If A = begin{bmatrix} 0 & 1 0 & 0 end{bmatri | vedclass

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(B) Given $A = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$. Note that $A^2 = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \ 0

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$a^n I + n a^{n-1} b A$

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(B) Given $A = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$. Note that $A^2 = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = O$.We use the Binomial Theorem for matrices. Since $I$ and $A$ commute $(IA = AI = A)$,we have:$(aI + bA)^n = \sum_{k=0}^{n} \binom{n}{k} (aI)^{n-k} (bA)^k$.Since $A^k = O$ for all $k \ge 2$,only the terms for $k=0$ and $k=1$ are non-zero:$(aI + bA)^n = \binom{n}{0} (aI)^n (bA)^0 + \binom{n}{1} (aI)^{n-1} (bA)^1$.$(aI + bA)^n = 1 \cdot a^n I \cdot I + n \cdot a^{n-1} I \cdot bA$.$(aI + bA)^n = a^n I + n a^{n-1} b A$.

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,then $(aI + bA)^n$ is (where $I$ is the identity matrix of order $2$)

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