If $A=\left[\begin{array}{rrr}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]$, show that $A B=B A=O_{3 \times 3}$.
Here,
$A B=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc}-2-3+5 & 6+9-15 & 10+15-25 \\ 1+4-5 & -3-12+15 & -5-20+25 \\ -1-3+4 & 3+9-12 & 5+15-20\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$\Rightarrow A B=0_{3 \times 3} \quad \ldots(1)$
$B A=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{ccc}-2-3+5 & 3+12-15 & 5+15-20 \\ 2+3-5 & -3-12+15 & -5-15+20 \\ -2-3+5 & 3+12-15 & 5+15-20\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$\Rightarrow B A=0_{3 \times 3} \quad \ldots(2)$
$\Rightarrow A B=B A=0_{3 \times 3}$ [From eqs. (1) and (2)]
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