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

Toolbox:
• 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 $c$.
• 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.
(d) a(C-A)=aC-aA
Step1:
Given
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$C= \begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
a=4
LHS:
C-A=C+(-1)A
$\Rightarrow \begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}+(-1)\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}+\begin{bmatrix}-1 & -2\\1 & -3\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2-1 & 0-2\\1+1 & -2-3\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1 & -2\\2 & -5\end{bmatrix}$
$a(C-A)=4\begin{bmatrix}1 & -2\\2 & -5\end{bmatrix}$
$\;\;\;\;\;\qquad=4\begin{bmatrix}4 & -8\\8 & -20\end{bmatrix}$
Step2:
RHS:-
aC-aA.
$aC\Rightarrow 4\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\qquad= \begin{bmatrix}8 & 0\\4 & -8\end{bmatrix}$
$aA\Rightarrow 4\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$\;\qquad= \begin{bmatrix}4 & 8\\-4 & 12\end{bmatrix}$
$aC-aA=\begin{bmatrix}8 & 0\\4 & -8\end{bmatrix}+(-1)\begin{bmatrix}4 & 8\\-4 & 12\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}8 & 0\\4 & -8\end{bmatrix}+\begin{bmatrix}-4 &- 8\\4 & -12\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}8-4 & 0-8\\4+4 & -8-12\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}4 & -8\\8 & -20\end{bmatrix}$
$\Rightarrow RHS=LHS$
$\Rightarrow a(C-A)=aC-aA$