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# If $A=\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix},B=\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}\;and\;C=\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix},verify:(ii)\;A(B+C)=AB+AC$

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

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Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• 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:
Given
$A=\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}$
$B=\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}$
$C=\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix}$
Consider the LHS:-
$A(B+C)=\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}(\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}+\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix})$
$(B+C)=\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}+\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}2+1 & 3+0\\3-1 & -4+0\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}3 & 3\\2 & -4\end{bmatrix}$
A(B+C)=$\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}\begin{bmatrix}3 & 3\\2 & -4\end{bmatrix}$
$\qquad=\begin{bmatrix}1(3)+2(2) & 1(3)+2(-4)\\-2(3)+1(2) & -2(3)+1(-4)\end{bmatrix}$
$\qquad=\begin{bmatrix}3+4 & 3-8\\-6+2 & -6-4\end{bmatrix}$
$\qquad=\begin{bmatrix}7 & -5\\-4 & -10\end{bmatrix}$
Step2:
RHS:-
AB=$\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}$
$\qquad=\begin{bmatrix}1(2)+2(3) & 1(3)+2(-4)\\-2(2)+1(3) & -2(3)+1(-4)\end{bmatrix}$
$\qquad=\begin{bmatrix}2+6 & 3-8\\-4+3 & -6-4\end{bmatrix}$
$\qquad=\begin{bmatrix}8 & -5\\-1 & -10\end{bmatrix}$
AC=$\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix}$
$\qquad=\begin{bmatrix}1(1)+2(-1) & 1(0)+2(0)\\-2(1)+1(-1) & -2(0)+1(0)\end{bmatrix}$
$\qquad=\begin{bmatrix}1-2 & 0+0\\-2-1 & 0+0\end{bmatrix}$
$\qquad=\begin{bmatrix}-1 & 0\\-3 & 0\end{bmatrix}$
$AB+AC=\begin{bmatrix}8 &-5\\-1 & -10\end{bmatrix}+\begin{bmatrix}-1 & 0\\-3 & 0\end{bmatrix}$
$\qquad\;\;\;\;\;=\begin{bmatrix}7 & -5\\-4 & -10\end{bmatrix}$
$\Rightarrow A(B+C)=AB+AC.$
$\Rightarrow LHS=RHS.$
answered Mar 23, 2013

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