Find values for k in a system of equations

[jberg074]

Hi, the following linear algebra question is causing trouble for me:

For which values of k does the following system of equations have a) a unique solution, b) infinite solutions, and c) no solutions?

[ 1 0 3 | 0 ]
[ 0 1 1 | 1 ]
[ -1 1 k | k ]

I know that having a unique solution implies a linear independence, infinite solutions implies a dependence with consistency, and no solutions implies no consistency.

I have tried getting this matrix into REF, but this didn't help me get any closer to the solution. Any pointers as to where I should start?

 

[gb7nash]

I have tried getting this matrix into REF, but this didn't help me get any closer to the solution. Any pointers as to where I should start?

What is the resulting matrix? (Or the closest matrix that you can get)

 

[jberg074]

Using the following steps:
R3=R3+R1, and
R3=R3-R2, I get:

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

Unless we have to get it in this form?

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

 

[gb7nash]

Ok, let's break this up into different cases:

a) In the third column, third row, what operation do we perform to obtain a 1? (assuming k+1 != 0)

b) What would (2+k) and (k-1) have to equal in order to get infinite solutions? Is this possible?

c) Think about this one. What would (2+k) (and consequently (k-1)) have to be equal to in order to have no solution? After you have the equation set up, what do you get for k?

 

[jberg074]

a) To get 1, I'm guessing we'd have to divide by (2+k)? Since (2+k)/(2+k)=1. However, this operation would have to carry through the whole row, right?

b) I'm not sure at all. They can't each equal zero, because that would yield k=-2 and k=1, which would turn it into an inconsistent system.

c) Would it have to be equal to (k-1)? If so, then all I can get from the equation is k = 3+k. This is also an inconsistency, which would indicate no solutions?

 

[gb7nash]

a) To get 1, I'm guessing we'd have to divide by (2+k)? Since (2+k)/(2+k)=1. However, this operation would have to carry through the whole row, right?

Correct. That is the next step.

b) I'm not sure at all. They can't each equal zero, because that would yield k=-2 and k=1, which would turn it into an inconsistent system.

You're partially correct. You want to set them both to 0. A row of 0's is the only way to get infinite solutions (You can verifying this by plugging in 0 for both and reducing the matrix even more) However, as you said, it's impossible to set both to 0, since you get k = -2 and k = 1. Therefore, there does not exist a k that gives infinite solutions.

c) Would it have to be equal to (k-1)? If so, then all I can get from the equation is k = 3+k. This is also an inconsistency, which would indicate no solutions?

No. For there to be an inconsistent system, you would need a system that's impossible to solve. For instance:

1 2 3 4
5 6 7 8
0 0 0 1

is an inconsistent system. For the third row, this essentially says that 0x₁ + 0x₂ + 0x₃ = 1, which is impossible. Looking at your problem now, what would (2+k) need to equal? Solve for k. What does (k-1) equal? Is this inconsistent?

 

[jberg074]

Allright, thank you so much for your help!