HSC Commerce 12th Board ExamMaharashtra State Board
Account
It's free!

Share

Books Shortlist

# Solution - Find the inverse of the following matrix by elementary row transformations if it exists. A=[[1,2,-2],[0,-2,1],[-1,3,0]] - HSC Commerce 12th Board Exam - Mathematics and Statistics

#### Question

Find the inverse of the following matrix by elementary row transformations if it exists. A=[[1,2,-2],[0,-2,1],[-1,3,0]]

#### Solution

A=[[1,2,-2],[0,-2,1],[-1,3,0]]

therefore A=|[1,2,-2],[0,-2,1],[-1,3,0]|

=1|[-2,1],[3,0]|-2|[0,1],[-1,1]|-2|[0,-2],[-1,3]|

|A|=1(0-3)-2(0+1)-2(0-2)

=-3-2+4

=-1!=0

therefore A^(-1) " exist"

We have

A A^(-1)=I

[[1,2,-2],[0,-2,1],[-1,3,0]]A^(-1)=[[1,0,0],[0,1,0],[0,0,1]]

R_3->R_3+R_1

[[1,2,-2],[0,-2,1],[0,5,-2]]A^(-1)=[[1,0,0],[0,1,0],[1,0,1]]

R_3->R_3+2R_2

[[1,2,-2],[0,-2,1],[0,1,-0]]A^(-1)=[[1,0,0],[0,1,0],[1,2,1]]

R_2 harr R_3

[[1,2,-2],[0,1,0],[0,-2,1]]A^(-1)=[[1,0,0],[1,2,1],[0,1,0]]

R_1->R_1-2R_2 "   "  R3->R_3+2R_2

[[1,0,-2],[0,1,0],[0,0,1]]A^(-1)=[[-1,-4,-2],[1,2,1],[2,5,2]]

R_1->R_1+2R_3

[[1,0,0],[0,1,0],[0,0,1]]A^(-1)=[[3,6,2],[1,2,1],[2,5,2]]

A^(-1)=[[3,6,2],[1,2,1],[2,5,2]]

Is there an error in this question or solution?

#### APPEARS IN

2014-2015 (March) (with solutions)
Question 2.2.1 | 4 marks

#### Video TutorialsVIEW ALL [1]

Solution for question: Find the inverse of the following matrix by elementary row transformations if it exists. A=[[1,2,-2],[0,-2,1],[-1,3,0]] concept: Inverse of Matrix. For the courses HSC Commerce, HSC Commerce (Marketing and Salesmanship)
S