Matrices - finding critical values

[steelphantom]

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 .

Homework Equations

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 &amp; -1 &amp; 2 &amp; 1 \\ 0 &amp; k+1 &amp; -1 &amp; 2 \\ 0 &amp; 0 &amp; 1 &amp; 3 \end {array} \right]

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

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

 

[desA]

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

Have you learned about eigenvalues?

 

[steelphantom]

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

Have you learned about eigenvalues?

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

 

[desA]

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

 

[steelphantom]

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 &amp; -1 &amp; 2 &amp; 1 \\ 0 &amp; k+1 &amp; -1 &amp; 2 \\ 0 &amp; 0 &amp; 1 &amp; 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 &amp; -1 &amp; 2 &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; -2 \\ 0 &amp; 0 &amp; 1 &amp; 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?

 

[desA]

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.

 

[steelphantom]

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

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

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

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

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

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

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

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 &amp; -1 &amp; 2 &amp; 1 \\ 0 &amp; k+1 &amp; -1 &amp; 2 \\ 0 &amp; 0 &amp; 1 &amp; 3 \end {array} \right]
If the "pivot", k+ 1, is 0 then you cannot have a single solution.

 

[steelphantom]

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 &amp; -1 &amp; 2 &amp; 1 \\ 0 &amp; k+1 &amp; -1 &amp; 2 \\ 0 &amp; 0 &amp; 1 &amp; 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!

 

[radou]

Hence, it's always necessary to create a 'discussion' before dividing with k+1, as in post #7, since, for k = -1 the term is undefined.