If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$, then the value of $(A+I)^3-4 A$ is

$4 I$

The correct answer is Option (2) - $4 I$

$A^2=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=I$

so $(A+I)^3-4A$

$=(A+I)^2+(A+I)-4A$

$(A^2+I^2+AI+IA)(A+I)-4A$

$2(A+I)^2-4A$

$4(A+I)-4A=4I$