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Let $A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]$ and $B^{-1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]$. If $\left(A B^{-1}\right)^{-1}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then $2 b+5 c+10 d=$
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$1$
$A B^{-1}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}1 & 5 \\ -2 & 0\end{array}\right]$
Now, $\left|A B^{-1}\right|=0-(-10)=10$
and $\operatorname{adj}\left(A B^{-1}\right)=\left[\begin{array}{cc}0 & -5 \\ 2 & 1\end{array}\right]$
$\because \quad\left(A B^{-1}\right)^{-1}=\frac{1}{\left|A B^{-1}\right|} \operatorname{adj}\left(A B^{-1}\right)$
$\therefore \quad\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\frac{1}{10}\left[\begin{array}{cc}0 & -5 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}0 & -\frac{1}{2} \\ \frac{1}{5} & \frac{1}{10}\end{array}\right]$
$\begin{aligned} & \Rightarrow \quad a=0, b=-\frac{1}{2}, c=\frac{1}{5}, d=\frac{1}{10} \\ & \therefore \quad 2 b+5 c+10 d=-1+1+1=1\end{aligned}$
Now, $\left|A B^{-1}\right|=0-(-10)=10$
and $\operatorname{adj}\left(A B^{-1}\right)=\left[\begin{array}{cc}0 & -5 \\ 2 & 1\end{array}\right]$
$\because \quad\left(A B^{-1}\right)^{-1}=\frac{1}{\left|A B^{-1}\right|} \operatorname{adj}\left(A B^{-1}\right)$
$\therefore \quad\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\frac{1}{10}\left[\begin{array}{cc}0 & -5 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}0 & -\frac{1}{2} \\ \frac{1}{5} & \frac{1}{10}\end{array}\right]$
$\begin{aligned} & \Rightarrow \quad a=0, b=-\frac{1}{2}, c=\frac{1}{5}, d=\frac{1}{10} \\ & \therefore \quad 2 b+5 c+10 d=-1+1+1=1\end{aligned}$
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