If matrix $A = [a_{ij}]_{2 \times 2}$, where $a_{ij} = 1$, if $i \neq j$ and $a_{ij} = 0$ if $i = j$, then $A^2$ is equal to

$I$

The correct answer is Option (1) → $I$ ##

We have, $A = [a_{ij}]_{2 \times 2}$, where $a_{ij} = 1$ if $i \neq j$ and $a_{ij} = 0$ if $i = j$

Let $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$

So, $a_{11} = 0$ [since, $i=j$], $a_{12} = 1$ [since, $i \neq j$], $a_{21} = 1$ [since, $i \neq j$], $a_{22} = 0$ [since, $i=j$]

$∴A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

and $A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$