MCQ
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....................

  • $5$
  • B
    $6$
  • C
    $7$
  • D
    $9$

Answer

Correct option: A.
$5$
a
$\begin{aligned} & \text { Let } A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right] \\ & {\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right], \mathrm{ad}-\mathrm{b}^2=1}\end{aligned}$

$ a+b=3, b+d=7,(3-b)(7-b)-b^2=1 $

$ 21-10 b=1 \rightarrow b=2, a=1, d=5$

$A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]$

$\mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$

$\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]$

$\alpha=-1, \beta=6 \rightarrow \alpha+\beta=5$

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