Let $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ , show that $(aI + bA)^n = a^nI+na^{n-1}bA,$ where I is the identity matrix of order zero $n \in N.$ If $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ prove that : $\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}, n \in N.$

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