Let
Question:

Let $A=\left[\begin{array}{rrr}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]$ and $B=\left[\begin{array}{rrr}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]$, compute $A^{2}-B^{2}$.

Solution:

Given : $A=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}1+3-5 & -1-3-5 & 1+3-5 \\ -3-9+15 & 3+9+15 & -3-9+15 \\ -5+15+25 & 5-15+25 & -5+15+25\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}-1 & -9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35\end{array}\right]$

$B^{2}=B B$

$\Rightarrow B^{2}=\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]\left[\begin{array}{ccc}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]$

$\Rightarrow B^{2}=\left[\begin{array}{ccc}0+4-3 & 0-12+12 & 0-12+12 \\ 0-3+3 & 4+9-12 & 3+9-12 \\ 0+4-4 & -4-12+16 & -3-12+16\end{array}\right]$

$\Rightarrow B^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$A^{2}-B^{2}$

$\Rightarrow A^{2}-B^{2}=\left[\begin{array}{ccc}-1 & -9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35\end{array}\right]-\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-B^{2}=\left[\begin{array}{ccc}-1-1 & -9-0 & -1-0 \\ 3-0 & 27-1 & 3-0 \\ 35-0 & 15-0 & 35-1\end{array}\right]$

$\Rightarrow A^{2}-B^{2}=\left[\begin{array}{ccc}-2 & -9 & -1 \\ 3 & 26 & 3 \\ 35 & 15 & 34\end{array}\right]$