# Deleting all zero columns.

1. Apr 3, 2014

### pondzo

If you have an augmented matrix and one column in your matrix is all zero (but not the answer column) are you able to 'delete' the all zero column and the corresponding variable? Because when you write out the equations, the co-efficient's of the variable corresponding to the all zero column are of course, all zero. Doesn't this mean that that variable is not important in these equations, and that no other variables are dependent upon this variable.

And on another note, can it even be considered a variable since nothing is dependent on this 'variable' and it can take any value without affecting it at all.

Thank you!

2. Apr 3, 2014

### SteamKing

Staff Emeritus
A more basic question is this: if you have a 'variable' which is not used in your system of equations, how did this 'variable' become included in the coefficient matrix in the first place?

I believe an example of what you are talking about is needed for clarification.

3. Apr 3, 2014

### pondzo

Yeah thats what i was wondering my self. how did this variable even get incorporated into the matrix if it doesn't affect it. its almost as if the variable doesn't exist, ghost variable?

Any way the question was;

$$\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & | & 0\\ 0 & 0 & 6 & 3 & 1 & | & π\\ 0 & 0 & 0 & 4 & 3 & | & 7\\ 0 & 0 & 0 & 0 & k^2 - 1 & | &0 \end{pmatrix}$$

which statement is correct;
A) The only values for k in which there are infinitely many solns are k=1 and k=-1
B) when k=1 or -1 there are no solutions
C) There are always infinitely many solutions
D) when k=1 or -1 there is a unique solution
E) when k≠1 or -1 there are no solutions

I thought it was between either A or C but i wasnt sure if an all zero column meant infinite solutions (in which case it would be C) or whether the all zero column could be excluded (in which case it would be A).

Last edited by a moderator: Apr 3, 2014
4. Apr 3, 2014

### pondzo

dont know why that matrix didnt come out right, but its meant to be an augmented matrix, hopefully you can see it anyway.

1,0,0,0,0|0
0,0,6,3,1|π
0,0,0,4,3|7
0,0,0,0,k2-1|0

5. Apr 3, 2014

### Staff: Mentor

I fixed your LaTeX in post #3. It wasn't displaying correctly because you had a [ sup ] tag in it. These BB tags aren't compatible with LaTeX.

Also, you were missing an & after the vertical bar.

With regard to your question, the system it came from must have looked like this:
x1 = 0
6x3 + 3x4 + x5 = $\pi$
etc.

x2 doesn't appear in the system, which means it is arbitrary.

6. Apr 4, 2014

### pondzo

Thanks for the edit.

And If X2 is arbitrary does that mean we can 'delete' its all zero column and X2? Or must we leave the matrix as is?

7. Apr 4, 2014

### Staff: Mentor

I would leave the column in. If you delete it, the effect is to change the names of all variables after the deleted column, with the old x3 becoming x2, the old x4 becoming x3, and so on.