If A= 0&c&-b -c&0&a &-a&0 and B= a^2&ab&ac &b^2&bc &bc&c^2 , show that AB = BA = O_ 3 3

Here, AB= 0&c&-b -c&0&a &-a&0 a^2&ab&ac &b^2&bc &bc&c^2 = 0+abc-abc&0+b^2c-b^2c&0+bc^2-bc^2 -a^2c+0+a^2c&-abc+0+abc&-ac^2+0+ac^2 ^2b-a^2b+0&ab^2-ab^2+0&abc-abc+0 = 0&0&0 0&0&0 0&0&0 =O_ 3 3 (1) BA= a^2&ab&ac &b^2&bc &bc&c^2 0&c&-b -c&0&a &-a&0 = 0-abc+abc&a^2c+0-a^2c&-a^2b+a^2b+0 0-b^2c+b^2c&abc+0-abc&-ab^2+ab^2+0 0-bc^2+bc^2&ac^2+0-ac^2&-abc+abc+0 = 0&0&0 0&0&0 0&0&0 =O_ 3 3 (2) =BA=0_ 3 3 [From eqs. (1) and (2)]

If A= 0&c&-b -c&0&a &-a&0 and B= a^2&ab&ac &b^2&bc &bc&c^2 , show…

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If $\text{A}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix},$ show that $AB = BA = O_{3 \times 3}$

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Answer

Here,
$\text{AB}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0+\text{abc}-\text{abc}&0+\text{b}^2\text{c}-\text{b}^2\text{c}&0+\text{bc}^2-\text{bc}^2\\-\text{a}^2\text{c}+0+\text{a}^2\text{c}&-\text{abc}+0+\text{abc}&-\text{ac}^2+0+\text{ac}^2\\\text{a}^2\text{b}-\text{a}^2\text{b}+0&\text{ab}^2-\text{ab}^2+0&\text{abc}-\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{AB}=\text{O}_{3\times3}\ \dots(1)$
$\text{BA}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0-\text{abc}+\text{abc}&\text{a}^2\text{c}+0-\text{a}^2\text{c}&-\text{a}^2\text{b}+\text{a}^2\text{b}+0\\0-\text{b}^2\text{c}+\text{b}^2\text{c}&\text{abc}+0-\text{abc}&-\text{ab}^2+\text{ab}^2+0\\0-\text{bc}^2+\text{bc}^2&\text{ac}^2+0-\text{ac}^2&-\text{abc}+\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{BA}=\text{O}_{3\times3}\ \dots(2)$
$ \Rightarrow\text{AB}=\text{BA}=0_{3\times3}$ [From eqs. (1) and (2)]

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