# Homogeneous, underdetermined equation system

• MHB
• ToastIQ
If this were a $0$, you would have to exchange the second and third row, but thankfully the $-1$ makes this unnecessary.)
ToastIQ
Hi!

Just started with linear algebra Could someone help me with this problem?

$$\displaystyle 2x_1 + x_2 - x_3 + 3x_4 - 3x_5 = 0\\ 3x_1 + 2x_2 + x_3 + 2x_4 + 2x_5 = 0\\ -4x_1 + 3x_2 + 2x_3 + x_4 - 4x_5 = 0$$

(Sorry, I don't know how to do these big brackets for equation systems in Latex.)

So it's a homogeneous system with more unknown variables than equations. I know I'll have to use two parameters. But I just can't get it right. What's the best way to start? Add the first and second equation to eliminate $$\displaystyle x_3$$?

$$\displaystyle x_1 = -4t_1 - t_2\\x_2 = -35t_1 + 2t_2\\x_3 = 32t_1 - 3t_2\\x_4 = 25t_1\\x_5 = t_2$$

ToastIQ said:
Hi!
What's the best way to start?

The ONLY "best way to start" is the way that makes sense to you and you can execute accurately.

Please demonstrate a first step. You may wish to take a moment and learn just enough LaTeX it get the idea on the screen.

ToastIQ said:
Hi!

Just started with linear algebra Could someone help me with this problem?

$$\displaystyle 2x_1 + x_2 - x_3 + 3x_4 - 3x_5 = 0\\ 3x_1 + 2x_2 + x_3 + 2x_4 + 2x_5 = 0\\ -4x_1 + 3x_2 + 2x_3 + x_4 - 4x_5 = 0$$

(Sorry, I don't know how to do these big brackets for equation systems in Latex.)

So it's a homogeneous system with more unknown variables than equations. I know I'll have to use two parameters. But I just can't get it right. What's the best way to start? Add the first and second equation to eliminate $$\displaystyle x_3$$?

$$\displaystyle x_1 = -4t_1 - t_2\\x_2 = -35t_1 + 2t_2\\x_3 = 32t_1 - 3t_2\\x_4 = 25t_1\\x_5 = t_2$$
Try to remember the goal. Don't you want a reduced diagonal form? It would seem prudent to get x1 with a coefficient of 1 as soon as possible,

Subtract #1 from #2 -- Done
Add twice the original #1 to #3 -- Okay, we're well on our way and we have not yet been forced to use any division.
Subtract 3 times the new #1 from #2. -- We are now done with x1.

Try to be consistent. Don't behave or think in such a way that every problem is a new experience. Relate what you are doing now to what you already have done. No need to rethink the whole process with every new problem.

ToastIQ said:
Hi!

Just started with linear algebra Could someone help me with this problem?

$$\displaystyle 2x_1 + x_2 - x_3 + 3x_4 - 3x_5 = 0\\ 3x_1 + 2x_2 + x_3 + 2x_4 + 2x_5 = 0\\ -4x_1 + 3x_2 + 2x_3 + x_4 - 4x_5 = 0$$

(Sorry, I don't know how to do these big brackets for equation systems in Latex.)

So it's a homogeneous system with more unknown variables than equations. I know I'll have to use two parameters. But I just can't get it right. What's the best way to start? Add the first and second equation to eliminate $$\displaystyle x_3$$?

No, that would not be my first choice.

Is your book already using matrix notation at this point? For solving linear systems this is not necessary, but it is convenient. In any case, here is a way to get those "big brackets" in $\LaTeX$. (There are better ones, but this works here.) Right-click on the formula, choose "Show Math As" and then "TeX Commands".

$\left( \begin{array}{r} 2& 1& -1& 3& -3\\ 3& 2& 1& 2& 2&\\ -4& 3& 2& 1& -4 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right),$

but usually you would use the abbreviations $\mathbf{x}$ and $\mathbf{0}$ for the unknown and the right-hand side.

tkhunny said:
The ONLY "best way to start" is the way that makes sense to you and you can execute accurately.

I agree to some degree. For solving linear systems like this by hand, I think the best way is to stick strictly to Gaussian elimination. This can be done either with back-substitution or with the Jordan variant. Both have their merits, but which one of the two you use should at this point depend on what your book or teacher wants you to do. Please do look up what choice is made in your class and stick with it.

From the looks of what the answer should be, it wants you to do plain Gaussian elimination with back-substitution.

tkhunny said:
Please demonstrate a first step. You may wish to take a moment and learn just enough LaTeX it get the idea on the screen.

Yes, learning $\LaTeX$ is an investment that will pay you rich dividends. There are a ton of tutorials out there. WikiBooks has a reasonable one. Here is its section on matrices and arrays.

tkhunny said:
Try to remember the goal. Don't you want a reduced diagonal form? It would seem prudent to get x1 with a coefficient of 1 as soon as possible,

Subtract #1 from #2 -- Done
Add twice the original #1 to #3 -- Okay, we're well on our way and we have not yet been forced to use any division.
Subtract 3 times the new #1 from #2. -- We are now done with x1.

I don't entirely agree. If I understand you correctly, then after the first two steps the associated augmented matrix would look like:

$\left( \begin{array}{rrrrr|r} 2 & 1 & -1 & 3 & -3 &0\\ 1 & 1 & 2 &-1 & 5 & 0\\ 0 & 5 & 0 & 7 &-10 & 0 \end{array} \right).$

(OP: Since your right-hand side is $\mathbf{0}$, you might as well leave the last column out, but it does not hurt to include it.) For Gaussian elimination it is now sensible to exchange the first and second row (i.e. the first and second equation) and then subtract two times the new first row from the new second row. This completes elimination below the first pivot, leaving you with

$\left( \begin{array}{rrrrr|r} 1 & 1 & 2 &-1 & 5 & 0\\ 0 &-1 & -5 & 5 & -13 & 0\\ 0 & 5 & 0 & 7 &-10 & 0 \end{array} \right)$

and you can continue with the next pivot, which will be the first nonzero entry in the second row, i.e. $-1$.

Last edited:
Krylov said:
I agree to some degree. For solving linear systems like this by hand, I think the best way is to stick strictly to Gaussian elimination. This can be done either with back-substitution or with the Jordan variant. Both have their merits, but which one of the two you use should at this point depend on what your book or teacher wants you to do. Please do look up what choice is made in your class and stick with it.

Which is the long way to say what I said. It is hoped that the sensible and accurate way to proceed has something to do with classroom presentation, teacher demands, and perhaps some personal variation.

I don't entirely agree. If I understand you correctly, then after the first two steps the associated augmented matrix would look like:

$\left( \begin{array}{rrrrr|r} 2 & 1 & -1 & 3 & -3 &0\\ 1 & 1 & 2 &-1 & 5 & 0\\ 0 & 5 & 0 & 7 &-10 & 0 \end{array} \right).$
No. That's a different set of instructions. My instructions lead to exactly what you produced later.

We seem to be on the same page. There is no disagreement. Different language, perhaps.

tkhunny said:
Which is the long way to say what I said.

No, it is not. What you wrote is different, namely:

tkhunny said:
The ONLY "best way to start" is the way that makes sense to you and you can execute accurately.

On the other hand, I wrote that In my view, it is not what makes sense to the OP, it is what makes sense mathematically and according to well-established algorithms such as Gauss'. To the OP it made sense to eliminate $x_3$ first:

ToastIQ said:
What's the best way to start? Add the first and second equation to eliminate $$\displaystyle x_3$$?

and that is certainly not the best way to start.

tkhunny said:
No. That's a different set of instructions.

No, I carried out exactly the first two steps in your instructions in your post #3.

Last edited:
Krylov said:
No, it is not. What you wrote is different, namely:
On the other hand, I wrote that In my view, it is not what makes sense to the OP, it is what makes sense mathematically and according to well-established algorithms such as Gauss'. To the OP it made sense to eliminate $x_3$ first:
The mathematical accuracy is of little value if the OP cannot make sense of it. They must both exist. Eliminating $x_3$ did not make sense to the OP. The OP was guessing. I really mean "make sense". "Can execute" also means appropriate and perhaps well-established.

No, I carried out exactly the first two steps in your instructions in your post #3.

That's a good point. I did not use sufficiently clear language. My instructions lead to:
$\left( \begin{array}{rrrrr|r} 1 & 1 & 2 & -1 & 5 &0\\ 0 & 5 & 0 &7 & -10 & 0\\ 0 & -1 & -5 & 5 &-13 & 0 \end{array} \right).$
This is approaching the same goal with the same well-established pattern.

Hi!

First of all, I am REALLY sorry for replying so late, I've had a really bad influenza.

tkhunny said:
You may wish to take a moment and learn just enough LaTeX it get the idea on the screen.

I wasn't asking about Latex in general, I just meant those big brackets that look like wings. I really don't know, I've googled but some things that seem to work on other sites don't seem to work on here.

tkhunny said:
Don't behave or think in such a way that every problem is a new experience. Relate what you are doing now to what you already have done. No need to rethink the whole process with every new problem.

I really do appreciate help but I don't see how this is relevant here. I am a complete beginner on this topic, and yes, many problems look like a new experience to me when I just started out with something new. It's called being a beginner. It takes practice to get to thinking the way you recommend me to, and I pretty much had 0 practice when I posted this question.
Every time I post a question I spend days trying to figure it out on my own. I do try to see similarities to other problems I've dealt with before and when I post a question here it means I couldn't solve it relating to problems I've done in the past. I NEVER post before doing a lot of thinking before, just for the record. Sorry if this was a stupid question to ask, all I wanted to know really was if I should start with eliminating $$\displaystyle x_3$$ or not.

Krylov said:
Krylov said:
In any case, here is a way to get those "big brackets" in $\LaTeX$. (There are better ones, but this works here.) Right-click on the formula, choose "Show Math As" and then "TeX Commands".

$\left( \begin{array}{r} 2& 1& -1& 3& -3\\ 3& 2& 1& 2& 2&\\ -4& 3& 2& 1& -4 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right),$

but usually you would use the abbreviations $\mathbf{x}$ and $\mathbf{0}$ for the unknown and the right-hand side.

Wow, thanks a lot!

Krylov said:
From the looks of what the answer should be, it wants you to do plain Gaussian elimination with back-substitution.

Correct.

Krylov said:
Yes, learning $\LaTeX$ is an investment that will pay you rich dividends. There are a ton of tutorials out there. WikiBooks has a reasonable one. Here is its section on matrices and arrays.

Thanks a lot, again! I spent a lot of time trying to find out how to do those "wing" brackets, it just didn't work on this site so it's not that I'm lazy. Really thankful for the link!

tkhunny said:
Eliminating $x_3$ did not make sense to the OP. The OP was guessing.

Eliminating $$\displaystyle x_3$$ did make sense to me. I was trying to solve this problem by relation to the few others I'd solved before. Guessing? Of course. I hadn't had a similar problem before, so all I could possibly do was make guesses, which were based on what I'd learned before (which wasn't a lot, as I said, I just started out).

But it did not lead me to the right result. Neither did many other ways I tried, ways that had worked with other, similar problems. That's why I wanted to try it step by step and get some help along the way to see where my error was, and that's why I wanted to know if eliminating $$\displaystyle x_3$$ was a good way to start off, because I certainly thought it did. I was obviously wrong.

I solved this problem using Gauss and substitution, but it took such a long time and I could have solved this the same way before I started linear algebra, and I don't think I was supposed to solve it like this. I know I'm supposed to use "tricks" from linear algebra, not the basic knowledge from normal algebra that took way too much time.

Anyways, thanks for the help and feel free to point out if I'm misunderstanding something.

ToastIQ said:
I really do appreciate help but I don't see how this is relevant here. I am a complete beginner on this topic, and yes, many problems look like a new experience to me when I just started out with something new. It's called being a beginner. It takes practice to get to thinking the way you recommend me to, and I pretty much had 0 practice when I posted this question.

That's all I was recommending. Start early and learn good habits. It is often very frustrating to have to start from scratch with every problem faced. If you don't see a reasonable sequence of progression from one thing to the next (more complicated, maybe), then maybe you are missing something or possibly you have a horrible textbook.

Every time I post a question I spend days trying to figure it out on my own. I do try to see similarities to other problems I've dealt with before and when I post a question here it means I couldn't solve it relating to problems I've done in the past. I NEVER post before doing a lot of thinking before, just for the record.

"0 practice" and "spend days" aren't quite consistent statements. Maybe they are. Anyway, this is great and not so great. if you "spend days", that may begin to be more frustrating than useful. However, you may be blessed with infinite patience with yourself. This is an advantage. Absolutely give it a good go before asking, but don't be afraid to ask BEFORE it gets frustrating.

Sorry if this was a stupid question to ask, all I wanted to know really was if I should start with eliminating $$\displaystyle x_3$$ or not.

Quite a valid question. The designation of x1, x2 or x3 probably is arbitrary. It should not matter if you start with one or the other. Typically, we shoot for the upper-triangular look. This suggests that you start on the left - no matter what. On the other hand, there is absolutely no harm in moving x3 to the far left if that's what you feel like doing. (x3, x1, x2 will not cause the universe to explode.) Sometimes, after a little experience, you can see that some variable (or column) or another might simply be more convenient - not always. Just be consistent and careful. "Mathematically sound" and "rigorous" do NOT mean that you must follow a completely prescribed pattern. Many students hold this fallacy and go into hiding, not knowing what really important, spiritually mandated step to take next. Play with it a little bit. Enjoy its company. Get the same answer a couple different ways. You'll be better for having done the exploration.

ToastIQ said:
Hi!

First of all, I am REALLY sorry for replying so late, I've had a really bad influenza.

It surely is appreciated that you came back to reply. I hope your illness has passed by now.

ToastIQ said:
.

Matrix notation is very convenient when solving linear systems, but not essential. One of my favorite books on linear algebra also starts by explaining Gaussian elimination and introduces matrix notation only a little bit later.

At some point you will see that there is a lot more to matrices than notational convenience, and that can be a lot of fun, but this will come in time.

ToastIQ said:
Thanks a lot, again! I spent a lot of time trying to find out how to do those "wing" brackets, it just didn't work on this site so it's not that I'm lazy. Really thankful for the link!

I'm happy it helped. This forums is using MathJax. It implements a large subset - but not all - of the usual $\LaTeX$ macros for online purposes.

You can also install $\LaTeX$ locally on your computer, so it becomes more easily accessible for school work. It's available for free for all current operating systems. Probably there are pointers in the tutorial I linked to, but if you need help with it, you could perhaps also ask https://mathhelpboards.com/latex-help-discussion-26/.
ToastIQ said:
Eliminating $$\displaystyle x_3$$
I solved this problem using Gauss and substitution, but it took such a long time and I could have solved this the same way before I started linear algebra, and I don't think I was supposed to solve it like this. I know I'm supposed to use "tricks" from linear algebra, not the basic knowledge from normal algebra that took way too much time.

Anyways, thanks for the help and feel free to point out if I'm misunderstanding something.

So, ultimately you did solve it using Gaussian elimination and back-substitution? That is good, well done. Is the basic algorithm clearly explained in your book? Meanwhile, did you have the chance to compare your solution with the solution(s) of some example problem(s)?

Note also that while Gaussian elimination is indeed the canonical way to solve linear systems (certainly when solving them by hand), the steps in Gaussian elimination are not unique. Different sequences of elementary row operations can lead to different row echelon forms of the original system, from which the solution set can then be obtained by back-substitution. Often, certain sequences are "handier" than others because they involve simpler arithmetic.

So, while different sequences of row operations may lead to different parametrizations, all solution sets obtained in this way are equivalent: They correspond to one and the same set of vectors $\mathbf{x}$.

## What is a homogeneous, underdetermined equation system?

A homogeneous, underdetermined equation system is a set of equations where the number of unknown variables is greater than the number of equations. This means that there are infinite solutions to the system.

## How is a homogeneous, underdetermined equation system different from a regular equation system?

In a regular equation system, the number of unknown variables is equal to the number of equations, resulting in a unique solution. However, in a homogeneous, underdetermined equation system, there are more unknown variables than equations, leading to infinite solutions.

## What is the role of null space in a homogeneous, underdetermined equation system?

The null space of a matrix represents the set of solutions to a homogeneous equation system. In a homogeneous, underdetermined equation system, the null space is non-trivial, meaning that there are infinite solutions to the system.

## How can a homogeneous, underdetermined equation system be solved?

Since there are infinite solutions to a homogeneous, underdetermined equation system, it cannot be solved using traditional methods. Instead, it can be reduced to a system of equations with a unique solution by adding additional constraints or using numerical methods.

## What are some real-world applications of homogeneous, underdetermined equation systems?

Homogeneous, underdetermined equation systems are commonly used in signal processing, image reconstruction, and data compression. They are also used in machine learning and statistics for solving regression and classification problems.

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