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  4. If A = beginpmatrix1&0 1&1 endpmatrix , then An + nI is equal to

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Q. If $A = \begin{pmatrix}1&0\\ 1&1\end{pmatrix}$ , then $A^n + nI$ is equal to

KEAMKEAM 2017Matrices

Solution:

We have
$A=\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$
$\therefore A^{2}=A \cdot A=\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}$
$A^{3}=A^{2} \cdot A=\begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 3 & 1\end{bmatrix}$
$\therefore A^{n}=\begin{bmatrix}1 & 0 \\ n & 1\end{bmatrix}$
Now,
$A^{n}+n I=\begin{bmatrix}1 & 0 \\ n & 1\end{bmatrix}+n\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$
$=\begin{bmatrix}1 & 0 \\ n & 1\end{bmatrix}+\begin{bmatrix}n & 0 \\ 0 & n\end{bmatrix}$
$=\begin{bmatrix}1+n & 0 \\ n & 1+n\end{bmatrix}$
Again, $I+n A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+n\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$
$=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+\begin{bmatrix}n & 0 \\ n & n\end{bmatrix}$
$=\begin{bmatrix}1+n & 0 \\ n & 1+n\end{bmatrix}$
$\therefore A^{n}+n I=I+ n A$