For M=left[begin{array}{ll}3 & -4 1 & -1end{ | vedclass

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(B) Given $M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight]$.First,calculate $M^2 = M 	imes M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\

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$\left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$

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(B) Given $M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight]$.First,calculate $M^2 = M 	imes M = \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight] \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight] = \left[\begin{array}{ll}9-4 & -12+4 \ 3-1 & -4+1\end{array}ight] = \left[\begin{array}{ll}5 & -8 \ 2 & -3\end{array}ight]$.Now,consider the expression $M^{n+1}-M^n = M^n(M-I)$.For $n=1$,$M^2-M = \left[\begin{array}{ll}5 & -8 \ 2 & -3\end{array}ight] - \left[\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight] = \left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$.Note that $M^2 - M = \left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$.Since $M^2 - 2M + I = 0$ (characteristic equation $\lambda^2 - 2\lambda + 1 = 0$),we have $M^2 = 2M - I$,which implies $M^2 - M = M - I$.Thus,$M^{n+1}-M^n = M^{n-1}(M^2-M) = M^{n-1}(M-I) = M^n - M^{n-1}$.By induction,$M^{n+1}-M^n = M^2-M = \left[\begin{array}{ll}2 & -4 \ 1 & -2\end{array}ight]$ for all $n \in N$.

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$

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