Question:
If matrix $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\left\{\begin{array}{ll}1, & \text { if } i \neq j \\ 0, & \text { if } i=j\end{array}\right.$, then $A^{2}$ is equal to
(a) $I$
(b) $A$
(c) 0
(d) $-1$
Solution:
Given: $a_{i j}= \begin{cases}1, & \text { if } i \neq j \\ 0, & \text { if } i=j\end{cases}$
$\therefore a_{11}=0$
$a_{12}=1$
$a_{21}=1$
$a_{22}=0$
Therefore, matrix $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$=\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=I$
Hence, the correct option is (a).