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# Let $A=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix},B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix},C=\begin{bmatrix}2 & 0\\1 & 2\end{bmatrix}\;,\;a=4,b=-2\;$.Show that:$\;(c)\;(a+b)B=aB+bB$

Toolbox:
• 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.
• 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:
Given
a=4 b=-2
B=$\begin{bmatrix}4 & 0\\1 &5\end{bmatrix}$
LHS:-
(a+b)=4-2=2.
(a+b)B=$2\begin{bmatrix}4 & 0\\1 &5\end{bmatrix}$
$\;\;\;\;\;\;=2\begin{bmatrix}4 & 0\\1 &5\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}8 & 0\\2 &10\end{bmatrix}$
Step2:
RHS:aB+bB
$aB=4\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;=\begin{bmatrix}16 & 0\\4 & 20\end{bmatrix}$
$bB=-2\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-8 & 0\\-2 & -10\end{bmatrix}$
$aB+bB=\begin{bmatrix}16 & 0\\4 & 20\end{bmatrix}+\begin{bmatrix}-8 & 0\\-2 & -10\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}16-8 & 0+0\\4-2 & 20-10\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}8 & 0\\2 & 10\end{bmatrix}$
Hence (a+b)B=aB+bB
LHS=RHS.