Question
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{bA})^{\text{T}}=\text{bA}^{\text{T}}.$

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{bA})^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}\ [\because\ \text{b}=-2]$
$=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}$
and $\text{A}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ $\therefore\ \text{bA}^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}=(\text{bA})^{\text{T}}$ Hence proved.

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