Question

If A = $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ , then det $(A+A^2+A^3+A^4+A^5)$ is

• A. $1$
• B. $32$
• C. $25$
• D. $30$

Answer 1

The correct option is C $25$

$A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 1+1\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

$A^3=A\times A^2=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 1+2\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

$A^4=A^2\times A^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 2+2\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]$

$A^5=A^3\times A^2=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 3+2\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}\right]$

$\therefore A+A^2+A^3+A^4+A^5$

$=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}5& 15\\ 0& 5\end{array}\right]$

$=5\times 5-0\times 15=25$