Using elementary transformations, find the inverse of the matrix : `(2 0-1 5 1 0 0 1 3)`

Explanation

Let `A=[(2,0,-1),(5,1,0),(0,1,3)]`
We kanow that `A=IA`. Therefore,
`[(2,0,-1),(5,1,0),(0,1,3)]=[(1,0,0),(0,1,0),(0,0,1)]A`
Applying `R_(1) rarr 1/2 R_(1)`, we have
`[(1,0,-1/2),(5,1,0),(0,1,3)]=[(1/2,0,0),(0,1,0),(0,0,1)]A`
Applying `R_(2) rarr R_(2)-5R_(1)`, we have
`[(1,0,-1/2),(0,1,5/2),(0,1,3)]=[(1/2,0,0),(-5/2,1,0),(0,0,1)]A`
Applying `R_(3) rarr R_(3)-R_(2)`, we have
`[(1,0,-1/2),(0,1,5/2),(0,0,1/2)]=[(1/2,0,0),(-5/2,1,0),(5/2,-1,1)]A`
Applying `R_(3) rarr 2R_(3)`, we have
`[(1,0,-1/2),(0,1,5/2),(0,0,1)]=[(1/2,0,0),(-5/2,1,0),(5,-2,2)]A`
Applying `R_(1) rarr R_(1)+1/2 R_(3)` and `R_(2) rarr R_(2) -5/2 R_(3)`, we have
`[(1,0,0),(0,1,0),(0,0,1)]=[(3,-1,1),(-15,6,-5),(5,-2,2)]A`
`A^(-1)=[(3,-1,1),(-15,6,-5),(5,-2,2)]`