# Row reduction help

1. Oct 25, 2008

### rock.freak667

1. The problem statement, all variables and given/known data
Find the condition of k such that the set of equations x+y-z=1, x+2y+kz=-1, x+ky-z=1,
has a unique soltuion,infinite sol'n or no solution.

2. Relevant equations

3. The attempt at a solution

In the augemented matrix form

[1 1 -1 1]
[1 2 k -1]
[1 k -1 -1]

R2-R1,R3-R1

[1 1 -1 1]
[0 1 (k+1) -2]
[0 (k-2) -(k+1) -2]

R3-(k-2)R2

[1 1 -1 1]
[0 1 (k+1) -2]
[0 0 -(k+1)(k+3) (2k-6)]

For a unique solution.

$-(k+1)(k+3) \neq 0$ so that $k \neq 1,3$

For infinite soltutions -(k+1)(k+3)=0 AND 2k-6=0
so that k=-1,-3 AND k=3

This doesn't make sense to me, as k can only be on value at a time, and if k=3, there will be no solution as the ranks of the augmented matrix and the initial matrix won't be the same.

SO where in my row reduction did I go wrong?

Last edited: Oct 25, 2008
2. Oct 25, 2008

### Staff: Mentor

You should have this:
[1 1 -1 1]
[0 1 (k+1) -2]
[0 (k-1) 0 -2]

To get the new 3rd row, you added (-1) times R1 to R3. I think you misread the entries in the 3rd row of your first matrix as 1 (k -1) ?? 1, when they actually are 1 k (-1) 1.

3. Oct 25, 2008

### rock.freak667

I got the 2nd matrix you put, and then interchanged row2 and row3.

Then did (k-1)R3-R2 to get

[1 1 -1 |-1]
[0 (k-1) 0 |-2]
[0 0 (k+1)(k-1)| -2(k-1)+1]

which would make no sense to me when I try to give the set of infinite solutions with paramters as it would mean that k should be either 1 or -1 AND -2(k-1)+1=0 at the same time,which can't occur.

4. Oct 26, 2008

### Staff: Mentor

So, clearly there's something going on if k = 1 or if k = -1.
If k = 1, the original system is:
x + y - z = 1
x + 2y + z = -1
x + y - z = 1

Notice that the 1st and 3rd equations are identical.
The augmented matrix is:

[1 1 -1 | 1]
[1 2 1 | -1]
[1 1 -1 | 1]

This row-reduces to
[1 0 -3 | 3]
[0 1 0 | -2]
[0 0 0 | 0]
Infinite number of solutions. Geometrically the two planes intersect in a line. What went wrong on your row-reduction is that when you multiplied R3 by (k - 1), you were multiplying by 0.

If k = -1, the original system looks like this:
x + y - z = 1
x + 2y - z = -1
x - y - z = 1

And the augmented matrix is like so:

[1 1 -1 | 1]
[1 2 -1 | -1]
[1 -1 -1 | 1]

This reduces to
[1 1 -1 | 1]
[0 1 0 | -2]
[0 0 0 | -4]

From the 3rd row, you can see that 0x + 0y + 0z = -4, which is impossible, so there are no solutions. Geometrically, the three planes don't intersect.

Finally, if k is any value other than 1 or -1, you get a unique solution for (x, y, z), with a different set of values for each value of k. Geometrically, for each value of k other than 1 or -1, the three planes intersect at a single point.

5. Oct 26, 2008

### HallsofIvy

Staff Emeritus
If there is no one value of k that makes all numbers in the last row 0, then there is no value of k that will give infinite solutions. Values of k that make all except the last number in the last row 0 give no solution. Values of k that make the next to last number in the last rwo non-zero give a unique solution.