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

Question

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

Aicharya · Accepted Answer

$A=\left[\begin{array}{lll}1 & 2 & 1 \ 2 & 1 & 0\end{array}ight]$ and $B=\left[\begin{array}{ll}1 & 2 \ 2 & 1 \ 0 & 1\end{array}ight]$ $	herefore A B$ $\begin{array}{l}(A B)^{-1} \ =\frac{1}{|A B|}\left[\begin{array}{cc}1+4+0 & 2+2+1 \ 2+2+0 & 4+1+0\end{array}ight]=\left[\begin{array}{ll}5 & 5 \ 4 & 5\end{array}ight]\end{array}$

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

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