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Using elementary transformation, find the inverse of the matrix `A=[(a,b),(c,((1+bc)/a))]`.

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The correct Answer is:
`[((1+bc)/a,-b),(-c,a)]`

`A=[(a,b),(c,((1+bc)/a))]`
We write,
`[(a,b),(c,((1+bc)/a))]=[(1,0),(0,1)]A`
`implies [(1,b/a),(c,((1+bc)/a))]=[(1/a,0),(0,1)]A" "(R_(1) rarr R_(1)/a)`
or `[(1,b/a),(0,1/a)]=[(1/a,0),((-c)/a,1)]A" "(R_(2) rarr R_(2)-cR_(1))`
or `[(1,b/a),(0,1)]=[(1/a,0),(-c,a)]A" "(R_(2)=aR_(2))`
or `[(1,0),(0,1)]=[((1+bc)/a,-b),(-c,a)]A" "(R_(1) rarr R_(1)- b/a R_(2))`
`implies A^(-1)=[((1+bc)/a,-b),(-c,a)]`
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