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# If $A=\begin{bmatrix}1 & 2\\4 & 1\\5 & 0\end{bmatrix},B=\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix},then\;verify\;that:(i)\quad(2A+B)'=2A'+B'$

Note: This is part 1 of a 2 part question, split as 2 separate questions here.

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.
Step1:
(i)LHS:-
$A=\begin{bmatrix}1 & 2\\4 & 1\\5 & 6\end{bmatrix}$
$B=\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
$(2A+B)'$
$\Rightarrow 2A+B=2\begin{bmatrix}1 & 2\\4 & 1\\5 & 6\end{bmatrix}+\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}2 & 4\\8 & 2\\10 & 12\end{bmatrix}+\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2+1 & 4+2\\8+6 & 2+4\\10+7 & 12+3\end{bmatrix}$
$\Rightarrow \begin{bmatrix}3 & 6\\14 & 6\\17 & 15\end{bmatrix}$
$(2A+B)'=\begin{bmatrix}3 & 14& 17\\6 & 6 &15\end{bmatrix}$
Step2:
RHS:-
$2A'+B'$
$A=\begin{bmatrix}1 & 2\\4 & 1\\5 & 6\end{bmatrix}$
$A'=\begin{bmatrix}1 & 4&5 \\2 &1& 6\end{bmatrix}$
$2A'=\begin{bmatrix}2 & 8&10 \\4 &2& 12\end{bmatrix}$
$B=\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
$B'=\begin{bmatrix}1 & 6 & 7\\2& 4& 3\end{bmatrix}$
$2A'+B'=\begin{bmatrix}2 & 8&10 \\4 &2& 12\end{bmatrix}+\begin{bmatrix}1 & 6 & 7\\2& 4& 3\end{bmatrix}$
$\;\;\;\;\qquad=\begin{bmatrix}2+1 & 8+6 & 10+7\\4+2& 4+2&12+ 3\end{bmatrix}$
$\;\;\;\;\qquad=\begin{bmatrix}3 & 14 & 17\\6& 6&15\end{bmatrix}$
$\Rightarrow (2A+B)'=2A'+B'$
$\Rightarrow LHS=RHS.$