If $A=\left[\begin{array}{rr}1 & 2 \\ 3 & -5\end{array}\right]$, then $A^{-1}$ is equal to
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If $A=\left[\begin{array}{rr}1 & 2 \\ 3 & -5\end{array}\right]$, then $A^{-1}$ is equal to .

(A) $\left[\begin{array}{rr}-5 & -2 \\ -3 & 1\end{array}\right]$

(B) $\left[\begin{array}{cc}\frac{5}{11} & \frac{2}{11} \\ \frac{3}{11} & -\frac{1}{11}\end{array}\right]$

(C) $\left[\begin{array}{rr}-\frac{5}{11} & -\frac{2}{11} \\ -\frac{3}{11} & -\frac{1}{11}\end{array}\right]$

(D) $\left[\begin{array}{rr}5 & 2 \\ 3 & -1\end{array}\right]$

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SOLUTION —

$\therefore|A|=\left|\begin{array}{rr}1 & 2 \\ 3 & -5\end{array}\right|=-11$  $\operatorname{adj}(A)=\left|\begin{array}{rr}-5 & -2 \\ -3 & 1\end{array}\right|$

$\therefore \quad A^{-1}=\frac{1}{-11}\left|\begin{array}{rr}-5 & -2 \\-3 & 1\end{array}\right|=\left|\begin{array}{rr}5 / 11 & 2 / 11 \\3 / 11 & -1 / 11\end{array}\right|$

So, The correct option will be (B).

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