logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Matrices
0 votes

If $A=\begin{bmatrix}0 & x\\x & 0\end{bmatrix},B=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\;and\;x^2=-1,then\;show\;that\;(A+B)^2=A^2+B^2.$

Can you answer this question?
 
 

1 Answer

0 votes
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:-
$A=\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}$
$B=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$A+B=\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0+0 & -x+1\\x+1 & 0+0\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}0 & 1-x\\x+1 & 0\end{bmatrix}$
Step2:
$(A+B)^2=\begin{bmatrix}0 & 1-x\\x+1 & 0\end{bmatrix}+\begin{bmatrix}0 & 1-x\\x+1 & 0\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}0+(1-x)(x+1) & 0(1-x)+(1-x)(0)\\x+1(0)+0(x+1) & (x+1)(1-x)+0\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}(1-x)(x+1) & 0\\0 & (x+1)(1-x)\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}x+1-x^2-x & 0\\0 & x+1-x^2-x\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}1-x^2 & 0\\0 & 1-x^2\end{bmatrix}$
Replace $x^2=-1.$
$\;\;\;\;\;\;=\begin{bmatrix}1-(-1) & 0\\0 & 1-(-1)\end{bmatrix}$
$\;\;\;\;\;\;=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$
Step3:
RHS:-
$A^2+B^2$
$A^2=\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}0(0)+(-x)(x) & 0(-x)+(-x)(0)\\x(0)+0(x) & x(-x)+0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}0-x^2 & 0+0\\0 & -x^2\end{bmatrix}$t >>
$\;\;\;=\begin{bmatrix}-x^2 & 0\\0 & -x^2\end{bmatrix}$
Replace $x^2=-1$
$\;\;\;=\begin{bmatrix}-(-1) & 0\\0 & -(-1)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1& 0\\0 & 1\end{bmatrix}$
Step4:
$B^2=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0+1 & 0+0\\0+0 & 1+0\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$A^2+B^2=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\;\;\;\;\;\qquad=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$
$\Rightarrow LHS=RHS.$
$(A+B)^2=A^2+B^2$
answered Mar 25, 2013 by sharmaaparna1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...