Solve the system of linear equations

[grand arrow]

I managed to only make the matrix in this form, I am not sure how to proceed now or what conclusion should I make

[plain sorrel]

1. Do not ping the Moderators, unless someone is breaking the rules.
2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
3. Do not ping other members randomly for help.
4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
5. Wait patiently for a helper to come along.
6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[rose flower]

If you are expected to solve the system, then I think you can directly proceed with Jordan-Gauss elimination

[rose flower]

So assuming you did, and the picture on the right is the resulting matrix

first of all you can see the bottom row is completely irrelevant

but just in case you should check that the RHS is also not zero on that coordinate

[grand arrow]

the RHS?

[grand arrow]

as far as i know, this means that this either has no solution, or countless solutions?

[rose flower]

you have a matrix equation in the form Ax = b

which you reformulated in the form A' x' = b'

with your new system

so you have to check if your system of equations is even compatible

because the last row is full of zeros on the LHS

but the last row of the RHS might not be zero, you need to check

[rose flower]

This is correct

[grand arrow]

i am really new to this, i am not sure how i should check that

[rose flower]

so you did a bunch of transformations on the LHS right?

you need to do the same transformations on the RHS

which is: $\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}$

[devout wing]

Miguel

[grand arrow]

oh, i see

wouldn't it become $\begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$ then?

[devout wing]

𝐜𝐚𝐭𝐱𝐫𝐢𝐧𝐞

[rose flower]

If so then the bottom row is completely irrelevant

So you'd be left with the following system:

\begin{align}
&x_1 &-x_2 &+x_3 &+2x_4 &= 1\\
& &3x_2 &-3x_3 &-5x_4 &= 0\\
& & & &0 &= 0
\end{align}

[devout wing]

Miguel

[rose flower]

See that line (3) is irrelevant

but the first two rows basically establish the dependencies between your variables

[grand arrow]

wait, i think i misunderstood you, you wanted me to apply the exact same transformations?

[rose flower]

so to no one's surprise you'll have infinite solutions

[rose flower]

What did you understand?

[grand arrow]

ah, i am sorry, give me a second

[rose flower]

I'm not trying to chastise you for not understanding, I just want to know what you had in mind so we can be on the same page

[grand arrow]

i thought that i had to transform it using the gauss method as well, but not the same operations as the previous one

i actually tried to do this earlier, but i'll do it again just to be sure it is correct

[rose flower]

Sure. Please upload all the steps here later so that we can check together

[grand arrow]

it's late here and i am just slow and tired, i apologize for this

[grand arrow]

sure thing

Alright, this is what I got

[rose flower]

First step is good

Second step is good too

Yep that's good

I think now you can try to eliminate more variables

[grand arrow]

hmm

should i R1-R2? Since we will be still left with variables, but smaller numbers

[rose flower]

according to the algorithm, you should do R1 -> R1 + (2/5) R2

alternatively, do R1 <- 5 R1 first

then R1 <- R1 + 2 R2

this will help you keep integer coefficients everywhere

[grand arrow]

are there specific steps that i have to follow? Up to now i was just doing what i thought would be right and wont violate the rules

[rose flower]

Well the Jordan-Gauss elimination is an algorithm

you have three valid operations, which are rescaling, add/substract or swap rows

and the algorithm tells you in which sequence to use them

[grand arrow]

i really need to revise the chapter better

this is the result after the operation:

[rose flower]

yeah it's a bit hard to read isn't it

just multiply that first row by 5

also I think you did a mistake

the coefficient at row 1, column 2

I found it to be 1/5

[grand arrow]

should be, since the one next to it is the same but with a different sign

so it becomes:

[rose flower]

Now based on that you can try to get a conclusion

x_4 depends on x_3 and x_2, and x_1 does too

but after that, there are infinite solutions

[grand arrow]

how should i format this answer?

[rose flower]

Can you express x_4 in function of x_3 and x_2?

[grand arrow]

i am not sure?

i don't think so

[rose flower]

you can

check the second row

it basically tells you: $3 x_2 - 3x_3 - 5x_4 = -2$

[devout wing]

Miguel

[grand arrow]

oh, yes

what should i do with it then?

[rose flower]

Write x_4 in function of the other two variables

Do the same with x_1

then you basically know what the solutions look like

[grand arrow]

is this what you mean i have to do?

[rose flower]

Sorry for the wait

Well then now you know what the solutions are like

they're in the form (x1, x2, x3, x4) where x1 and x4 are like that

[grand arrow]

so this is the final answer?

[grand arrow]

no worries!

should i formulate it better in some way, or is what's written in the image given is enough?

[rose flower]

So what you showed here is

If (x_1, x_2, x_3, x_4) is a solution of the system, then x_1 and x_4 are in the form above

Now you need the other way around

[grand arrow]

for x_2 and x_3?

or how should it be the other way around?

[rose flower]

If x_1 and x_4 are in the above form, then (x_1, x_2, x_3, x_4) satisfies the system of equations

you need to prove that

all you have to do is just plug that into the system

actually wait

never mind, the Jordan-Gauss elimination gives you an equivalence already

For short, the algorithm guarantees now that $(x_1, x_2, x_3, x_4)$ satisfies the system of equations \textbf{if and only if}:
\begin{align*}
x_1 = \frac{1}{5} - \frac{1}{5} x_2 + \frac{1}{5} x_3\
x_4 = \frac{2}{5} + \frac{3}{5} x_2 - \frac{3}{5} x_3
\end{align*}

[devout wing]

Miguel

[grand arrow]

oh i see

so that was it?

[rose flower]

now you can write the solution set

based on that information

It would be very nice if you could rewrite it as an affine subspace of R^4

[grand arrow]

i am not sure what this means :']

[rose flower]

isn't this in the context of a linear algebra course?

[grand arrow]

i guess so? I am still not in college and i needed to go through the topic of matrices and what not for a couple of days

i have a lot of misses

[rose flower]

I had assumed because this help thread is in the university section

my bad

Then yeah you're probably not expected to write anything else, other than x2 and x3 being free (so they can be whatever)

and x1 and x4 are bound to x2 and x3

[grand arrow]

i just know that those are tasks from college, that's why i posted it here

[grand arrow]

the tasks are for someone who is in fact in college, i am just the one trying to solve them

so i think they will need it

[rose flower]

Later I would typically expect you to write the solution set as:

$$S = \frac{1}{5}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 2 \end{bmatrix} + \mathrm{Span}\left( 
\begin{bmatrix} - 1 \\ 5 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 5 \\ -3 \end{bmatrix}
\right)$$

[devout wing]

Miguel

[grand arrow]

so this is the form of the answer that is expected?

[rose flower]

more like

after a linear algebra course, this is the kind of answer I would expect from you on a test

[rose flower]

This is fine for the time being

[grand arrow]

what does the Span means?

[rose flower]

oh

the span of a family of vectors is the set of all linear combinations you can form with these vectors

or equivalently, the smallest subspace that contains all those vectors

so for two vectors, span(v_1, v_2) = { a v_1 + b v_2, a and b are scalars}

[grand arrow]

all of this sounds very interesting, even though i am not aware of most of the stuff, if it wasn't for the deadlines i would've spend a good amount of time practicing

i really appreciate that you tried to explain everything, even tho i am clearly not very aware of the subject

[rose flower]

I don't quite understand your circumstances so unfortunately I cannot help you better

[grand arrow]

it's okay! You really helped me a lot

[grand arrow]

+close

[covert basalt]

Please thank your Helpers before closing!

Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.

[covert basalt]

Thank you for your feedback! Miguel has been awarded 1 . They now have 76 . They have 2 daily left for today.