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 4 Algebra of Matrices Exercise Very Short Asnwer Question 67 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 4 Algebra of Matrices Exercise Very Short Asnwer Question 67 Maths Textbook Solution.

Answers (1)

Answer: P=\begin{bmatrix} 3 &6 \\ 6& 9 \end{bmatrix}

Hint: Here we use basic of matrix

Given: A=\begin{bmatrix} 3 &5 \\ 7& 9 \end{bmatrix}A=P+Q

Solution:

A=P+Q \: \: \: \: \: \cdot \cdot \cdot \cdot (1)

Let's take tranpose of equation (1)

\begin{aligned} A^{T} &=(P+Q)^{T} \\ A^{T} &=P^{T}+Q^{T} \\ A^{T} &=P-Q \quad \text { ......(2) } \quad\left(P^{T}=\mathrm{P} \text { and } Q^{T}=-\mathrm{Q}\right) \end{aligned}

From eqn (1) and (2)

\begin{aligned} P &=\frac{1}{2}\left(A+A^{T}\right) \\ P &=\frac{1}{2}\left(\left[\begin{array}{ll} 3 & 5 \\ 7 & 9 \end{array}\right]+\left[\begin{array}{ll} 3 & 7 \\ 5 & 9 \end{array}\right]\right)=\frac{1}{2}\left[\begin{array}{cc} 6 & 12 \\ 12 & 18 \end{array}\right]=\left[\begin{array}{ll} 3 & 6 \\ 6 & 9 \end{array}\right] \end{aligned}

 

Posted by

infoexpert27

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

CBSE Class 10, 12 re-evaluation, verification LOT 3 results 2023 declared at cbseresults.nic.in
anam-khan

CBSE asks schools to keep LOC, registration data of students ready for board exams 2024
anam-khan

CBSE 12th supplementary result verification application from tomorrow, board announces revaluation details

Latest Question