For two matrices A and B, A= 2&1&3 4&1&0 ,B= 1&-1 0&2 5&0 verify that (AB)^T = B^TA^T.

Given, A= 2&1&3 4&1&0 ,B= 1&-1 0&2 5&0 (AB)^T=B^TA^T 2&1&3 4&1&0 1&-1 0&2 5&0 ^T = 1&-1 0&2 5&0 ^T 2&1&3 4&1&0 ^T 2+0+15&-2+20 4+0+0&-4+2+0 ^T = 1&0&5 -1&2&0 2&4 1&1 3&0 17&0 4&-2 ^T= 2+0+15&4+0+0 -2+2+0&-4+2+0 17&4 0&-2 = 17&4 0&-2 =RHS So, (AB)^T=B^TA^T

For two matrices A and B, A= 2&1&3 4&1&0 ,B= 1&-1 0&2 5&0 verify …

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Question

For two matrices $A$ and $B, \text{A}=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix},\text{B}=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$ verify that $(AB)^T = B^TA^T.$

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Answer

Given,
$\text{A}=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix},\text{B}=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}$
$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}^\text{T}\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2+0+15&-2+20\\4+0+0&-4+2+0\end{bmatrix}^\text{T}$ $=\begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix}\begin{bmatrix}2&4\\1&1\\3&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}17&0\\4&-2\end{bmatrix}^\text{T}=\begin{bmatrix}2+0+15&4+0+0\\-2+2+0&-4+2+0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}17&4\\0&-2\end{bmatrix}=\begin{bmatrix}17&4\\0&-2\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So, $(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$

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