JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Matrices

Write $A=I+M$ with $M=\left[\begin{array}{cc}2& -4\\ 1& -2\end{array}\right]$. Then $M^2=\left[\begin{array}{cc}2& -4\\ 1& -2\end{array}\right]\left[\begin{array}{cc}2& -4\\ 1& -2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$, so $M^r=0$ for every $r\ge 2$. Because $I$ and $M$ commute, the binomial theorem gives $A^{100}=(I+M{)}^{100}=I+100M$, as all higher powers of $M$ vanish. Comparing with $A^{100}=100B+I$ yields $100B=100M$, so $B=M$. Since $B^2=M^2=0$, $B^{100}=0$, the null matrix. The sum of all the elements of $B^{100}$ is $0$.