Let $M=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}\right]$ and $N=\left[\begin{array}{ll}1 & 0 \ 0 & 2\end{array}\right]$. Then $N M^{10} N^{-1}=$

Question

Let $M=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}\right]$ and $N=\left[\begin{array}{ll}1 & 0 \ 0 & 2\end{array}\right]$. Then $N M^{10} N^{-1}=$, $\left[\begin{array}{ll}1 & 5 \ 0 & 1\end{array}\right]$, $\left[\begin{array}{cc}1 & -5 \ 0 & 1\end{array}\right]$, $\left[\begin{array}{cc}1 & -10 \ 0 & 1\end{array}\right]$, $\left[\begin{array}{cc}1 & 10 \ 0 & 1\end{array}\right]$

MARKS App · Accepted Answer

$\begin{aligned} M & =\left[\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}\right] \ M^2 & =\left[\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}\right] \ M^2 & =\left[\begin{array}{ll} 1 & 2 \ 0 & 1 \end{array}\right] \ M^3 & =\left[\begin{array}{ll} 1 & 3 \ 0 & 1 \end{array}\right] \ & \vdots \ M^{10} & =\left[\begin{array}{ll} 1 & 10 \ 0 & 1 \end{array}\right] \ N & =\left[\begin{array}{ll} 1 & 0 \ 0 & 2 \end{array}\right] \ |N| & =2-0=2 \ \operatorname{Adj}^N & =\left[\begin{array}{ll} 2 & 0 \ 0 & 1 \end{array}\right] \ N^{-1} & =\frac{1}{|N|} \text { Adj A } \ N^{-1} & =\frac{1}{2}\left[\begin{array}{ll} 2 & 0 \ 0 & 1 \end{array}\right] \ N^{-1} & =\left[\begin{array}{ll} 1 & 0 \ 0 & \frac{1}{2} \end{array}\right] \end{aligned}$ Consider, $\begin{aligned} \mathbf{N} \cdot \mathbf{M}^{10} \cdot \mathbf{N}^{-1} & =\left[\begin{array}{ll} 1 & 0 \ 0 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 10 \ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \ 0 & \frac{1}{2} \end{array}\right] \ & =\left[\begin{array}{cc} 1 & 10 \ 0 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \ 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ll} 1 & 5 \ 0 & 1 \end{array}\right] \end{aligned}$ Hence, option (a) is correct.

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