If A= [ ll 2 & 3 1 & 2 ], B= [ ll 1 & 0 3 & 1 ], find A B and (A B)^ -1

& A B= [ ll 2 & 3 1 & 2 ] [ ll 1 & 0 3 & 1 ] . = [ ll 2+9 & 0+3 1+6 & 0+2 ] = [ ll 11 & 3 7 & 2 ] & Now, |A B|= | ll 11 & 3 7 & 2 | =22-21 =1 0 (A B)^ -1 exists. Consider, (A B)(A B)^ -1 =1 [ ll 11 & 3 7 & 2 ](A B)^ -1 = [ ll 1 & 0 0 & 1 ] Applying R_1 2 R_1, [ cc 22 & 6 7 & 2 ](A B)^ -1 = [ ll 2 & 0 0 & 1 ] Applying R_1 R_1-3 R_2, & [ ll 1 & 0 7 & 2 ](A B)^ -1 = [ cc 2 & -3 0 & 1 ] Applying R_2 R_2-7 R_ 1^ [ ll 1 & 0 0 & 2 ](A B)^ -1 = [ cc 2 & -3 -14 & 22 ] Applying R_2 (1 2 ) R_2 & [ ll 1 & 0 0 & 1 ](A B)^ -1 = [ cc 2 & -3 -7 & 11 ] (A B)^ -1 = [ cc 2 & -3 -7 & 11 ] (1) |A|= | ll 2 & 3 1 & 2 |=4-3=1 0 A^ -1 exists. Consider, A A^ -1 =1 Applying R_2 3 R_1, & [ ll 1 & 0 0 & 1 ] A^ -1 = [ cc 1 & 0 -3 & 1 ] B^ -1 = [ cc 1 & 0 -3 & 1 ] B^ -1 A^ -1 = [ cc 1 & 0 -3 & 1 ], [ cc 2 & -3 -1 & 2 ] = [ cc 2-0 & -3+0 -6-1 & 9+2 ] = [ cc 2 & -3 -7 & 11 ] .(2) From (1) and (2), (A B)^ -1 =B^ -1 A^ -1

If A= [ ll 2 & 3 1 & 2 ], B= [ ll 1 & 0 3 & 1 ], find A B and (A …

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Question

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

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Answer

$\begin{aligned} & A B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right] .\end{aligned} $
$ =\left[\begin{array}{ll}2+9 & 0+3 \\ 1+6 & 0+2\end{array}\right] $
$ =\left[\begin{array}{ll}11 & 3 \\ 7 & 2\end{array}\right]$
$\begin{aligned} & \text { Now, }|A B|=\left|\begin{array}{ll}11 & 3 \\ 7 & 2\end{array}\right|\end{aligned} $
$ =22-21$
$ =1 \neq 0$
$ \therefore(A B)^{-1} \text { exists. }$
$ \text { Consider, }(A B)(A B)^{-1}=1$
$ \therefore\left[\begin{array}{ll} 11 & 3 \\ 7 & 2 \end{array}\right](A B)^{-1}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Applying $R_1 \rightarrow 2 R_1$,
$\left[\begin{array}{cc} 22 & 6 \\ 7 & 2 \end{array}\right](A B)^{-1}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right]$
Applying $R_1 \rightarrow R_1-3 R_2,$
$\begin{aligned} & {\left[\begin{array}{ll}1 & 0 \\ 7 & 2\end{array}\right](A B)^{-1}=\left[\begin{array}{cc}2 & -3 \\ 0 & 1\end{array}\right]} \end{aligned}$
Applying $R_2 \rightarrow R_2-7 R_{1^{\prime}} $
$ {\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right](A B)^{-1}=\left[\begin{array}{cc}2 & -3 \\ -14 & 22\end{array}\right]} $
Applying $R_2 \rightarrow\left(\frac{1}{2}\right) R_2$
$\begin{aligned} & {\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right](A B)^{-1}=\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]} \end{aligned} $
$ \therefore(A B)^{-1}=\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right] \ldots(1) $
$ |A|=\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=4-3=1 \neq 0 $
$ \therefore A^{-1} \text { exists. } $
$ \text { Consider, } A A^{-1}=1$
Applying $R_2 \leftrightarrow 3 R_1$,
$ \begin{aligned} & {\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A^{-1}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1\ \end{array}\right]} \end{aligned} $
$ \therefore B^{-1}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right] $
$ \therefore B^{-1} \cdot A^{-1}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right],\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right] $
$ =\left[\begin{array}{cc} 2-0 & -3+0 \\ -6-1 & 9+2 \end{array}\right] $
$ =\left[\begin{array}{cc} 2 & -3 \\ -7 & 11 \end{array}\right] \quad \ldots \ldots .(2) $
From $(1)$ and $(2), (A B)^{-1}=B^{-1} \cdot A^{-1}$