If and , show that AB = BA = 03×3.

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

If $A = [\begin{array}{ccc} 0 & c & - b \\ - c & 0 & a \\ b & - a & 0 \end{array}]$ and $B = [\begin{array}{ccc} a^{2} & ab & ac \\ ab & b^{2} & bc \\ ac & bc & c^{2} \end{array}]$ , show that AB = BA = 0_3×3.

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$AB = [\begin{array}{ccc} 0 & c & - b \\ - c & 0 & a \\ b & - a & 0 \end{array}] [\begin{array}{ccc} a^{2} & ab & ac \\ ab & b^{2} & bc \\ ac & bc & c^{2} \end{array}]$

$= {[\begin{array}{ccc} 0 + abc - abc & 0 + b^{2} c - b^{2} c & 0 + {bc}^{2} - {bc}^{2} \\ - {ca}^{2} + 0 + {ca}^{2} & - abc + 0 + abc & - {ac}^{2} + 0 + {ac}^{2} \\ {ba}^{2} - a^{2} b + 0 & {ab}^{2} - {ab}^{2} + 0 & bac - abc + 0 \end{array}]}_{3 \times 3}$

$= {[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]}_{3 \times 3} = 0_{3 \times 3}$

$BA = [\begin{array}{ccc} a^{2} & ab & ac \\ ab & b^{2} & bc \\ ac & bc & c^{2} \end{array}] [\begin{array}{ccc} 0 & c & - b \\ - c & 0 & a \\ b & - a & 0 \end{array}]$

$BA = [\begin{array}{ccc} 0 - abc + abc & a^{2} c + 0 - a^{2} c & - a^{2} b + a^{2} b + 0 \\ 0 - b^{2} c + b^{2} c & abc + 0 - abc & - {ab}^{2} + 0 + {ab}^{2} \\ 0 - {bc}^{2} + {bc}^{2} & {ac}^{2} + 0 - {ac}^{2} & - abc + abc + 0 \end{array}]$

$= {[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]}_{3 \times 3} = 0_{3 \times 3}$

Hence AB = Ba = 0_3×3 is proved.

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