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# Let $A=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix},B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix},C=\begin{bmatrix}2 & 0\\1 & 2\end{bmatrix}\;and\;a=4,b=-2.Show\;that$:$(f)\;(bA)^T=bA^T.$

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.
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
Step1:
( f)Given:
b=-2
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
LHS:-
$bA=-2\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-2 & -4\\2 & -6\end{bmatrix}$
$(bA)^T=\begin{bmatrix}-2 & 2\\-4 & -6\end{bmatrix}$
Step2:
RHS:-
${bA}^T$
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$A^T=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
$bA^T=-2\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-2 & 2\\-4 & -6\end{bmatrix}$
$LHS=RHS.$
$\Rightarrow (bA)^T=bA^T$