Solve this
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).

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