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First, we need to find $A^2$, which is $A \times A$. $$A^2 = \begin{bmatrix}2&3\\ -1&2\end{bmatrix} \begin{bmatrix}2&3\\ -1&2\end{bmatrix}$$ $$A^2 = \begin{bmatrix}(2\times2 + 3\times(-1))&(2\times3 + 3\times2)\\ (-1\times2 + 2\times(-1))&(-1\times3 + 2\times2)\end{bmatrix}$$ $$A^2 = \begin{bmatrix}(4 - 3)&(6 + 6)\\ (-2 - 2)&(-3 + 4)\end{bmatrix}$$ $$A^2 = \begin{bmatrix}1&12\\ -4&1\end{bmatrix}$$
Next, we calculate $4A$ by multiplying the matrix $A$ by the scalar 4. $$4A = 4\begin{bmatrix}2&3\\ -1&2\end{bmatrix} = \begin{bmatrix}8&12\\ -4&8\end{bmatrix}$$
Now, we calculate $7I$, where $I$ is the identity matrix of order 2. $$7I = 7\begin{bmatrix}1&0\\ 0&1\end{bmatrix} = \begin{bmatrix}7&0\\ 0&7\end{bmatrix}$$
Now, we substitute the calculated matrices into the expression $A^2 - 4A + 7I$. $$A^2 - 4A + 7I = \begin{bmatrix}1&12\\ -4&1\end{bmatrix} - \begin{bmatrix}8&12\\ -4&8\end{bmatrix} + \begin{bmatrix}7&0\\ 0&7\end{bmatrix}$$ $$A^2 - 4A + 7I = \begin{bmatrix}1-8+7&12-12+0\\ -4-(-4)+0&1-8+7\end{bmatrix}$$ $$A^2 - 4A + 7I = \begin{bmatrix}0&0\\ 0&0\end{bmatrix}$$ $$A^2 - 4A + 7I = 0$$
Final Answer: $A^{2}-4A+7I=0$
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