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