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

<p>Let $$A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]$$</p> <p>$|A|=1 \Rightarrow a c-b^2=0 \quad \text{... (i)}$</p> <p>$$\text { Given }\left[\begin{array}{ll} a & b \\ b & c \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 7 \end{array}\right]$$</p> <p>$$\begin{aligned} & \Rightarrow \quad a+b=3 \quad \text{... (ii)}\\ & \text { and } b+c=7 \quad \text{... (iii)} \end{aligned}$$</p> <p>from (i), (ii) and (iii) $a=1, b=2, c=5$</p> <p>$$\begin{aligned} \Rightarrow & A=\left[\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right] \Rightarrow A^{-1}=\left[\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right] \\ & \text { Given } A^{-1}=\alpha A+\beta I \\ \Rightarrow & {\left[\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right]=\alpha\left[\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right]+\beta\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] } \\ \Rightarrow & \alpha=-1 \text { and } \beta=6 \\ & \alpha+\beta=5 \end{aligned}$$</p>