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

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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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
(b)A(BC)=(AB)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(BC)
(BC)=$\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;=\begin{bmatrix}4(2)+0(1) & 4(0)+0(-2)\\1(2)+5(1) & 1(0)+5(-2)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}8+0& 0+0\\2+5& 0-10\end{bmatrix}$
$\;\;\;=\begin{bmatrix}8& 0\\7& -10\end{bmatrix}$
$A(BC)=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\begin{bmatrix}8 & 0\\7 & -10\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}1(8)+2(7) & 1(0)+2(-10)\\-1(8)+3(7) & -1(0)+3(-10)\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}8+14 & 0-20\\-8+21 & 0-30\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}22 & -20\\13 & -30\end{bmatrix}$
Step2:
RHS:-
(AB)=$\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1(4)+2(1) & 1(0)+2(5)\\-1(4)+3(1) & -1(0)+3(5)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}4+2 & 0+10\\-4+3 & 0+15\end{bmatrix}$
$\;\;\;=\begin{bmatrix}6 & 10\\-1 & 15\end{bmatrix}$
$(AB)C=\begin{bmatrix}6 & 10\\-1 & 15\end{bmatrix}\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}6(2)+10(1) & 6(0)+10(-2)\\-1(2)+15(1) & 11(0)+15(-2)\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}12+10 & 0-20\\-2+15 & 0-30\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}22 & -20\\13 & -30\end{bmatrix}$
$\Rightarrow A(BC)=(AB)C.$
$\Rightarrow LHS=RHS.$
answered Mar 24, 2013 by sharmaaparna1
 

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