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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix},B=\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}\;and\;C=\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix},verify:(i)\;(AB)C=A(BC)$

Note: This is part 1 of a 2 part question, split as 2 separate questions here.
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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.
  • 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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Step1:
Given
$A=\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}$
$B=\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}$
$C=\begin{bmatrix}1 & 0\\-1 & 0\end{bmatrix}$
(i) Consider LHS:-
(AB)C
$\Rightarrow AB=\begin{bmatrix}1 & 2\\-2 & 1\end{bmatrix}\begin{bmatrix}2 & 3\\3 & -4\end{bmatrix}$
$\qquad=\begin{bmatrix}1(2)+2(3) & 1(3)+2(-4)\\-2 (2)+1(3)& -2(3)+1(-4)\end{bmatrix}$
$\qquad=\begin{bmatrix}2+6 & 3-8\\-4+3& -6-4\end{bmatrix}$
$\qquad=\begin{bmatrix}8 & -5\\-1& -10\end{bmatrix}$
$(AB)C=\begin{bmatrix}8 & -5\\-1& -10\end{bmatrix}\begin{bmatrix}1 & 0\\-1& 0\end{bmatrix}$
$\qquad=\begin{bmatrix}8(1)+(-5)(-1) & 8(0)+(-5)(0)\\-1 (1)+(-10)(-1)& -1(0)+(-10)(0)\end{bmatrix}$
$\qquad=\begin{bmatrix}8+5 & 0+0\\-1+10& 0+0\end{bmatrix}$
$\qquad=\begin{bmatrix}13 & 0\\9& 0\end{bmatrix}$
Step2:
RHS:-
A(BC)
BC=$\begin{bmatrix}2 & 3\\3& -4\end{bmatrix}\begin{bmatrix}1 & 0\\-1& 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2(1)+3(-1) & 2(0)+3(0)\\3(1)+(-4)(-1)& 3(0)+(-4)(0)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2-3 & 0+0\\3+4&0+0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-1 & 0\\7&0\end{bmatrix}$
$A(BC)=\begin{bmatrix}1 & 2\\-2& 1\end{bmatrix}\begin{bmatrix}-1 & 0\\7& 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1(-1)+2(7) & 1(0)+2(0)\\-2(-1)+1(7)& -2(0)+1(0)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-1+14 & 0+0\\2+7&0+0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}13 & 0\\9&0\end{bmatrix}$
$\Rightarrow LHS=RHS.$
$\Rightarrow (AB)C=A(BC).$
answered Mar 23, 2013 by sharmaaparna1
 

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