# Infinite solution to system with no free variables?

• CynicusRex
In summary, the assignment was to find all values of k (in R) for which the system of linear equations has 0 solutions, 1 solution, or infinite solutions with a certain number of free variables. Through row reduction, it was determined that the system is inconsistent when k = 0 and has infinite solutions and one free variable when k = -1. However, when plugging in k = -1, the solutions obtained do not match the given system. This is because in the row reduction process, dividing by k is illegal as it is possible for k to be 0. Therefore, the final solution is incorrect and the system actually has an infinite number of solutions when k = -1.
CynicusRex
Gold Member
Member warned about posting with no template, and with incomplete problem statement

## Homework Statement

The assignment is to find all values of k (in R) for which the system has 0 solutions, 1 solution and infinite solutions. If there are infinite solutions, find the amount of free variables.

The system of linear equations:
kx + (k+1)y + z = 0
kx + y + (k+1)z = 0
2kx + y + z = k+1

Now k=-1

## The Attempt at a Solution

So I was correct that this system is undefined when k=0.
Then I thought this system had infinite solution when k =/= 0. But here I'm wrong.
Apparently if k=-1 then there are infinite solution with one free variable. However, when I plug in -1 then all solutions are -1/2.

I don't understand why this is infinite.

I did my row operations as practise.

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The assignment was to find all values of k (in R) for which the system has 0 solutions, 1 solution and infinite solutions. If there are infinite solutions, find the amount of free variables.

So I was correct that this system is undefined when k=0.
The proper term is that the system is inconsistent; i.e., there are no solutions.
Then I thought this system had infinite solution when k =/= 0. But here I'm wrong.
At this point, you can assume that ##k \ne 0##, so continue row reduction from here. The possibilties now are that for some value(s) of k, the system has an infinite number of solutions, and for some other value(s), the system has a single solution.
Apparently if k=-1 then there are infinite solution with one free variable. However, when I plug in -1 then all solutions are -1/2.
That's not what I get. If k = -1, there are an infinite number of solutions. Check your arithmetic.

Also, it wasn't clear until well into your hand-written work that your matrix is an augmented matrix, and that the system is three equations in three unknowns. You should have included the system of equations in the problem statement of the template (which you deleted).
I don't understand why this is infinite.

Now k=-1

I did my row operations as practise.

Mark44 said:
That's not what I get. If k = -1, there are an infinite number of solutions. Check your arithmetic.

Well that's what I don't understand. In the row reduced echelon form there is only one k left. If k=-1 then you get -1/2.
I do realize if k=-1 from the start, then you'll get a system of infinite solutions, but how do you figure out -1? I shouldn't have row reduced the matrix?

Well that's what I don't understand. In the row reduced echelon form there is only one k left. If k=-1 then you get -1/2.
I do realize if k=-1 from the start, then you'll get a system of infinite solutions, but how do you figure out -1? I shouldn't have row reduced the matrix?
Check back through your working during the row reduction. Look for each place you divided through by an expression with k in it. What do you need to consider?

By row reduction, I get the augmented matrix down to this:
##\begin{bmatrix}k & k + 1 & 1 & | &0\\ 0 & k & -k & | & 0 \\ 0 & 0 & 2k^2 + 2k & | & -k^2 - k \end{bmatrix}##

Where we're going to have trouble is in the last row. The third term in that row is 0 if k = 0 or k = -1 (which you have already said).

Go back to the original matrix, and plug in k = 0, and row reduce this matrix. This should tell you that the system is inconsistent if k = 0. Again, using the original matrix, plug in k = -1, and row reduce this matrix. You should see that you have one free variable in this situation.

Mark44 said:
By row reduction, I get the augmented matrix down to this:
##\begin{bmatrix}k & k + 1 & 1 & | &0\\ 0 & k & -k & | & 0 \\ 0 & 0 & 2k^2 + 2k & | & -k^2 - k \end{bmatrix}##

Where we're going to have trouble is in the last row. The third term in that row is 0 if k = 0 or k = -1 (which you have already said).

Go back to the original matrix, and plug in k = 0, and row reduce this matrix. This should tell you that the system is inconsistent if k = 0. Again, using the original matrix, plug in k = -1, and row reduce this matrix. You should see that you have one free variable in this situation.

I get something a bit different: if the equations are written as (just by reordering the variables and the equations)
y + k*x + (k+1)*z = 0
y + 2*k*x + z = k+1
(k+1)*y + k*x + z = 0
the augmented matrix is
$$\left|\begin{matrix} y & x & z & | &\text{RHS} \\ \hline 1 & k & k+1 &|& 0 \\ 1 & 2k & 1 &|& k+1 \\ k+1 & k & 1 &|& 0 \end{matrix}\right|$$
After row-reducing on columns 1 and 2 we have
$$\left|\begin{matrix} y & x & z& | & \text{RHS} \\ \hline 1 & k & k+1 &|& 0 \\ 0 & k & -k &|& k+1 \\ 0 & 0 & -2k(k+1) &|& k(k+1) \end{matrix} \right|$$
This is a row-reduced form, but not a RREF.

Note that when ##k = 0## the second row gives an inconsistent equation ##0y+0x+0z = 1##. The cases ##k = -1## and ##k \neq -1,0## can be treated similarly, starting from the row-reduced form.

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haruspex said:
Check back through your working during the row reduction. Look for each place you divided through by an expression with k in it. What do you need to consider?

That I cancel out / reduce the possibilities for finding free variables?

That I cancel out / reduce the possibilities for finding free variables?
No. In algebra, what is it illegal to divide by?

haruspex said:
No. In algebra, what is it illegal to divide by?

Zero. But dividing by k is not dividing by zero?

As far as I know, rewriting the system in a different form does not change the original system? But apparently it does because when k=-1 in the first image you get a free variable and infinite solutions, whereas in the second image you get one solution. {-1/2,-1/2,-1/2}

Okay, the question I must ask myself is why. I'll try and figure it out.
=

PS It has been a real long time since I came into contact with math and went further than memorizing formulas. I really want to understand why things work now, so sorry for asking stupid questions.

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Zero. But dividing by k is not dividing by zero?
It is when k is zero.
Given an equation like x f(x)= x g(x) you should argue " either x=0 or ...".

CynicusRex
I did not realize it wasn't allowed to divide by an unknown which could be zero. I did so many wrong calculations doing that.

I did not realize it wasn't allowed to divide by an unknown which could be zero. I did so many wrong calculations doing that.

That is not quite right: you ARE allowed to divide by k as long as k ≠ 0. So you can split things up into cases, for example:
Case (1): k = 0 ---> proceed.
Case (2): k ≠ 0 ---> proceed. Dividing by k is OK in this case.
In Case (2) you may run into further problems if k = -1, so you can have:
Case (2a) k = -1 ---> proceed; and
Case (2b) k ≠ -1 (and, f course, k ≠ 0). Dividing by k and k+1 are both OK in this case.

Sorry, I had to be more specific. I meant when there is a leading k in the matrix, just leave it as k. Don't multiply the row by 1/k just to get a one. In all my exercises I did stuff like that because Wolfram|Alpha (I use it as verification) did it also. So I applied it everywhere.

Btw, don't you mean: Case (1): k = 0 ---> don't proceed.

Sorry, I had to be more specific. I meant when there is a leading k in the matrix, just leave it as k. Don't multiply the row by 1/k just to get a one. In all my exercises I did stuff like that because Wolfram|Alpha (I use it as verification) did it also. So I applied it everywhere.

Btw, don't you mean: Case (1): k = 0 ---> don't proceed.

No, I meant "proceed", but not necessarily in the same way you would if k were nonzero. There can be several ways to proceed, depending on conditions.

## What does it mean when a system has an infinite solution?

When a system has an infinite solution, it means that there are an infinite number of solutions that satisfy the equations in the system. This can occur when the equations are redundant or when the equations are not linearly independent.

## What is a system with no free variables?

A system with no free variables is a system of equations where all variables have a unique value and there are no variables that can be chosen arbitrarily. This means that the system has a unique solution.

## How can a system have an infinite solution if all variables have a unique value?

A system can have an infinite solution if the equations are not linearly independent. This means that one or more equations in the system can be derived from the other equations, making them redundant. As a result, there are an infinite number of solutions that satisfy the system of equations.

## What is the difference between an inconsistent system and a system with an infinite solution?

An inconsistent system is one that has no solution, meaning that the equations are contradictory and cannot be satisfied simultaneously. On the other hand, a system with an infinite solution has an infinite number of solutions that satisfy the equations. This occurs when the equations are redundant or not linearly independent.

## How can I determine if a system has an infinite solution?

You can determine if a system has an infinite solution by solving the system of equations and checking if there are any redundant or linearly dependent equations. If so, then the system has an infinite solution. You can also check for an infinite solution by examining the coefficient matrix and seeing if it has a row of zeros or a row that is a multiple of another row.

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