# Please help: determining infinitely many solutions in a matrix

• kdubb22
In summary, by using the determinant and row reduction method, it is possible to find the value of k that would result in infinitely many solutions for the given system of equations. However, the process may require some trial and error in finding the correct value of k.
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!

kdubb22 said:
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]

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?

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!

I would approach this problem by first understanding the concept of infinitely many solutions in a matrix. In a system of equations, infinitely many solutions occur when there are more variables than equations, meaning that there are not enough equations to uniquely determine the values of all the variables. In this case, there can be an infinite number of combinations of values for the variables that satisfy the equations.

To determine if this system has infinitely many solutions, we can use the concept of determinant in a matrix. The determinant of a matrix is a value that can be calculated based on the coefficients of the variables in a system of equations. If the determinant is equal to 0, then the system has infinitely many solutions.

In this case, the determinant of the matrix would be 2k-9. To have infinitely many solutions, we need this determinant to be equal to 0. Therefore, we can solve for k by setting 2k-9=0, which gives us k=4.5. This means that if we choose k=4.5, the system will have infinitely many solutions.

In conclusion, to determine if a system of equations has infinitely many solutions, we can use the concept of determinant in a matrix. In this case, we can choose k=4.5 to make the system have infinitely many solutions.

## 1. What is a matrix and why is it important in determining solutions?

A matrix is a rectangular array of numbers or symbols arranged in rows and columns. It is important in determining solutions because it represents a system of equations and allows us to solve for unknown variables.

## 2. How can I determine if a matrix has infinitely many solutions?

A matrix has infinitely many solutions if the number of variables is greater than the number of equations, or if the equations are linearly dependent, meaning one equation can be obtained by combining other equations in the system.

## 3. What is the difference between no solution and infinitely many solutions in a matrix?

No solution means there is no set of values that satisfies all of the equations in the system. Infinitely many solutions means that there are multiple sets of values that satisfy all of the equations in the system.

## 4. How do I represent infinitely many solutions in a matrix?

Infinitely many solutions are represented by variables with no numerical values. For example, if we have two equations and three variables, we would represent infinitely many solutions as x = y + z, where x, y, and z can take on any value.

## 5. Can a matrix have both infinitely many solutions and no solutions?

No, a matrix cannot have both infinitely many solutions and no solutions. It can only have one or the other. If there are infinitely many solutions, there cannot be no solutions and vice versa.

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