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 2 sub question (iii) maths textbook solution.

Please solve RD Sharma class 12 Chapter Algebra of Matrices  exercise 4.3 question 2 sub question (iii) maths textbook solution.

 

Answers (1)

Hence proved AB \neq BA

\begin{bmatrix} 3 &1 &0 \\ 1 &1 & 0\\ 1& 4 &0 \end{bmatrix}\neq \begin{bmatrix} 1 &1 &0 \\ 1& 3 & 0\\ 9& 6 &0 \end{bmatrix}

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

Given: A = \begin{bmatrix} 1 &3 &0 \\ 1& 1 & 0\\ 4& 1 &0 \end{bmatrix} and B = \begin{bmatrix} 0&1 &0 \\ 1& 0 & 0\\ 0& 5 &0 \end{bmatrix}

Consider,

AB = \begin{bmatrix} 1 &3 &0 \\ 1& 1 & 0\\ 4& 1 &0 \end{bmatrix}\begin{bmatrix} 0&1 &0 \\ 1& 0 & 0\\ 0& 5 &0 \end{bmatrix}

A B=\left[\begin{array}{lll} 1 \times 0+3 \times 1+0 \times 0 & 1 \times 1+3 \times 0+0 \times 5 & 1 \times 0+3 \times 0+0 \times 1 \\ 1 \times 0+1 \times 1+0 \times 0 & 1 \times 1+1 \times 0+0 \times 5 & 1 \times 0+1 \times 0+0 \times 1 \\ 4 \times 0+1 \times 1+0 \times 0 & 4 \times 1+1 \times 0+0 \times 5 & 4 \times 0+1 \times 0+0 \times 1 \end{array}\right]

A B=\left[\begin{array}{lll} 0+3+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 4+0+0 & 0+0+0 \end{array}\right]

A B=\begin{bmatrix} 3 &1 &0 \\ 1& 1 & 0\\ 1& 4 &0 \end{bmatrix}                ...(i)

Now again consider,

BA = \begin{bmatrix} 0&1 &0 \\ 1& 0 & 0\\ 0& 5 &0 \end{bmatrix}\begin{bmatrix} 1 &3 &0 \\ 1& 1 & 0\\ 4& 1 &0 \end{bmatrix}

B A=\left[\begin{array}{lll} 0 \times 1+1 \times 1+0 \times 4 & 0 \times 3+1 \times 1+0 \times 1 & 0 \times 0+1 \times 0+0 \times 0 \\ 1 \times 1+0 \times 1+0 \times 4 & 1 \times 3+0 \times 1+0 \times 1 & 1 \times 0+0 \times 0+0 \times 0 \\ 0 \times 1+5 \times 1+1 \times 4 & 0 \times 3+5 \times 1+1 \times 1 & 0 \times 0+5 \times 0+1 \times 0 \end{array}\right]

B A=\left[\begin{array}{lll} 0+1+0 & 0+1+0 & 0+0+0 \\ 1+0+0 & 3+0+0 & 0+0+0 \\ 0+5+4 & 0+5+1 & 0+0+0 \end{array}\right]

B A=\begin{bmatrix} 1 & 1 &0 \\ 1& 3 & 0\\ 9& 6 &0 \end{bmatrix}             ...(ii)

From equation (i) and (ii), it is clear that AB \neq BA

Posted by

infoexpert21

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 Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question