Mathematics - Matrices Question with Solution | TestHub

Given,

$A=\left(\begin{array}{ll}2& -2\\ 1& -1\end{array}\right)$ So, $A^2=\left(\begin{array}{ll}2& -2\\ 1& -1\end{array}\right)\left(\begin{array}{ll}2& -2\\ 1& -1\end{array}\right)=\left(\begin{array}{ll}2& -2\\ 1& -1\end{array}\right)=A$

Also, $B=\left(\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right)$

$B^2=\left(\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right)\left(\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right)=B$

So, $A+B=\left(\begin{array}{ll}2& -2\\ 1& -1\end{array}\right)+\left(\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$

Therefore the equation $n{A}^n+m{B}^m=I$ is true for $n=1 \& m=1$ so, only one set is possible.