Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 21

Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 21

Answers (1)

Answer: Hence proved,

\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]

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

w^3 = 1 when w is a complex cube root of unity.

Given: A= \left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]

w is a complex cube root of unity

Prove: LHS=RHS

          \left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]

Consider the LHS\

 

            =\left[\begin{array}{ccc} 1+w & w+w^{2} & w^{2}+1 \\ w+w^{2} & w^{2}+1 & 1+w \\ w^{2}+w & 1+w^{2} & w+1 \end{array}\right]\left[\begin{array}{c} 1 \\ w \\ w^{2} \end{array}\right]

We know that 1+w+w^{2}=0 and w^{3}=1

=\left[\begin{array}{ccc} -w^{2} & -1 & -w \\ -1 & -w & -w^{2} \\ -1 & -w & -w^{2} \end{array}\right]\left[\begin{array}{c} 1 \\ w \\ w^{2} \end{array}\right]

Now by simplifying we get,

\begin{array}{l} \\\\\\\\\\ =\left[\begin{array}{l} -w^{2}-w-w^{3} \\ -1-w^{2}-w^{4} \\ -1-w^{2}-w^{4} \end{array}\right] \\ =\left[\begin{array}{l} -w\left(w+1+w^{2}\right) \\ -1-w^{2}-w^{3} w \\ -1-w^{2}-w^{3} w \end{array}\right] \end{array}

Again by substituting 1+w+w^{2}=0 and w^3 = 1 in above matrix we get,

=\left[\begin{array}{c} -w(0) \\ -1-w^{2}-w \\ -1-w^{2}-w \end{array}\right]=\left[\begin{array}{c} 0 \\ -\left(1+w+w^{2}\right) \\ -\left(1+w+w^{2}\right) \end{array}\right]\\\\ =\left[\begin{array}{c} 0 \\ -(0) \\ -(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]

Therefore LHS=RHS hence proved.

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

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question