Let A= 1&-1&0 2&1&3 1&2&1 and B= 1&2&3 2&1&3 0&1&1 , Find A^T, B^T and verify that.(AB)^T=B^T+A^T

Given: A= 1&-1&0 2&1&3 1&2&1 and B= 1&2&3 2&1&3 0&1&1 A^T= 1&2&1 -1&1&2 0&3&1 and B^T= 1&2&0 2&1&1 3&3&1 AB= 1&-1&0 2&1&3 1&2&1 1&2&3 2&1&3 0&1&1 AB= 1-2+0&2-1+0&3-3+0 2+2+0&4+1+3&6+3+3 1+4+0&2+2+1&3+6+1 = -1&1&0 4&8&12 5&5&10 (AB)^T= -1&4&5 1&8&5 0&12&10 (1) Now, B^TA^T= 1&2&0 2&1&1 3&3&1 1&2&1 -1&1&2 0&3&1 ^AA^T= 1-2+0&2+2+0&1+4+0 2-1+0&4+1+3&2+2+1 3-3+0&6+3+3&3+6+1 ^TA^T= -1&4&5 1&8&5 0&12&10 (2) (AB)^T=B^TA^T [from eqs. (1) and (2)]

Let A= 1&-1&0 2&1&3 1&2&1 and B= 1&2&3 2&1&3 0&1&1 , Find A^T, B^…

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Question

Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}+\text{A}^\text{T}$

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Answer

Given: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$ $\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$ $\text{AB}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$ $\Rightarrow\ \text{AB}=\begin{bmatrix}1-2+0&2-1+0&3-3+0\\2+2+0&4+1+3&6+3+3\\1+4+0&2+2+1&3+6+1\end{bmatrix}$$\Rightarrow\text{AB}=\begin{bmatrix}-1&1&0\\4&8&12\\5&5&10\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(1)$
Now, $\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ $\Rightarrow\text{B}^\text{A}\text{A}^\text{T}=\begin{bmatrix}1-2+0&2+2+0&1+4+0\\2-1+0&4+1+3&2+2+1\\3-3+0&6+3+3&3+6+1\end{bmatrix}$ $\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(2)$ $\Rightarrow(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [from eqs. (1) and (2)]

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