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यदि $A =\left[\begin{array}{r}-2 \\ 4 \\ 5\end{array}\right], B =\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]$ है तो सत्यापित कीजिए $( AB )^{\prime}= B ^{\prime} A ^{\prime}$ है
Solution
We have
$A=\left[\begin{array}{r}-2 \\ 4 \\ 5\end{array}\right], B=\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]$
then $A B=\left[\begin{array}{c}-2 \\ 4 \\ 5\end{array}\right]\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]=\left[\begin{array}{ccc}-2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30\end{array}\right]$
Now ${A^\prime } = [ – 2\,\,\,4\,\,\,5],{B^\prime } = \left[ {\begin{array}{*{20}{r}}
1 \\
3 \\
{ – 6}
\end{array}} \right]$
$B^{\prime} A^{\prime}=\left[\begin{array}{r}1 \\ 3 \\ -6\end{array}\right]\left[\begin{array}{lll}-2 & 4 & 5\end{array}\right]$ $=\left[\begin{array}{rrr}-2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30\end{array}\right]=(\mathrm{AB})^{\prime}$
Clearly $(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}$