Question

If $A=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ , then $A^n+nI$ is equal to

• A. I
• B. nA
• C. I + nA
• D. I - nA
• E. nA - I

Answer 1

Answer: (C)

We have
$A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$
$\therefore A^2=A\cdot A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]$
$A^3=A^2\cdot A=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]$
$\therefore A^n=\left[\begin{array}{cc}1& 0\\ n& 1\end{array}\right]$
Now,
$A^n+nI=\left[\begin{array}{cc}1& 0\\ n& 1\end{array}\right]+n\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}1& 0\\ n& 1\end{array}\right]+\left[\begin{array}{cc}n& 0\\ 0& n\end{array}\right]$
$=\left[\begin{array}{cc}1+n& 0\\ n& 1+n\end{array}\right]$
Again, $I+nA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+n\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$
$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}n& 0\\ n& n\end{array}\right]$
$=\left[\begin{array}{cc}1+n& 0\\ n& 1+n\end{array}\right]$
$\therefore A^n+nI=I+nA$