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