Find the inverse using elementary row transformations: A=`[[-1, 1, 2],[ 1, 2, 3],[ 3 ,1 ,1]]`

Explanation

Given,A=`[[-1, 1, 2],[ 1, 2, 3],[ 3 ,1 ,1]]`

Let `A=I_3A`

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

Applying `r_1->(-1)r_1` and `r_2->r_2-r_1` and `r_3->r_3-3r_1`,

`[[1, -1, -2],[ 0,3,5],[ 0,4,7]]=[[1, 0, 0],[ 1, 1, 0],[3,1,0]]A`

Applying `r_2->1/2r_2` and `r_1->r_1+r_2` and `r_3->r_3-4r_2`,

=`[[1, 0, -1/3],[ 0,1,5/3],[ 0,0,1]]=[[-2/3, 1/3, 0],[ 1/3, 1/3, 0],[5,-4,3]]A`

Applying `r_3->3r_3` and `r_1->r_1+1/3r_3` and `r_2->r_2-5/3r_2`,

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