Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 51 maths textbook solution

Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 51 maths textbook solution

Answers (1)

Answer: Hence proved,

(A+B)^{2}=A^{2}+B^{2}

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right], x^{2}=-1

To prove:

(A+B)^{2}=A^{2}+B^{2}

Consider RHS

\\\\ A^{2}=A A \\\\ B^{2}=B B

Then,

\\\\ \Rightarrow A^{2}+B^{2}=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]+\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}0(0)+(-x)(x) & 0(-x)+(-x) 0 \\ x(0)+0(x) & x(-x)+0(0)\end{array}\right]+\left[\begin{array}{ll}0(0)+1(1) & 0(1)+1(0) \\ 1(0)+0(1) & 1(1)+0(0)\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}0-x^{2} & 0-0 \\ 0+0 & -x^{2}+0\end{array}\right]+\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2} & 0 \\ 0 & -x^{2}\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2}+1 & 0+0 \\ 0+0 & -x^{2}+1\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2}+1 & 0 \\ 0 & -x^{2}+1\end{array}\right]

Now put

\\\\x^{2}=-1

\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-(-1)+1 & 0 \\ 0 & -(-1)+1\end{array}\right]=\left[\begin{array}{cc}1+1 & 0 \\ 0 & 1+1\end{array}\right] \\\\\\A^{2}+B^{2}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]

Now taking LHS side

\\\\(A+B)^{2}=(A+B)(A+B)\\\\ A+B=\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{cc} 0+0 & -x+1 \\ x+1 & 0+0 \end{array}\right]=\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0(0)+(-x+1)(x+1) & 0(-x+1)+(-x+1)(0) \\ (x+1)(0)+0(x+1) & (x+1)(-x+1)+0(0) \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0+(-x+1)(x+1) & 0+0 \\ 0+0 & (x+1)(-x+1)+0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} -x^{2}-x+x+1 & 0 \\ 0 & -x^{2}+x-x+1 \end{array}\right]

(A+B)^{2}=\left[\begin{array}{cc} -x^{2}+1 & 0 \\ 0 & -x^{2}+1 \end{array}\right]

Now put

x^2 = -1

\\\\ (A+B)^{2}=\left[\begin{array}{cc} -(-1)+1 & 0 \\ 0 & -(-1)+1 \end{array}\right]=\left[\begin{array}{cc} 1+1 & 0 \\ 0 & 1+1 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\\\\\\

Hence proved, LHS=RHS

(A+B)^{2}=A^{2}+B^{2}

Posted by

infoexpert22

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question