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Question: Answered & Verified by Expert
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 _______
MathematicsMatricesJEE MainJEE Main 2024 (04 Apr Shift 2)
Solution:
1117 Upvotes Verified Answer
The correct answer is: 5
$\begin{aligned} & \text { Let } A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right] \\ & {\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right], a d-b^2=1} \\ & a+b=3, b+d=7,(3-b)(7-b)-b^2=1 \\ & 21-10 b=1 \rightarrow b=2, a=1, d=5 \\ & A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right] \\ & A^{-1}=\alpha A+\beta I \\ & {\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]} \\ & \alpha=-1, \beta=6 \rightarrow \alpha+\beta=5\end{aligned}$

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