Let A=[0100], show that (aI+bA)^n=a^nI+nan−1 bA, where, I is the ide – InterviewSolution

1.	Let A=[0100], show that (aI+bA)^n=a^nI+nan−1 bA, where, I is the identity matrix of order 2 and n∈N.
Answer» Let $A = [\begin{matrix} 0 & 1 0 & 0 \end{matrix}]$ , show that (aI+bA)^n=a^nI+ $n a^{n - 1}$ bA, where, I is the identity matrix of order 2 and $n \in N .$

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