JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Matrices

Compute $A^2=\left(\begin{array}{cc}2& -2\\ 1& -1\end{array}\right)\left(\begin{array}{cc}2& -2\\ 1& -1\end{array}\right)=\left(\begin{array}{cc}2& -2\\ 1& -1\end{array}\right)=A$. Similarly, $B^2=B$. So $A^n=A$ and $B^m=B$ for all $n,m\ge 1$. The equation becomes $nA+mB=I$. Substituting the matrices: $n\left(\begin{array}{cc}2& -2\\ 1& -1\end{array}\right)+m\left(\begin{array}{cc}-1& 2\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$. This gives $2n-m=1$, $-2n+2m=0$, $n-m=0$, $-n+2m=1$. From the third equation, $n=m$. Substituting into the first: $2n-n=1$, so $n=1$. Thus $(n,m)=(1,1)$ is the only solution, giving 1 element.