Finding values of k in a system of linear equations.

In summary: For k = -5, how many solutions does the linear system have?For k ≠ -5, how many solutions does the linear system have?In summary, the system of linear equations given has a unique solution for all values of k except k = -5. For k = -5, the system is undefined and has no solution.
  • #1
thatguythere
91
0

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.
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
thatguythere said:
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.
 
  • #6
thatguythere said:
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.
 
  • #7
Yes, your matrix was wrong.
thatguythere said:
2x+(k-1)y = 6
3x+(2k+1)y= 9
Below, there is no gain in expanding the terms with k.
thatguythere said:
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.
thatguythere said:
[2 y -1 6] 1/2R1
[3 2y 1 9]
 
  • #8
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]
 
  • #9
thatguythere said:
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?

thatguythere said:
[1 (k-1)/2 3]
[0 1 0]
 
  • #10
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.
 
  • #11
thatguythere said:
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?
 
  • #12
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?
 
  • #13
thatguythere said:
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?
thatguythere said:
, 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?
 
  • #14
The "it" I was referring to was the division if k = -5 that Dick was pointing out.
 
  • #15
Also, I'm really uncertain what you were attempting to demonstrate with my second quote.
 
  • #16
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.
 
  • #17
For any value of k ≠ 5, the second equation will not equal 0.
 
  • #18
No worries, we are bearing with you. Can you answer the questions I asked in post #13?
 
  • #19
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...
 
  • #20
thatguythere said:
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?
 
  • #21
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?
 
  • #22
thatguythere said:
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?
 
  • #23
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.
 
  • #24
thatguythere said:
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.
 
  • #25
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.
 

FAQ: Finding values of k in a system of linear equations.

1. How do you solve for k in a system of linear equations?

The first step in solving for k is to set up the system of linear equations. This can be done by writing out each equation with the variables and their corresponding coefficients. Then, use algebraic methods such as substitution or elimination to solve for k. Once k is isolated, plug it back into the equations to check for consistency.

2. Can you use matrices to find the value of k in a system of linear equations?

Yes, matrices can be used to solve for k in a system of linear equations. This method involves setting up the equations in matrix form and using row operations to reduce the matrix to an upper triangular form. The value of k can then be easily determined from the reduced matrix.

3. Is there a shortcut or trick to finding the value of k in a system of linear equations?

There is no one specific shortcut or trick to finding the value of k in a system of linear equations. However, some methods such as substitution or elimination may be faster depending on the given equations. It is important to understand the concepts and apply them appropriately in order to solve for k efficiently.

4. How many solutions are there for a system of linear equations when solving for k?

The number of solutions for a system of linear equations when solving for k depends on the number of equations and variables in the system. If there are the same number of equations and variables, there will be a unique solution for k. If there are more equations than variables, the system may have no solution or infinitely many solutions.

5. Can you use a graph to find the value of k in a system of linear equations?

Graphing can be a helpful tool in visualizing a system of linear equations. However, it is not a reliable method for finding the value of k. Graphs can give an approximate solution, but algebraic methods should be used to find the exact value of k.

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