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Question:

If $A=\left[\begin{array}{r}-2 \\ 4 \\ 5\end{array}\right], B=\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]$, verify that $(A B)^{T}=B^{T} A^{T}$

Solution:

Given : $A=\left[\begin{array}{c}-2 \\ 4 \\ 5\end{array}\right]$

$A^{T}=\left[\begin{array}{lll}-2 & 4 & 5\end{array}\right]$

$B=\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]$

$B^{T}=\left[\begin{array}{c}1 \\ 3 \\ -6\end{array}\right]$

Now,

$A B=\left[\begin{array}{c}-2 \\ 4 \\ 5\end{array}\right]\left[\begin{array}{lll}1 & 3 & -6\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{ccc}-2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30\end{array}\right]$

$\Rightarrow(A B)^{T}=\left[\begin{array}{ccc}-2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30\end{array}\right]$       ...(1)

$B^{T} A^{T}=\left[\begin{array}{c}1 \\ 3 \\ -6\end{array}\right]\left[\begin{array}{lll}-2 & 4 & 5\end{array}\right]$

$\Rightarrow B^{T} A^{T}=\left[\begin{array}{ccc}-2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30\end{array}\right]$    ...(2)

$\therefore(A B)^{T}=B^{T} A^{T}$  [From eqs. (1) and (2)] 

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