If and , show that AB = BA = 03×3.

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

If $A = [\begin{array}{ccc} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{array}]$ and $B = [\begin{array}{ccc} - 1 & 3 & 5 \\ 1 & - 3 & - 5 \\ - 1 & 3 & 5 \end{array}]$ , show that AB = BA = 0_3×3.

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L.H.S $= AB = {[\begin{array}{ccc} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{array}]}_{3 \times 3} {[\begin{array}{ccc} - 1 & 3 & 5 \\ 1 & - 3 & - 5 \\ - 1 & 3 & 5 \end{array}]}_{3 \times 3}$

$= {[\begin{array}{ccc} - 2 - 3 + 5 & 6 + 9 - 15 & 10 + 15 - 25 \\ 1 + 4 - 5 & - 3 - 12 + 15 & - 5 - 20 + 25 \\ - 1 - 3 + 4 & 3 + 9 - 12 & 5 + 15 - 20 \end{array}]}_{3 \times 3}$

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

R.H.S $= BA = {[\begin{array}{ccc} - 1 & 3 & 5 \\ 1 & - 3 & - 5 \\ - 1 & 3 & 5 \end{array}]}_{3 \times 3} {[\begin{array}{ccc} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{array}]}_{3 \times 3}$

$= {[\begin{array}{ccc} - 2 - 3 + 5 & 3 + 12 - 15 & 5 + 15 - 20 \\ 2 + 3 - 5 & - 3 - 12 + 15 & - 5 - 15 + 20 \\ - 2 - 3 + 5 & 3 + 12 - 15 & 5 + 15 - 20 \end{array}]}_{3 \times 3}$

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

Hence,L.H.S = R.H.S = 0_3×3.

Hence proved.

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