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

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 sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
(i)$(A-B)^T=A^T-B^T$
Step1:
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
LHS:-
$(A-B)^T$
$(A-B)=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}+(-1)\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}+\begin{bmatrix}-4 & 0\\-1 & -5\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}1-4 & 2+0\\-1-1 & 3-5\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}-3 & 2\\-2 & -2\end{bmatrix}$
$(A-B)^T=\begin{bmatrix}-3 & -2\\2 & -2\end{bmatrix}$
Step2:
RHS:-
$A^T-B^T$
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$A^T=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
$B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$B^T=\begin{bmatrix}4 & 1\\0 & 5\end{bmatrix}$
$A^T-B^T=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}+(-1)\begin{bmatrix}4 & 1\\0 & 5\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}+\begin{bmatrix}-4 & -1\\0 & -5\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}1-4 & -1-1\\2+0 & 3-5\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}-3 & -2\\2 & -2\end{bmatrix}$
$\Rightarrow LHS=RHS.$
$\Rightarrow (A-B)^T=A^T-B^T.$