Question

Let A denote the matrix $\begin{bmatrix}0&i\\ i&0\end{bmatrix}$, where $i^{2}=-1$, and let I denote the identity matrix $\begin{bmatrix}1&0\\ 0&1\end{bmatrix}$ Then, $I+A+A^{2}+\dots+A^{2010}$ is

• A. $\begin{bmatrix}0&0\\ 0&0\end{bmatrix}$
• B. $\begin{bmatrix}0&i\\ i&0\end{bmatrix}$
• C. $\begin{bmatrix}1&i\\ i&1\end{bmatrix}$
• D. $\begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}$

Answer 1

Answer: (B)

We have,
$A=\begin{bmatrix}0&i\\ i&0\end{bmatrix}$,
$A^{2}=\begin{bmatrix}0&i\\ i&0\end{bmatrix}\begin{bmatrix}0&i\\ i&0\end{bmatrix}$
$=\begin{bmatrix}i^{2}&0\\ 0&i^{2}\end{bmatrix}=\begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}$
$=-\begin{bmatrix}1&0\\ 0&1\end{bmatrix}=-I$
$A^{3}=A^{2}\cdot A =\begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}0&i\\ i&0\end{bmatrix}$
$=\begin{bmatrix}0&-i\\ -i&0\end{bmatrix}$
$=-\begin{bmatrix}0&i\\ i&0\end{bmatrix}=-A $
and $ A^{4}=A^{2}\cdot A^{2}=\left(-I\right)\left(-I\right)=I=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}$
$\therefore I+A+A^{2}+A^{3}+A^{4}+A^{5}+\ldots+A^{2008}+A^{2009}+A^{2010}$
$=I+A+A^{2}+A^{3}+A^{4}\left[I+A+A^{2}+A^{3}\right]+\ldots+A^{2008}\left[I+A+A^{2}\right]$
$=0+0+\ldots+\left[I+A+A^{2}\right]$
$\left[\because I+A+A^{2}+A^{3}=0\right]$
$\therefore I+A+A^{2}=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}+\begin{bmatrix}0&i\\ i&0\end{bmatrix}+\begin{bmatrix}-1&0\\ 0&-1\end{bmatrix} $
$=\begin{bmatrix}0&i\\ i&0\end{bmatrix}$