1} = \begin{bmatrix} 4 & 0 \ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -1 & 2 \end{bmatrix}$.

Show that the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ satisfies the equation $A^{2} - 4A + I = O$,where $I$ is the $2 \times 2$ identity matrix and $O$ is the $2 \times 2$ zero matrix. Using this equation,find $A^{-1}$.

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