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Solution for 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

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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 | 4.00 marks

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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)
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