# Matrices - finding critical values

1. Homework Statement

Find k so that the system has exactly one solution.

$$\left{ \begin {array} {rcl} x - y + 2z = 1 \\ -x + y - z = 2 \\ x + ky + z = 3 \end {array} \right .$$

2. Homework Equations

3. The Attempt at a Solution

Ok, so I create an augmented matrix here:
$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ -1 & 1 & -1 & 2 \\ 1 & k & 1 & 3 \end {array} \right]$$

Here I do the following: R2+R1 and R3-R1.
$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & k+1 & -1 & 2 \end {array} \right]$$

R2 <=> R3.
$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right]$$

(1/k+1)R2.
$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 1 & -1/(k+1) & 2/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right]$$

At this point, I'm stuck. How do I know which k will give the system exactly one solution?

Last edited:

Related Precalculus Mathematics Homework Help News on Phys.org
Is your first equation correct? It does not match with your matrix.

Is your first equation correct? It does not match with your matrix.

Sorry; I just fixed it. Everything should be correct now. We haven't learned about eigenvalues yet.

Would reducing your matrix all the way to the identity matrix help?

Would reducing your matrix all the way to the identity matrix help?
We haven't even gone over the identity matrix yet. I think I may have come up with a solution:

$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right]$$

OK, in this form, if k = -1, the second and third rows would be contradictory (i.e. no solution):

$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & 3 \end {array} \right]$$

However, if k were anything else, there should be exactly one solution, right? So is the answer all real numbers except -1, or am I totally off-base here?

Looks like 'a' solution. How would you know if there were others, or not?

The identity matrix has ones on the diagonal, & zeros everywhere else. Basically continue your reduction method, or back-substitute from where you are, to obtain complete expressions for x,y,z in terms of k & then make a decision.

This will give you the full solution for x, y, z & you could then re-check your current solution.

Ok, I've reduced my matrix to the identity matrix. Here are my steps:

$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right]$$

$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & 0 & 5 \\ 0 & 0 & 1 & 3 \end {array} \right]$$

$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right]$$

$$\left[ \begin {array} {ccc|r} 1 & -1 & 0 & -5 \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right]$$

$$\left[ \begin {array} {ccc|r} 1 & 0 & 0 & 5/(k+1) - 5 \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right]$$

After all these steps, my final identity matrix looks like this:
$$\left[ \begin {array} {ccc|r} 1 & 0 & 0 & -5k/(k+1) \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right]$$

So now it looks like the possible values of k are all real numbers except -1 and 0. Is this correct?

HallsofIvy
Homework Helper
Why could k not be 0? Isn't x= 0, y= 5, z= 3 an acceptable solution?

The matrix equation, Ax= b, will have a single solution as long as det(A) is not 0. Since there is only one k in the matrix, det(A)= 0 will be a linear equation in k. There should be only one value of k for which this system does not have a single solution.

You could have seen that back in your original reduction:
$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right]$$
If the "pivot", k+ 1, is 0 then you cannot have a single solution.

Why could k not be 0? Isn't x= 0, y= 5, z= 3 an acceptable solution?

The matrix equation, Ax= b, will have a single solution as long as det(A) is not 0. Since there is only one k in the matrix, det(A)= 0 will be a linear equation in k. There should be only one value of k for which this system does not have a single solution.

You could have seen that back in your original reduction:
$$\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right]$$
If the "pivot", k+ 1, is 0 then you cannot have a single solution.
Wow, how dumb of me... Of course k=0 is a solution! For some reason I was looking at it like 1=0. So it looks like the only value that won't work is -1, as I thought earlier. Thanks!