Solution:
Take 
![\begin{array}{l} C=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right]\left[\begin{array}{lll} -2 & -1 & -4 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{array}\right] \\ L H S=C^{\prime} \\ \Rightarrow\left[\begin{array}{ccc} 2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12 \end{array}\right] \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31926ff823faa25526f01986fedf1443_l3.png "Rendered by QuickLaTeX.com")
To find RHS we will find transpese of matrix 
![\mathrm{A}^{\prime}=\left[\begin{array}{lll}-1 & 2 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c5aa67677871d508165746e70e791494_l3.png "Rendered by QuickLaTeX.com")
![\mathrm{B}^{*}=\left[\begin{array}{c}-2 \\ -1 \\ -4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f117bc25f843b7a8db5346ae2ef6e0c4_l3.png "Rendered by QuickLaTeX.com")
RHS 
![\Rightarrow\left[\begin{array}{l}-2 \\ -1 \\ -4\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e6f3eaf0e2b11ca58c87521da4402260_l3.png "Rendered by QuickLaTeX.com")
![\Rightarrow\left[\begin{array}{ccc}2 & -4 & -6 \\ 1 & -2 & -3 \\ 4 & -8 & -12\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-80bd9a6e772ef89e1562f04c30797420_l3.png "Rendered by QuickLaTeX.com")
LHS = RHS.
As a result, hence proved.
For each of the following pairs of matrices 
![\mathrm{A}=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right] \text { and } \mathrm{B}=[-2 \cdot 1-4]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-86cfd44f3d05fdc26f7ec6f9a464ad20_l3.png "Rendered by QuickLaTeX.com")
Solution:
Take
To find RHS we will find transpese of matrix
RHS
LHS = RHS.
As a result, hence proved.
