Let $M$ and $N$ be square matrices of the same order satisfying $MN = M$ and $NM = N$. Then $(M^{2024} + N^{2024})^{2025}$ is equal to
Step-by-Step Solution
Key Concept: Derive $M^2 = M$ and $N^2 = N$ (both are idempotent). Then compute powers of $M$ and $N$, and use the binomial theorem on the sum.
$M^2=M$, $N^2=N$, $MN+NM=M+N$. $(M+N)^2=2(M+N) \Rightarrow (M+N)^{2025}=2^{2024}(M+N)$. Also $M^{2024}=M$, $N^{2024}=N$.
Correct Answer: (C) $2^{2024}(M+N)$