Let M = begin{bmatrix} 1 & 1 0 & 1 end{bmatr | vedclass

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(A) Given $M = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$.Calculating powers of $M$:$M^2 = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmat

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$\begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$

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(A) Given $M = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$.Calculating powers of $M$:$M^2 = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$$M^3 = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}$By induction,$M^n = \begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix}$. Thus,$M^{10} = \begin{bmatrix} 1 & 10 \ 0 & 1 \end{bmatrix}$.Given $N = \begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix}$.Determinant $|N| = (1 	imes 2) - (0 	imes 0) = 2$.Inverse $N^{-1} = \frac{1}{|N|} 	ext{adj}(N) = \frac{1}{2} \begin{bmatrix} 2 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1/2 \end{bmatrix}$.Now,calculate $N M^{10} N^{-1}$:$N M^{10} = \begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 10 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 10 \ 0 & 2 \end{bmatrix}$.$(N M^{10}) N^{-1} = \begin{bmatrix} 1 & 10 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 0 & 1/2 \end{bmatrix} = \begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$.Therefore,the correct option is $A$.

Let $M = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $N = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$. Then $N M^{10} N^{-1} =$

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