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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Let $A=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix},B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix},C=\begin{bmatrix}2 & 0\\1 & 2\end{bmatrix}\;and\;a=4,b=-2.Show\;that:$\[(a)\;A+(B+C)=(A+B)+C\]

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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.
(a)A+(B+C)=(A+B)+C.
Step1:
Given:
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$C=\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
LHS:-
A+(B+C)
$(B+C)=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}+\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}4+2 & 0+0\\1+1 & 5-2\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}6 & 0\\2 & 3\end{bmatrix}$
$A+(B+C)=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}+\begin{bmatrix}6 & 0\\2 & 3\end{bmatrix}$
$\;\;\;\qquad\qquad=\begin{bmatrix}1+6 & 2+0\\-1+2 & 3+3\end{bmatrix}$
$\;\;\;\qquad\qquad=\begin{bmatrix}7 & 2\\1 & 6\end{bmatrix}$
Step2:
RHS:-
(A+B)+C
$(A+B)=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}+\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}1+4 & 2+0\\-1+1 & 3+5\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}5 & 2\\0 & 8\end{bmatrix}$
$(A+B)+C=\begin{bmatrix}5 & 2\\0 & 8\end{bmatrix}+\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;\qquad\qquad=\begin{bmatrix}5+2 & 2+0\\0+1 & 8-2\end{bmatrix}$
$\;\;\;\qquad\qquad=\begin{bmatrix}7 & 2\\1 & 6\end{bmatrix}$
$\Rightarrow A+(B+C)=(A+B)+C.$
LHS=RHS.
answered Mar 24, 2013 by sharmaaparna1
 

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