Solutions to systems of linear equations

[Robb]

Homework Statement

Determine whether of not each system has a nonzero solution.

x + 3y - 2z = 0
2x - 3y + z = 0
3x - 2y + 2z = 0

Homework Equations

The Attempt at a Solution

Using a ti-84 I put the matrix in rref and I get a nonzero solution for x and y and zero for z but when I find the determinant of the coefficient matrix I get zero, indicating there is not a unique solution. I'm unsure what to make of this. Please advise.

1 0 .727
rref= 0 1 -.909
0 0 0

det = 1 3 -2
2 -3 1 = 0
3 -2 2

 

[Mark44]

Homework Statement

Determine whether of not each system has a nonzero solution.

x + 3y - 2z = 0
2x - 3y + z = 0
3x - 2y + 2z = 0

Homework Equations

The Attempt at a Solution

Using a ti-84 I put the matrix in rref and I get a nonzero solution for x and y and zero for z but when I find the determinant of the coefficient matrix I get zero, indicating there is not a unique solution. I'm unsure what to make of this. Please advise.

1 0 .727
rref= 0 1 -.909
0 0 0

det = 1 3 -2
2 -3 1 = 0
3 -2 2

Check your calculation of the determinant. I did it two ways and got -17 both ways.

I am also suspect of what you entered into your calculator. Your reduced matrix isn't right, at least based on the equations you wrote. Also, your reduced matrix does not mean that z = 0.

 

[Robb]

My apologies, mixed up my problems

rref = 1 0 0 0
0 1 0 0
0 0 0 0

det = -17

 

[Mark44]

My apologies, mixed up my problems

rref = 1 0 0 0
0 1 0 0
0 0 0 0

Still not right. What this says is that x = 0, y = 0, and z is arbitrary.

Also, you have too many columns. You don't need to include the constants on the right sides of the equations.

det = -17

 

[Robb]

I realize I don't need them for the determinant but why not for the augmented matrix? Is that not in rref with 4 columns? Also, my head must be up my you know what today. Here is the correct rref (finally):
1 0 0 0
0 1 0 0
0 0 1 0

 

[Mark44]

I realize I don't need them for the determinant but why not for the augmented matrix? Is that not in rref with 4 columns? Also, my head must be up my you know what today. Here is the correct rref (finally):
1 0 0 0
0 1 0 0
0 0 1 0

Yes, that's more like it.
For your system of equations, you don't need an augmented matrix, because all the constant terms are zero. It doesn't hurt, but it isn't necessary, since no row operation can possibly change the values on the right sides of the equations.

You would need an augmented matrix if one or more of the constant terms happened to be nonzero.

 

[Robb]

Gotcha! But, what is this telling me? x, y, z are all zero yet the determinant is -17. Doesn't the determinant say that I should have a unique solution? I guess I'm not sure what the determinant is telling me here.

 

[Ray Vickson]

Homework Statement

Determine whether of not each system has a nonzero solution.

x + 3y - 2z = 0
2x - 3y + z = 0
3x - 2y + 2z = 0

Homework Equations

The Attempt at a Solution

Using a ti-84 I put the matrix in rref and I get a nonzero solution for x and y and zero for z but when I find the determinant of the coefficient matrix I get zero, indicating there is not a unique solution. I'm unsure what to make of this. Please advise.

1 0 .727
rref= 0 1 -.909
0 0 0

det = 1 3 -2
2 -3 1 = 0
3 -2 2

If you are serious about wanting to learn the subject I suggest you lay aside your calculator and just do the problem by hand, maybe using the calculator to add, subtract, multiply and divide numbers, but nothing else. After you have grasped the material (which requires practice to do properly), then you can let the calculator do some more of the work for you.

 

[Mark44]

Gotcha! But, what is this telling me? x, y, z are all zero yet the determinant is -17. Doesn't the determinant say that I should have a unique solution? I guess I'm not sure what the determinant is telling me here.

x = y = z = 0 is the unique solution. This solution is consistent with the determinant being -17 (or in fact any nonzero value).

 

[Dick]

Gotcha! But, what is this telling me? x, y, z are all zero yet the determinant is -17. Doesn't the determinant say that I should have a unique solution? I guess I'm not sure what the determinant is telling me here.

Yes, the determinant being nonzero tells you there is a unique solution. $x=0, y=0, z=0$ is that unique solution.