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]]`