Please help: determining infinitely many solutions in a matrix

[kdubb22]

Given the system of equations in x and y:
2x-3y=12
3x+ky=10
Could you choose k so the system had infinitely many solutions? If so, give k. If not, why not?

I'm pretty sure that it CAN have infinitely many solutions because the number of equations is equal to the number of variables in the system. I'm just not sure how to find the value of k that would make this true. If anyone has any helpful ideas I'd REALLY appreciate it!

 

[chiro]

Given the system of equations in x and y:
2x-3y=12
3x+ky=10
Could you choose k so the system had infinitely many solutions? If so, give k. If not, why not?

I'm pretty sure that it CAN have infinitely many solutions because the number of equations is equal to the number of variables in the system. I'm just not sure how to find the value of k that would make this true. If anyone has any helpful ideas I'd REALLY appreciate it!

Hello kdubb22 and welcome to the forums.

A quick way of determining if a system of equations with n equations and n variables has a unique solution is if the determinant is non-zero.

However you have to be careful about using the determinant result.

In a linear system like this, you can have three outcomes: unique solution, infinitely many solutions or no solutions.

The unique case is straightforward. However for testing if something has no solutions you will end with a row reduced system that looks something like this:

[ 1 3 | 2]
[ 0 0 | 1]

The bottom row is full of zeros, but the right hand side is non-zero: in this case the system has no solutions. If however after you row reduce your linear system and you get a corresponding zero on the right hand side, and you have a non-zero determinant, you know that you must have infinitely many solutions.

As for your question, your best bet is to start from an augmented matrix description of your linear system, and then row reduce. After you put it in this form, you will be able to find the right k that will give you the unique, infinite, or null solution case.

So to start you off here is your starting matrix:

[2 -3 | 12]
[3 k | 10]

 

[kdubb22]

Thank you for your response! Yeah the first part of the problem was to find the value of k that made the system have no solution. I found it to be k=-4.5.
This is where I'm stumped. I start working through the matrix but when I get to the step where I have to convert k into 1 I don't know how to continue. This is as far as I get:

[2 -3 l 12] 1/2 R1 ->[1 -3/2 l 6] R2+(-3)R1 -> [1 -3/2 l 6] 1/k R2-> [1 -3/2 l 6]
[3 k l 10] [3 k l 10] [0 k l -26] [0 0 l -26/k]

I hope that makes sense. I actually get a further than this but its hard to type it in because its abnormal. I didn't type in my actual work btw steps but if you're confused I will. So obviously instead of making k a 1, I would divide the row by a number that makes k 0 because I want to find its value in an infinitely many solution. After the last step I showed I would then make the -3/2 in row 1 equal 0 by multiplying row 2 by 3/2 and adding that to row 1 (R1 + 3/2R2). After all that, I can't complete the problem because I can't calculate the solution. The solution to R2 ends up being (3/2)-26/k, so 3/2 times -26 divided by k.

So how am I supposed to find the number that will make -26 do all those things and end up equalling 0? do i just guess and randomly plug in numbers?

 

[kdubb22]

Wow my row reductions posted really off. If you just space over the bottom ones, each bracket set goes with its corresponding one above. Sorry about that!