Question
If $A=\left[\begin{array}{c}-2 \\ 4 \\ 5\end{array}\right], B=\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]$; Verify that (AB)' = $B^{\prime} A^{\prime}$
✓
Answer
$AB = \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 6}&{12} \\ 4&{12}&{ - 24} \\ 5&{15}&{ - 30} \end{array}} \right]$
$\left( {AB} \right)' = \left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \\ { - 6}&{12}&{15} \\ {12}&{ - 24}&{ - 30} \end{array}} \right]$
$A' = \left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \end{array}} \right]$
$B' = \left[ {\begin{array}{*{20}{c}} 1 \\ 3 \\ { - 6} \end{array}} \right]$
$B'A' = \left[ {\begin{array}{*{20}{c}} 1 \\ 3 \\ { - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \\ { - 6}&{12}&{15} \\ {12}&{ - 24}&{ - 30} \end{array}} \right]$
Hence (AB)' = B'. A'
$\left( {AB} \right)' = \left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \\ { - 6}&{12}&{15} \\ {12}&{ - 24}&{ - 30} \end{array}} \right]$
$A' = \left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \end{array}} \right]$
$B' = \left[ {\begin{array}{*{20}{c}} 1 \\ 3 \\ { - 6} \end{array}} \right]$
$B'A' = \left[ {\begin{array}{*{20}{c}} 1 \\ 3 \\ { - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} { - 2}&4&5 \\ { - 6}&{12}&{15} \\ {12}&{ - 24}&{ - 30} \end{array}} \right]$
Hence (AB)' = B'. A'
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