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Answer: $\left ( AB \right )^{T}=B^{T}A^{T}$

Given: $A=-\begin{bmatrix} -2\\ 4\\ 5 \end{bmatrix}, B=\begin{bmatrix} 1 & 3 &- 6 \end{bmatrix}$

To prove: $\left ( AB \right )^{T}=B^{T}A^{T}$

Hint: The  $A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.

Solution:

$A^{T}=\begin{bmatrix} -2 & 4 &5 \end{bmatrix}, B^{T}=\begin{bmatrix} 1\\ 3\\ -6 \end{bmatrix}$

$\left ( AB \right )^{T}=B^{T}A^{T}$

$\left ( \begin{bmatrix} -2\\ 4\\ 5 \end{bmatrix}\begin{bmatrix} 1 &3 & -6 \end{bmatrix} \right )^{T}=\begin{bmatrix} 1\\ 3\\ -6 \end{bmatrix}\begin{bmatrix} -2 &4 & 5 \end{bmatrix}$

$\begin{bmatrix} -2 & -6 &12 \\ 4& 12 & -24\\ 5 & 15 & -30 \end{bmatrix}=\begin{bmatrix} -2 & 4 &5 \\ -6 &12 & 15\\ 12 &-24 & -30 \end{bmatrix}$

$\begin{bmatrix} -2 & 4 &5 \\ -6 &12 & 15\\ 12 &-24 & -30 \end{bmatrix}=\begin{bmatrix} -2 & 4 &5 \\ -6 &12 & 15\\ 12 &-24 & -30 \end{bmatrix}$

∴LHS=RHS

Hence, $\left ( AB \right )^{T}=B^{T}A^{T}$ is proved.

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