Question

If `A=|[2,0,-1],[5,1,0],[0,1,3]|` , then find A^-1 using elementary row operations

Answer 1

`|A|=|[2,0,1],[5,1,0],[0,1,3]|`

=2(3-0)-0(15-0)-1(5-0)

=6-0-5

=1

≠0

Hence A^-1 exists.

A^-1A=1

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

Applying `R_1 ->(1/2)R_1`

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

Applying `R_2->R_2+(-5)R_1`

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

Applying `R_3->R3+(-1)R_2`

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

Applying `R_3->(2)R_3`

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

Applying `R_1->R_1+(1/2)R_3 and R_2->R_2+(-5/2)R_3`

`A^(-1)[[1,0,0],[0,1,0],[0,0,1]]=[[3,-1,1],[-15,6,-5],[5,-2,2]]`

`A^-1 =[[3,-1,1],[-15,6,-5],[5,-2,2]]`