The answer to this question is: Homogenous System: Ax=b Must Have Many Solutions

[shiri]

If a homogenous system Ax = 0 has infinitely many solutions, then for a non-zero vector b, the associated system Ax = b ____ have _______In my assignment, the answer I wrote on this question is must have many solutions. However, what I got is wrong.

Can anybody tell me why it is wrong?

 

[aPhilosopher]

if Ay = b, then what can you say about any vector of the form x + y where Ax = 0?

 

[shiri]

if Ay = b, then what can you say about any vector of the form x + y where Ax = 0?

Well I am assuming there has to be many solutions since Ax=0 gives infinite solutions.

Plus, no solution or/and one solution sounds less appropriate than many solutions at the moment I answer the question.

 

[aPhilosopher]

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Also, I might be being pedantic but I would say many is a broader term than infinite. I think that I've seen books using 'many' though so if your book or teacher does, just ignore me.

 

[shiri]

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Also, I might be being pedantic but I would say many is a broader term than infinite. I think that I've seen books using 'many' though so if your book or teacher does, just ignore me.

Well I don't understand why many solutions is wrong since this tell me how many variables and how many unknowns.

So, I ran out of options. Please help me.

 

[aPhilosopher]

Must is surely wrong as the example shows. The homogeneous equation has infinite solutions. Can you find a solution to the non-homogeneous equation?

Like I said, I was probably just being pedantic with the whole many/infinite thing.

 

[shiri]

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Also, I might be being pedantic but I would say many is a broader term than infinite. I think that I've seen books using 'many' though so if your book or teacher does, just ignore me.

Teacher gave me these choices:

A. may have exactly one solution
B. must have many solutions
C. must have either one solution or no solution
D. may have no solution
E. need not satisfy any of the above

 

[aPhilosopher]

Well, you shouldn't have chosen B. Work my example to find out why.

 

[shiri]

Well, you shouldn't have chosen B. Work my example to find out why.

I would say many solutions and no solutions. B and D

 

[aPhilosopher]

B is wrong in general. Do you see why? If you want, we can talk about when it's right.

 

[shiri]

B is wrong in general. Do you see why? If you want, we can talk about when it's right.

So B is wrong in any possible answers?

 

[aPhilosopher]

It's wrong without conditions that weren't assumed in the question. Have you studied dimension yet?

The key thing here is the use of the word "must". I've given you an example already that doesn't so think about it until you realize that it doesn't "must" have infinite solutions.

 

[shiri]

It's wrong without conditions that weren't assumed in the question. Have you studied dimension yet?

The key thing here is the use of the word "must". I've given you an example already that doesn't so think about it until you realize that it doesn't "must" have infinite solutions.

So the correct answer is D only? May have no solutions?

 

[shiri]

It's wrong without conditions that weren't assumed in the question. Have you studied dimension yet?

The key thing here is the use of the word "must". I've given you an example already that doesn't so think about it until you realize that it doesn't "must" have infinite solutions.

the answer is C? is it?

 

[aPhilosopher]

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

 

[shiri]

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$


$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

I see x has a solution and y has no solution.

Am I right?

 

[aPhilosopher]

The second one has no solution, you are correct. How many solutions does the first one have again?

 

[shiri]

The second one has no solution, you are correct. How many solutions does the first one have again?

the first one has one solution?

 

[aPhilosopher]

No. (1 0) is a solution. (1 1) is a solution. (1 45/77) is also a solution.

How many solutions does $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ have? You can take anyone of them, add it to (1 0) and get a solution to

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Right?

 

[shiri]

No. (1 0) is a solution. (1 1) is a solution. (1 45/77) is also a solution.

How many solutions does $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ have? You can take anyone of them, add it to (1 0) and get a solution to

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Right?

so are you saying that there must have either one solution or no solution?

 

[aPhilosopher]

I have to go to work now. Maybe somebody else can help you.

 

[aPhilosopher]

Take an homogeneous system, Ax = 0. For example in the following equation,


$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

The matrix $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ would be called A.

The vector $$\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}$$ would be called x.

We want to find all x₁, x₂ such that Ax = 0. Towards that end, expand it out into

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.

For any real number x₂ that we choose, $$x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

This means that $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ so that the value of x₂ doesn't affect the value of Ax at all.

So to solve the above equation, we need to solve 1x₁ = 0. But whenever we have two numbers a and b such that ab = 0, we know that either a or b must equal 0. Because we know that 1 is not equal to 0, we can conclude that x₁ must be.

This means that $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ when $$\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ x_{2} \end{bmatrix}$$ where x₂ is any real number that we choose.

We have shown that any solution of Ax = 0 must be of the form $$\begin{bmatrix} 0 \\ x_{2} \end{bmatrix}$$.


In other words, there is one solution for every real number x₂.

Non-Homogeneous System

Let's now consider a non-homogeneous system Ax = r where r is a non-zero vector. Let's take our example as,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ r_{1} \end{bmatrix}$$

Again, we expand this out into,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.


Again, $$x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.

So $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$.

Now, 0x₁ = 0. This has the consequence that we can only solve the equation if r₂ = 0 because that is all we are ever going to get from A. We must also have 1x₁ = r₁. So we can only solve the equation if

$$r = \begin{bmatrix} r_{1} \\ 0 \end{bmatrix}$$

Now, as you know, A(v₁ + v₂) = Av₁ + Av₂ so if Av₁ = r and Av₂ = 0 then A(v₁ + v₂) = Av₁ + Av₂ = r + 0 = r

This means that the solution set is all vectors of the form

$$\begin{bmatrix} r_{1} \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ x_{2} \end{bmatrix}$$

where r₁ is determined by r and x₂ is free.

In other words,

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ r_{2} \end{bmatrix}$$

has solutions only if r₂ = 0. In that case, we have

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} r_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ 0 \end{bmatrix}$$ for all real numbers, x₂.

Four more good matrices to solve and really look at would be $$\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} , \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

You probably want to use row reduction or some other method for more complicated systems but you should look at those three like we've looked at the one above.