Question

Solve the following system of linear equations:

-eq is posted in reply-

If the system has infinitely many solutions, your answer may use expressions involving the parameters r, s, and t.

Answer 1

4

Answer 2

how so?

Answer 3

don't know

Answer 4

\[ x _{1}+5x _{2}−5x _{3} = 5 .\]

Answer 5

I'll post in just one second.

Answer 6

First of all, let's be clear that this system has infinitely many solutions. Now I have a question: are you supposed to be finding the vectors that span the solution space (or whatever the correct linear algebra terminology is)? Or is it basically like simplifying the system as much as possible (reduced row echelon form) and then relating the variables? Because if it's the latter I think it'd be as simple as saying x1=5-5s+5t where s=x2 and t=x3.

Answer 7

Yes, you are suppose to simplify this to R.E.F and once you have it all expanded. it would have each x1, x2, x3= to a variable. I am having difficulties just putting it into R.E.F

Answer 8

Well good news: it's already in REF. lol

Answer 9

I'll post the matrix in just a sec.

Answer 10

lol, you are amazing

Answer 11

Which phrase best defines the term direct object?



A.
an adverb that modifies an action verb


B.
a word connected to the subject by a linking verb


C.
a noun, pronoun, or word group that receives the action of the verb


D.
a noun or pronoun before an action verb

Answer 12

please help

Answer 13

this is a math forum. please go learn english in another field

Answer 14

\[\left[\begin{matrix}1 & 5 & -5 \\ 0 & 0 & 0\\ 0 & 0 & 0\ &\end{matrix}\right]*x=\left[\begin{matrix}5 \\ 0 \\ 0\end{matrix}\right]\]
where x is a vector.

Answer 15

Notice that all rows start with leading ones, which means this is in REF.

Answer 16

Oh ya! that is right.

Answer 17

Do you care about the other thing I was talking about at all or that too complicated?

Answer 18

I would really like to know how would you solve for x1. x2. and x3. I understand now that i can physically see in a R.E.F, but would like how to continue from there

Answer 19

Well solve for x1: x1=-5x2+5x3+5

Answer 20

Now let x2 and x3 be equal to different parametric values: let x2=s and x3=t for instance. So x1=-5x2+5x3+5=-5s+5t+5

Answer 21

ohhhh. so would the x2= be just corresponding to s and x3= t

Answer 22

It is worth noting that you could solve for any of the 3 xn but typically we solve for the lowest aka the one corresponding to the leading 1 in REF. And yeah pretty much.

Answer 23

wow, this is much easier than what I assumed before

Answer 24

You are amazing! :) thank you so much.

Answer 25

I'll go ahead and post the vectors that span the solution space just for the heck of it. Just one second.

Answer 26

I have another question too regarding about this type of question. I hope you can help me out too! :)

Answer 27

\[x=\left[\begin{matrix} x _{1} \\ x _{2}\\ x _{3}\\ \end{matrix}\right]=\left[\begin{matrix} -5 \\ 1\\ 0\\ \end{matrix}\right]s+\left[\begin{matrix}5\\0\\1\\ \end{matrix}\right]t+\left[\begin{matrix} 5 \\0\\0\\ \end{matrix}\right]\]

Answer 28

I'll do my best. Go for it.

Answer 29

in another forum? or here?

Answer 30

Typically when you're looking for the span of a space though you have Ax=0 instead of Ax=b as we have here and that (5,0,0) wouldn't be there.

Answer 31

You can put it here if you want.

Answer 32

Sweet. Thank you so much!

Answer 33

Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions.
5x1+5x2+5x3 = 10
−3x1−2x2+ax3 = −4
x1+4x2+4x3 = 9

Answer 34

In that case, I am trying to find how this would be an unique solution, infintely many, and no solutions

Answer 35

\[\left[\begin{matrix}5 & 5 & 5 \\ -3 & -2 & a\\ 1 & 4 & 4 \end{matrix}\right]x=\left[\begin{matrix} 10 \\ -4\\ 9\end{matrix}\right]\]

Answer 36

One way is to put it in RREF and then solve for a to make those conditions true.

Answer 37

ok

Answer 38

Excuse me REF not RREF.

Answer 39

lol :)

Answer 40

Ok I'm getting\[\left[\begin{matrix}1 & 4 & 4 \\ 0 & 1 & 1\\ 0 & 0 & a+2\end{matrix}\right]x+\left[\begin{matrix} 9 \\ 7/3\\ 23-70/3\end{matrix}\right]\]

Answer 41

ok, i got that too despite fractions

Answer 42

So if it were going to be infinitely many solutions, you would find that you have less than 3 independent equations. So you have only 1 or 2 leading ones and that the last row is all 0. So we can let a+2=0 which means a=-2 for infinitely many solutions.

Answer 43

yep, got that too

Answer 44

Oh I know. A better way is to then take the determinant of the matrix.

Answer 45

a determiniant?

Answer 46

Have you not done that before?

Answer 47

nope.

Answer 48

Ok we won't do that then lol. Ok back to what we were doing then.

Answer 49

Ok recapping: we have that if a=-2, 0=23-70/3 which cannot be. This is the no solution case (not the infinitely many solutions case like I mistakenly said earlier). For infinitely many solutions we need 2 more more of the equations to be the same.

Answer 50

interesting

Answer 51

Yeah sorry about that XD.

Answer 52

haha no problem.

Answer 53

Grrr I see I messed up the REF let me change that too. My arithmetic is terrible today.

Answer 54

hahah same with mine!

Answer 55

this course really defines how good i am at arthmetic

Answer 56

\[\left[\begin{matrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0& 0 & a+2\end{matrix}\right]x=\left[\begin{matrix} 2 \\ 7/4\\ 1/4\end{matrix}\right]\] Is this the REF you got?

Answer 57

This however, won't change the fact that a=-2 makes this have no solution since it's still 0=something which can't happen.

Answer 58

yep, i corrected mine too.my row 3 was messed

Answer 59

Anyway, If we subtract the 2nd equation from the first:\[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1\\ 0 & 0 & a+2\end{matrix}\right]x=\left[\begin{matrix} 1/4 \\7/4\\ 1/4\end{matrix}\right]\]

Answer 60

Nah nevermind that's not so useful, after all.

Answer 61

hmmm

Answer 62

I'm not seeing the way to make this have infinitely many solutions. There's something I'm missing or there is a unique for solution for any value of a that isn't a=-2. I'll keep looking and get back to you in a bit.

Answer 63

arlighty

Answer 64

Here is what I got for RREF btw:\[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1\end{matrix}\right]x=\left[\begin{matrix} 1/4 \\ 7/4-\frac{ 1 }{ 4(a+2) }\\ \frac{ 1 }{ 4(a+2) }\ \end{matrix}\right]\]

Answer 65

woah. thats crazy

Answer 66

Actually I'm fairly certain the matrix can't be both no solutions and infinitely many solutions. Think about it in 2D: What does the graph look like?