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

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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:
(h)LHS:-
(A-B)C
A-B=A+(-1)B.
$\;\;\;\;=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}+(-1)\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}+\begin{bmatrix}-4 & 0\\-1 & -5\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}1-4 & 2+0\\-1-1 & 3-5\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}-3 & 2\\-2 & -2\end{bmatrix}$
$(A-B)C=\begin{bmatrix}-3 & 2\\-2 & -2\end{bmatrix}\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}-3(2)+2(1) & -3(0)+2(-2)\\-2(2)+(-2)(1) & -2(0)+(-2)(-2)\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}-6+2 & 0-4\\-4-2 & 0+4\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}-4 & -4\\-6 & 4\end{bmatrix}$
Step2:
RHS:-
AC-BC
$AC=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}1(2)+2(1) & 1(0)+2(-2)\\-1(2)+3(1) & -1(0)+3(-2)\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}2+2 & 0-4\\-2+3 & 0-6\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}4 & -4\\1 & -6\end{bmatrix}$
$BC=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}4(2)+0(1) & 4(0)+0(-2)\\1(2)+5(1) & 1(0)+5(-2)\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}8+0 & 0+0\\2+5 & 0-10\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}8 & 0\\7 & -10\end{bmatrix}$
AC-BC$=\begin{bmatrix}4 & -4\\1 & -6\end{bmatrix}+(-1)\begin{bmatrix}8 & 0\\7 & -10\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}4 & -4\\1 & -6\end{bmatrix}+\begin{bmatrix}-8 & 0\\-7 & 10\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}-4 & -4\\-6 & 4\end{bmatrix}$
$\Rightarrow LHS=RHS.$
(A-B)C=AC-BC.

 

answered Mar 19, 2013 by sharmaaparna1
edited Mar 25, 2013 by sharmaaparna1
 

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