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I have a doubt. When finding inverse of matrix, Let us take,

$A$ be a matrix, and $A^{-1}$ exists, then to find $A^{-1}$,

we write A=IA and we will apply a sequence of row operation(can we do column operation?) on $A=IA$ till we get, $I=BA$. The $B$ will be the inverse of $A$.

If we write $A=AI$ then then we will apply a sequence of column operation(can we do row operation?) on $A=AI$ till we get, $I=AB$.

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Make a row operation to matrix $A$ is multiplying this matrix on the left by a dilatation or transvection matrix and make a column operation to matrix $A$ is multiplying this matrix on the right by a dilatation or transvection matrix hence for example if you apply a sequence of row operations to find the inverse of $A$ you shouldn't at the same time apply a column operation.

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I ve taken $A=AI$ can we operate column operation to get $I=AB$? – SHREE6174 Jun 10 '13 at 16:27
Yes operate column operation is multiplying at the right by sequence of matrices up to find the identity matrix on the LRH – user63181 Jun 10 '13 at 16:30
No, Taking A=AI and operating column operations we will get different B(which is $A^{-1}$) from operating row operation. But correct B(which is $A^{-1}$) is operating row operation. – SHREE6174 Jun 10 '13 at 16:35

If you begin with row operations stick to it all the time, otherwise instead of multiplying your matrix from the left by elementary matrices you multiply it from the right and things go kapput:

$$E_n\cdot\ldots\cdot E_1A=I\iff A^{-1}=\left(E_n\cdot\ldots\cdot E_1\right)^{-1}=E_1^{-1}\cdot\ldots\cdot E_n^{-1}\;\;(**)$$

where each $\,E_i\,$ is an elementary matrix that performs an elementary operation on the rows of $\,A\,$ . If at some step $\,k\,$ you perform a column operation, then you get

$$E_{k-1}\cdot\ldots\cdot E_1AC$$

and then you can't get easily the nice form for $\,A^{-1}\,$ as in (**)

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