If $A=\left[\begin{array}{rrr}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]$ and $B=$ $\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]$, show that $A B=B A=O_{3 \times 3}$
Here,
$A B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc}0+a b c-a b c & 0+b^{2} c-b^{2} c & 0+b c^{2}-b c^{2} \\ -a^{2} c+0+a^{2} c & -a b c+0+a b c & -a c^{2}+0+a c^{2} \\ a^{2} b-a^{2} b+0 & a b^{2}-a b^{2}+0 & a b c-a b c+0\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$\Rightarrow A B=O_{3 \times 3} \quad \ldots(1)$
$B A=\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{lll}0-a b c+a b c & a^{2} c+0-a^{2} c & -a^{2} b+a^{2} b+0 \\ 0-b^{2} c+b^{2} c & a b c+0-a b c & -a b^{2}+a b^{2}+0 \\ 0-b c^{2}+b c^{2} & a c^{2}+0-a c^{2} & -a b c+a b c+0\end{array}\right]$
$\Rightarrow B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$\Rightarrow B A=O_{3 \times 3}$
$\Rightarrow A B=B A=O_{3 \times 3}$ [From eqs. (1) and (2)]