If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$, then the value of $(A+I)^3-4 A$ is
Answer & explanation
Correct answer: option 2
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 $(A+I)^3-4A$
$=(A+I)^2+(A+I)-4A$
$(A^2+I^2+AI+IA)(A+I)-4A$
$2(A+I)^2-4A$
$4(A+I)-4A=4I$