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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}0 & 1\\1& 1\end{bmatrix}\;and\;B=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix},\;show\;that\;(A+B)(A-B)\neq A^2-B^2$

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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
LHS:-
A+B=$\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0+0 & 1+1\\1+1 & 1+0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0 & 2\\2 & 1\end{bmatrix}$
A-B=A+(-1)(B)
$\;\;\;\;\;=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+(-1)\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+\begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0 & 1-1\\1-1 & 1+0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$
(A+B)(A-B)=$\begin{bmatrix}0 & 2\\2 & 1\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}0+0 & 0+2\\0+0 & 0+1\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}0 & 2\\0 & 1\end{bmatrix}$
Step2
RHS:-
$A^2-B^2$
$A^2=A.A\Rightarrow \begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}$
$\qquad\Rightarrow \begin{bmatrix}0+1 & 0+1\\0+1 & 1+1\end{bmatrix}$
$\qquad\Rightarrow \begin{bmatrix}1 & 1\\1 & 2\end{bmatrix}$
$B^2=B.B\Rightarrow \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$\qquad\Rightarrow \begin{bmatrix}0+1 & 0+0\\0+0 & 1+0\end{bmatrix}$
$\qquad\Rightarrow \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$A^2-B^2=A^2+(-1)(B^2)$
$\Rightarrow \begin{bmatrix}1 & 1\\1 & 2\end{bmatrix}+(-1)\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1 & 1\\1 & 2\end{bmatrix}+\begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1-1 & 1+0\\1+0 & 2-1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}$
$\Rightarrow (A+B)(A-B)=\begin{bmatrix}0 & 2\\0 & 1\end{bmatrix}$
$A^2-B^2=\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}$
$\Rightarrow (A+B)(A-B)\neq A^2-B^2$.
Hence proved.

 

answered Mar 19, 2013 by sharmaaparna1
edited Mar 21, 2013 by sharmaaparna1
 

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