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Question
If A = `[[0,c,-b],[-c,0,a],[b,-a,0]]`and B =`[[a^2 ,ab,ac],[ab,b^2,bc],[ac,bc,c^2]]`, show that AB = BA = O3×3.
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Solution
\[Here, \]
\[AB = \begin{bmatrix}0 & c & - b \\ - c & 0 & a \\ b & - a & 0\end{bmatrix}\begin{bmatrix}a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}0 + abc - abc & 0 + b^2 c - b^2 c & 0 + b c^2 - b c^2 \\ - a^2 c + 0 + a^2 c & - abc + 0 + abc & - a c^2 + 0 + a c^2 \\ a^2 b - a^2 b + 0 & a b^2 - a b^2 + 0 & abc - abc + 0\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\[ \Rightarrow AB = O_{3 \times 3} . . . \left( 1 \right)\]
\[\]
\[BA = \begin{bmatrix}a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2\end{bmatrix}\begin{bmatrix}0 & c & - b \\ - c & 0 & a \\ b & - a & 0\end{bmatrix}\]
\[ \Rightarrow BA = \begin{bmatrix}0 - abc + abc & a^2 c + 0 - a^2 c & - a^2 b + a^2 b + 0 \\ 0 - b^2 c + b^2 c & abc + 0 - abc & - a b^2 + a b^2 + 0 \\ 0 - b c^2 + b c^2 & a c^2 + 0 - a c^2 & - abc + abc + 0\end{bmatrix}\]
\[ \Rightarrow BA = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\]
\[ \Rightarrow BA = O_{3 \times 3} . . . \left( 2 \right)\]
\[ \]
`\ \Rightarrow AB=BA = O_{3 \times 3} ` [From eqs. (1) and (2)]
