Finding values of k in a system of linear equations.

[thatguythere]

Homework Statement

For what value(s) of k does the system of linear equations below have a unique solution?

2x+(k-1)y = 6
3x+(2k+1)y= 9

Homework Equations

The Attempt at a Solution

I'm uncertain how to even go about tackling this problem. Given how all the other problems beforehand were dealing with matrices, I suppose I should start there, however the variable k throws me for a loop.

2x+ky-y = 6
3x+2ky+y = 9

[2 y -1 6] 1/2R1
[3 2y 1 9]

[1 y/2 -1/2 3] -3R1+R2
[3 2y 1 9]

[1 y/2 -1/2 3] 2R2
[0 y/2 5/2 0]

[1 y/2 -1/2 3]
[0 y 5 0] 1/yR2

[1 y/2 -1/2 3]
[0 1 5/y 0]

That is as far as I get in my blind attempt before I get stuck and no longer have any idea what I am doing. :) Any help would be greatly appreciated.

 

[SteamKing]

Try re-writing the two equations in matrix form. Remember to include k in the matrix.
What must be true about the matrix equations in order to have a unique solution?

 

[thatguythere]

Alright, so I will assume from your response that my matrix was completely wrong. I am uncertain what you mean by "Try re-writing the two equations in matrix form." in that case, isn't that what I did? Should I not distribute beforehand?

 

[thatguythere]

Also, I found this in my textbook. a. If the number of equations is greater than or equal to the number of variables
in a linear system, then one of the following is true:
i. The system has no solution.
ii. The system has exactly one solution.
iii. The system has infinitely many solutions.
b. If there are fewer equations than variables in a linear system, then the system
either has no solution or it has infinitely many solutions.

 

[Dick]

Alright, so I will assume from your response that my matrix was completely wrong. I am uncertain what you mean by "Try re-writing the two equations in matrix form." in that case, isn't that what I did? Should I not distribute beforehand?

The two rows of your matrix are [2,(k-1),6] and [3,(2k+1),9]. There is no y in the matrix and there are three columns, not four.

 

[Dick]

Also, I found this in my textbook. a. If the number of equations is greater than or equal to the number of variables
in a linear system, then one of the following is true:
i. The system has no solution.
ii. The system has exactly one solution.
iii. The system has infinitely many solutions.
b. If there are fewer equations than variables in a linear system, then the system
either has no solution or it has infinitely many solutions.

You have two equations and two unknowns. k isn't really an 'unknown'. You just want to know how it's value affects the possibility for finding x and y.

 

[Mark44]

Yes, your matrix was wrong.

2x+(k-1)y = 6
3x+(2k+1)y= 9

Below, there is no gain in expanding the terms with k.

2x+ky-y = 6
3x+2ky+y = 9

What you have below is what is wrong. The coefficient of y in the first row is (k - 1), and similar for the 2nd row.

[2 y -1 6] 1/2R1
[3 2y 1 9]

 

[thatguythere]

Alright, let's see how badly I can get this wrong.

[2 (k-1) 6]
[3 (2k+1) 9] 1/2R1

[1 (k-1)/2 3]
[3 (2k+10 9] -3R1+R2

[1 (k-1)/2 3]
[0 (k+5)/2 0] 2/(k+5)R2

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

 

[Mark44]

Alright, let's see how badly I can get this wrong.

[2 (k-1) 6]
[3 (2k+1) 9] 1/2R1

[1 (k-1)/2 3]
[3 (2k+10 9] -3R1+R2

[1 (k-1)/2 3]
[0 (k+5)/2 0] 2/(k+5)R2

What if k = -5?

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

 

[SteamKing]

You are just interested in the matrix of coefficients.

Your equations have the form:

[A] * (x,y)^T = (6,9)^T

The RHS is irrelevant to determining the value of k which permits a unique solution to the system.

 

[Dick]

Alright, let's see how badly I can get this wrong.

[2 (k-1) 6]
[3 (2k+1) 9] 1/2R1

[1 (k-1)/2 3]
[3 (2k+1) 9] -3R1+R2

[1 (k-1)/2 3]
[0 (k+5)/2 0] 2/(k+5)R2

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

That's pretty good. But notice you can't divide by (k+5) if k=(-5). What do you say about the number of solutions if k=(-5) vs the number of solutions if k is not equal to -5?

 

[thatguythere]

Well, I suppose if k = -5 then it would be undefined, whereas if it is not equal to -5 there could be infinite solutions? So, where does that leave me? There are no values for which this system of linear equations has a unique solution?

 

[Mark44]

Well, I suppose if k = -5 then it would be undefined

See if you can write an explanation with using "it" - I don't know what you're referring to.

If k = -5, your matrix looks like this:

$$ \begin{bmatrix} 1 & -3 & | & 3 \\ 0 & 0 & | & 0\end{bmatrix}$$

How many solutions do you have with this augmented matrix?

What's the situation if k ≠ -5?

, whereas if it is not equal to -5 there could be infinite solutions? So, where does that leave me? There are no values for which this system of linear equations has a unique solution?

 

[thatguythere]

The "it" I was referring to was the division if k = -5 that Dick was pointing out.

 

[thatguythere]

Also, I'm really uncertain what you were attempting to demonstrate with my second quote.

 

[thatguythere]

I realize I may seem impossible, for this I apologize, however I am attempting to figure this all out by myself with no aid from a professor. Please bear with me.

 

[thatguythere]

For any value of k ≠ 5, the second equation will not equal 0.

 

[Mark44]

No worries, we are bearing with you. Can you answer the questions I asked in post #13?

 

[thatguythere]

For the augmented matrix you wrote I can see plenty of solutions (if it indeed it equates to x-3y=3). For instance, 6-3(1)= 3, or 3-3(0)= 3, 9-3(2)= 3, etc...

 

[Mark44]

For the augmented matrix you wrote I can see plenty of solutions (if it indeed it equates to x-3y=3). For instance, 6-3(1)= 3, or 3-3(0)= 3, 9-3(2)= 3, etc...

That's when k = -5. What happens if k ≠ -5?

 

[thatguythere]

Well let's see. First off, the second equation will not equal 0 0 0. Now, I think I'll choose a number arbitrarily and see what happens. So k = 4.

1 3/2 3
0 9/2 0

x+3/2y = 3
9/2y = 0

I can see that for the first equation, x = 3/2, y = 1 , x = 0, y = 2, x = 3, y = 0 would also work, safe to say many would. For the second equation, only y = 0 satisfies it yes? What is this telling me? If I can solve it with k = -5, and also with k = 4, then is that to say we have no unique solution?

 

[Dick]

Well let's see. First off, the second equation will not equal 0 0 0. Now, I think I'll choose a number arbitrarily and see what happens. So k = 4.

1 3/2 3
0 9/2 0

x+3/2y = 3
9/2y = 0

I can see that for the first equation, x = 3/2, y = 1 , x = 0, y = 2, x = 3, y = 0 would also work, safe to say many would. For the second equation, only y = 0 satisfies it yes? What is this telling me? If I can solve it with k = -5, and also with k = 4, then is that to say we have no unique solution?

If only y=0 satisfies the second equation then only x=3 satisfies the first one. So you have a unique solution in this case. In the case k=4 there is a unique solution. In the case k=(-5) there is NOT a unique solution. They want to know for what values of k there is a unique solution. Do you have to check e.g. k=3, k=2, k=1? Or can you figure out what will happen?

 

[thatguythere]

No, because no matter what we are only affecting y and y must = 0 for the second equation to be satisfied, therefore x must = 3, so k has a unique solution for all values not being -5.

 

[Dick]

No, because no matter what we are only affecting y and y must = 0 for the second equation to be satisfied, therefore x must = 3, so k has a unique solution for all values not being -5.

That's it. If the second equation has a nonzero number multiplying the y, there is a unique solution.

 

[thatguythere]

Thanks a lot. That was the one problem giving me headaches. Now I can hand in an entire unit tomorrow with confidence. Much obliged gentlemen.