# If $A = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \: and \: B = \begin{bmatrix} -2 & -1 & -4 \end{bmatrix}$verify that $$(AB)^T=B^TA^T$$

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Step1:
Given
A=$\begin{bmatrix}-1\\2\\3\end{bmatrix}\$
B=$\begin{bmatrix}-2 & -1& -4\end{bmatrix}$
AB=$\begin{bmatrix}-1\\2\\3\end{bmatrix}\begin{bmatrix}-2 & -1& -4\end{bmatrix}$
$\;\;\;=\begin{bmatrix}(-1)(-2) &-1(-1) & -1(-4)\\2(-2) & 2(-1)&2(-4)\\3(-2) &3(-1) &3(-4)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2 &1 & 4\\-4 & -2&-8\\-6 &-3 &-12\end{bmatrix}$
Transpose of a matrix can be obtained by changing the rows and the column.
$(AB)'=\begin{bmatrix}2 & -4 & 6\\1 & -2 & -3\\4 & -8 & -12\end{bmatrix}$-------(1)
Step2:
$B'=\begin{bmatrix}-2\\-1\\-4\end{bmatrix}$
$A'=\begin{bmatrix}-1&2&3\end{bmatrix}$
$B'A'=\begin{bmatrix}-2\\-1\\-4\end{bmatrix}\begin{bmatrix}-1&2&3\end{bmatrix}$
$B'A'=\begin{bmatrix}-2(-1)&-2(2)&(3)(-2)\\-1(-1) & -1(2) & -1(3)\\-4(-1)& -4(2) &-4(3)\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}2 & -4 & 6\\1 & -2 & -3\\4 & -8 & -12\end{bmatrix}$-------(2)
Hence equ(1)=equ(2)
$(AB)^T=B^TA^T$
answered Apr 8, 2013