Question

If matrix $\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]$, then prove that $A^{-1}=A^3$

Answer 1

Given,

$\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]=\left|\begin{array}{cc}1& -1\\ 2& -1\end{array}\right|$

$=-1+2$

$|A|=1\ne 0$

So, $A^{-1}$ exists.

On finding adjoint of matrix $A$,

$a_{11}=-1,a_{12}=-2,a_{21}=1,a_{22}=1$

Matrix formed by adjoint of $A$,

$B=\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]Now,adj.A=B^T={\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]}^{T}adj.A=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]\therefore A^{-1}=\frac{adj.A}{|A|}=\frac{1}{1}\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]So,A^{-1}=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]...\left(i\right)Now,A^3=A\times A\times A=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right].\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right].\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]=\left[\begin{array}{cc}1-2& -1+1\\ 2-2& -2+1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]=\left[\begin{array}{cc}-1+0& 1+0\\ 0-2& 0+1\end{array}\right]=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]So,A^3=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]..\left(ii\right)$

From$\left(i\right)$ and $\left(ii\right)$,

$A^{-1}=A^3$

Hence proved.