Finding values of c for no solutions in a 3×3 linear system

[azupol]

http://www.scribd.com/doc/66759813/mat223tut1"
Question 2 and 3

Homework Statement

For what values of c does the system have no solutions? I figured out the other ones, for c=0 we can assign a parameter to x₁ and it will then have infinite solutions. For c=1 the system will have a unique solution. I know that if the matrix is in reduced row echelon form, and if the last row is all zeroes except for the right hand side of the equality, it has no solutions, but how can I manipulate the augmented matrix to make it so?

Homework Equations

Here is the augmented matrix:
1 1 3|c
c 1 5|4
1 c 4|c

The Attempt at a Solution

I can use elementary row/column operations, but I end up with this matrix in trying to make the bottom row 0...

1 1 3 |c
0 (1-c) (5-3c) |-c²+4
0 (c-1) 1 |-c²

 

[Mark44]

http://www.scribd.com/doc/66759813/mat223tut1"
Question 2 and 3

Homework Statement

For what values of c does the system have no solutions? I figured out the other ones, for c=0 we can assign a parameter to x₁ and it will then have infinite solutions. For c=1 the system will have a unique solution. I know that if the matrix is in reduced row echelon form, and if the last row is all zeroes except for the right hand side of the equality, it has no solutions, but how can I manipulate the augmented matrix to make it so?

Homework Equations

Here is the augmented matrix:
1 1 3|c
c 1 5|4
1 c 4|c

The Attempt at a Solution

I can use elementary row/column operations, but I end up with this matrix in trying to make the bottom row 0...

1 1 3 |c
0 (1-c) (5-3c) |-c²+4
0 (c-1) 1 |-c²

You have a mistake in your bottom row. You should have this:
1 1 3 |c
0 (1-c) (5-3c) |-c²+4
0 (c-1) 1 |0