# Matrices(column operations)

1. Jan 3, 2008

### rock.freak667

If I have a matrix

$$A= \left( \begin{array}{ccc} 1 & 2 & 3\\ 0 & -1 & 4\\ 1 & 1 & 6 \end{array} \right)$$
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.)

2. Jan 4, 2008

### ice109

what are column operations?

3. Jan 4, 2008

### Hurkyl

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

4. Jan 4, 2008

### Hurkyl

Staff Emeritus
You mean reduced column echelon form.

5. Jan 4, 2008

### rock.freak667

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?Also how would I solve a system of equations using column operations only?

6. Jan 4, 2008

### ice109

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

7. Jan 4, 2008

### rock.freak667

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

8. Jan 4, 2008

### Hurkyl

Staff Emeritus
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.

9. Jan 4, 2008

### rock.freak667

oh okay then thanks