Question
Find $(A B)^{-1}$ if $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$
✓
Answer
$A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$
$A B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{cc}6 & -3 \\ -4 & 1\end{array}\right]$
$(A B)^{-1}(A B)=I$
$(A B)^{-1}\left[\begin{array}{cc}6 & -3 \\ -4 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Using $R_1 \rightarrow \frac{1}{6} R_1$
$(A B)^{-1}\left[\begin{array}{cc}1 & -\frac{1}{2} \\ -4 & 1\end{array}\right]=\left[\begin{array}{ll}\frac{1}{6} & 0 \\ 0 & 1\end{array}\right]$
Using $R_2 \rightarrow R_2+4 R_1$
$(A B)^{-1}\left[\begin{array}{ll}1 & -\frac{1}{2} \\ 0 & -1\end{array}\right]=\left[\begin{array}{ll}\frac{1}{6} & 0 \\ \frac{2}{3} & 1\end{array}\right]$
using $R_2 \rightarrow(-1) R_2$
$(A B)^{-1}\left[\begin{array}{cc}1 & -\frac{1}{2} \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6} & 0 \\ -\frac{2}{3} & -1\end{array}\right]$
Using $R_1 \rightarrow R_1+\left(\frac{1}{2}\right) R_2$
$(A B)^{-1}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}-\frac{1}{6} & -\frac{1}{2} \\ -\frac{2}{3} & -1\end{array}\right]$
$(A B)^{-1} I=\left[\begin{array}{ll}-\frac{1}{6} & -\frac{1}{2} \\ -\frac{2}{3} & -1\end{array}\right]$
$(A B)^{-1}=\left[\begin{array}{ll}-\frac{1}{6} & -\frac{1}{2} \\ -\frac{2}{3} & -1\end{array}\right]$
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