Question

If A = $\left[\begin{array}{ccc}0& c& -b\\ -c& 0& a\\ b& -a& 0\end{array}\right]$ and B = $\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]$, show that AB = BA = O_3×3.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ AB=\left[\begin{array}{ccc}0& c& -b\\ -c& 0& a\\ b& -a& 0\end{array}\right]\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}0+abc-abc& 0+b^{2}c-b^{2}c& 0+b{c}^2-b{c}^2\\ -a^{2}c+0+a^{2}c& -abc+0+abc& -a{c}^2+0+a{c}^2\\ a^{2}b-a^{2}b+0& a{b}^2-a{b}^2+0& abc-abc+0\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right] \\ \Rightarrow AB=O_{3\times 3}...\left(1\right) \\ BA=\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]\left[\begin{array}{ccc}0& c& -b\\ -c& 0& a\\ b& -a& 0\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}0-abc+abc& a^{2}c+0-a^{2}c& -a^{2}b+a^{2}b+0\\ 0-b^{2}c+b^{2}c& abc+0-abc& -a{b}^2+a{b}^2+0\\ 0-b{c}^2+b{c}^2& a{c}^2+0-a{c}^2& -abc+abc+0\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right] \\ \Rightarrow BA=O_{3\times 3}...\left(2\right) \\ \Rightarrow AB=BA=O_{3\times 3}\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right] \end{gathered}$$