Can Column Operations be Used to Find the Inverse of a Matrix?

[rock.freak667]

If I have a matrix

<br /> A= \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; 3\\<br /> 0 &amp; -1 &amp; 4\\<br /> 1 &amp; 1 &amp; 6<br /> \end{array}<br /> \right)<br />
and I need to find A^{-1} I would just augment with the identity matrix and then do row operations. But if I want to use column operations instead does it work in the same manner? because I think if use the column operations, the matrix A would be reduced to RRE form but nothing will happen to the identity matrix.
(Not too sure if I was clear about my problem.)

 

[ice109]

what are column operations?

 

[Hurkyl]

You need to adjoin an identity matrix below the original, not to its right.

 

[Hurkyl]

I think if use the column operations, the matrix A would be reduced to RRE form

You mean reduced column echelon form.

 

[rock.freak667]

You mean reduced column echelon form.

Yeah sorry about that.That is what I meant.

You need to adjoin an identity matrix below the original, not to its right.

Below it? So there is no way to both row and column operations at the same time to find the inverse or whatever?Also how would I solve a system of equations using column operations only?

 

[ice109]

wouldn't these column operations be the same thing as row reducing the transpose?

 

[rock.freak667]

wouldn't these column operations be the same thing as row reducing the transpose?

It is the same thing basically except that column operations uses the columns of the matrix,

 

[Hurkyl]

Yeah sorry about that.That is what I meant.


Below it? So there is no way to both row and column operations at the same time to find the inverse or whatever?

Not with this particular bookkeeping method. Of course you can use both kinds of operations to find an inverse: algebraically, if R is the matrix denoting your row operations and C is the one denoting your column operations, then if you reduce your matrix to the identity, that says

RAC = I​

which you can easily solve for A.

 

[rock.freak667]

oh okay then thanks