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(A) Let $A = \begin{bmatrix} a & b \ b & d \end{bmatrix}$.Given $A \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 7 \end{bmatrix}$,we ha

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$5$

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(A) Let $A = \begin{bmatrix} a & b \ b & d \end{bmatrix}$.Given $A \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 7 \end{bmatrix}$,we have $a + b = 3$ and $b + d = 7$.Thus,$a = 3 - b$ and $d = 7 - b$.The determinant $|A| = ad - b^2 = 1$.Substituting the values,$(3 - b)(7 - b) - b^2 = 1$.$21 - 3b - 7b + b^2 - b^2 = 1 \implies 21 - 10b = 1 \implies 10b = 20 \implies b = 2$.Then $a = 3 - 2 = 1$ and $d = 7 - 2 = 5$.So,$A = \begin{bmatrix} 1 & 2 \ 2 & 5 \end{bmatrix}$.The inverse $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \begin{bmatrix} 5 & -2 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \ -2 & 1 \end{bmatrix}$.Given $A^{-1} = \alpha A + \beta I$,we have $\begin{bmatrix} 5 & -2 \ -2 & 1 \end{bmatrix} = \alpha \begin{bmatrix} 1 & 2 \ 2 & 5 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\alpha \ 2\alpha & 5\alpha + \beta \end{bmatrix}$.Comparing elements,$2\alpha = -2 \implies \alpha = -1$.Substituting $\alpha = -1$ into $\alpha + \beta = 5$,we get $-1 + \beta = 5 \implies \beta = 6$.Therefore,$\alpha + \beta = -1 + 6 = 5$.

Let $A$ be a $2 \times 2$ symmetric matrix such that $A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$ and the determinant of $A$ is $1$. If $A^{-1} = \alpha A + \beta I$,where $I$ is an identity matrix of order $2 \times 2$,then $\alpha + \beta$ equals:

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