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$A$ तथा $B$ आव्युहों के लिए सत्यापित कीजिए कि $( AB )^{\prime}= B ^{\prime} A ^{\prime},$ जहाँ
$A =\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right], B =\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$
Solution
$A B=\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{array}\right]$
$\therefore(A B)^{\prime}=\left[\begin{array}{ccc}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]$
Now, $A^{\prime}=\left[\begin{array}{lll}1 & -4 & 3\end{array}\right], B^{\prime}=\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right]$
$\therefore $ ${B^\prime }{A^\prime } = \left[ {\begin{array}{*{20}{l}}
{ – 1} \\
2 \\
1
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
1&{ – 4}&3
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
{ – 1}&4&{ – 3} \\
2&{ – 8}&6 \\
1&{ – 4}&3
\end{array}} \right]$
Hence, we have verified $(A B)^{\prime}=B^{\prime} A^{\prime}$.
