Question:
If $A=\left[\begin{array}{rrr}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]$, show that $A^{2}=/ 3$
Solution:
Here,
$A^{2}=A A$
$\Rightarrow A^{2}=\left[\begin{array}{ccc}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]\left[\begin{array}{ccc}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ccc}16-3-12 & -4+0+4 & -16+4+12 \\ 12+0-12 & -3+0+4 & -12+0+12 \\ 12-3-9 & -3+0+3 & -12+4+9\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore A^{2}=I_{3}$