Question

If $A=\left[\begin{array}{cc}2& 3\\ 1& -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]$, then verify that $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

Answer 1

$A=\left[\begin{array}{cc}2& 3\\ 1& -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]$
$AB=\left[\begin{array}{cc}2& 3\\ 1& -4\end{array}\right]\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]$
$\Rightarrow AB=\left[\begin{array}{cc}2\left(1\right)+3(-1)& 2(-2)+3\left(3\right)\\ 1\left(1\right)+(-4)(-1)& 1(-2)+(-4)\left(3\right)\end{array}\right]$
$\Rightarrow AB=\left[\begin{array}{cc}2-3& -4+9\\ 1+4& -2-12\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 5& -14\end{array}\right]$
$\left|\begin{array}{c}AB\end{array}\right|=\left|\begin{array}{cc}-1& 5\\ 5& -14\end{array}\right|=(-1)(-14)-5\left(5\right)$
$\Rightarrow \left|\begin{array}{c}AB\end{array}\right|=14-25=-11$
$\left|\begin{array}{c}A\end{array}\right|=\left|\begin{array}{cc}2& 3\\ 1& -4\end{array}\right|=2(-4)-1\left(3\right)=-11$
$\left|\begin{array}{c}B\end{array}\right|=\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|=3-2=1$
Calculating $B^{-1}$
$B=\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]$
$adj\left(B\right)=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$
$B^{-1}=\frac{1}{\left|\begin{array}{c}B\end{array}\right|}adj\left(B\right)$
$\Rightarrow B^{-1}=\frac{1}{1}\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$
Calculating $A^{-1}$
$A=\left[\begin{array}{cc}2& 3\\ 1& -4\end{array}\right]$
$adj\left(A\right)=\left[\begin{array}{cc}-4& -3\\ -1& 2\end{array}\right]$
$A^{-1}=\frac{1}{\left|\begin{array}{c}A\end{array}\right|}adj\left(A\right)$
$\Rightarrow A^{-1}=\frac{1}{-11}\left[\begin{array}{cc}-4& -3\\ -1& 2\end{array}\right]$
$AB=\left[\begin{array}{cc}-1& 5\\ 5& -14\end{array}\right]$
$adj\left(AB\right)=\left[\begin{array}{cc}-14& -5\\ -5& -1\end{array}\right]$
Now, let's prove $(AB{)}^{-1}=B^{-1}{A}^{-1}$
Taking $L.H.S.$
$(AB{)}^{-1}=\frac{1}{\left|\begin{array}{c}AB\end{array}\right|}adj(AB)$
$\Rightarrow (AB{)}^{-1}=\frac{1}{(-11)}\left[\begin{array}{cc}-14& -5\\ -5& -1\end{array}\right]$
$\Rightarrow (AB{)}^{-1}=\frac{1}{11}\left[\begin{array}{cc}14& 5\\ 5& 1\end{array}\right]$
Taing R.H.S.
$B^{-1}{A}^{-1}$
$B^{-1}{A}^{-1}=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]\times \frac{1}{11}\left[\begin{array}{cc}4& 3\\ 1& -2\end{array}\right]$
$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}4& 3\\ 1& -2\end{array}\right]$
$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}3\left(4\right)+2\left(1\right)& 3\left(3\right)+2(-2)\\ 1\left(4\right)+1\left(1\right)& 1\left(3\right)+1(-2)\end{array}\right]$
$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}12+2& 9-4\\ 4+1& 3-2\end{array}\right]$
$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}14& 5\\ 5& 1\end{array}\right]$
$\therefore L.H.S.=R.H.S.$
Hence verified.
Therefore, its verified that $(AB{)}^{-1}=B^{-1}{A}^{-1}$