If $A=\left[\begin{array}{ll}2 & 3 \ 1 & 2\end{array} ight], B=\left[\begin{array}{ll}1 & 0 \ 3 & 1\end{array} ight]$, then $(A B)^{-1}=$

Question

If $A=\left[\begin{array}{ll}2 & 3 \ 1 & 2\end{array}ight], B=\left[\begin{array}{ll}1 & 0 \ 3 & 1\end{array}ight]$, then $(A B)^{-1}=$

Aicharya · Accepted Answer

$\begin{aligned} A B &=\left[\begin{array}{ll}2 & 3 \ 1 & 2\end{array}ight]\left[\begin{array}{ll}1 & 0 \ 3 & 1\end{array}ight]=\left[\begin{array}{cc}11 & 3 \ 7 & 2\end{array}ight] \Rightarrow|A B|=\left|\begin{array}{cc}11 & 3 \ 7 & 2\end{array}ight|=22-21=1 \ \operatorname{adj}(A B) &=\left[\begin{array}{cc}2 & -3 \ -7 & 11\end{array}ight] \Rightarrow(A B)^{-1}=\left[\begin{array}{cc}2 & -3 \ -7 & 11\end{array}ight] \end{aligned}$

If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 &…

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