Question

Let \(F:= \left[\begin{matrix}1\\1\\1\\1\\1\end{matrix}\right] +p\left[\begin{matrix}0\\0\\1\\1\\1\end{matrix}\right]+q\left[\begin{matrix}0\\1\\1\\1\\1\end{matrix}\right]+r\left[\begin{matrix}1\\1\\1\\1\\1\end{matrix}\right]\)
Find a matrix A and vector x, b such that \(F:={x \in R^? |Ax=b}\)
Please, help

Answer 1

Dimension of F is 3 , but \(\vec x \in \mathbb R^5\) How to go back to original system?
I know how to go forward but don't know how to go backward.

Answer 2

I meant if I have a system of equation, after doing row reduce form, I can get this parametric system. Now, the exercise gives me the parametric system, I have to go back ward to find the system. :(

Answer 3

.

Answer 4

Let \(\vec x=\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\\x_5\end{matrix}\right] \)

Answer 5

then, \(x_1=1+r\\x_2= 1+q+r\\x_3=x_4=x_5=1+p+q+r\)

Answer 6

then what?

Answer 7

It seems to me the way F is defined, the last three variables are not independent.
so something like:
1 2 5 5 5
2 7 1 1 1
3 4 2 2 2
5 4 3 3 3
7 1 4 4 4
gives the REF as
1 2 5 5 5
0 1 -3 -3 -3
0 0 1 1 1
0 0 0 0 0
0 0 0 0 0
and gives the solution as shown.

Answer 8

For b, you can try [13,12,13,18,20] (sum of all the coefficients)

Answer 9

How can you get it? the last three p, q , r vectors are independent. they span F, F is defined as \(F:= v_0 + pV_1+q V_2+rV_3\)

Answer 10

I think 'b' is the zero vector because the given particular solution is part of the null solution.

Answer 11

3 free variables in the general solution implies the rank is 5-3 = 2.

That means the number of independent columns is also 2

Answer 12

Try this :

A:
0 0 1 0 -1
0 0 0 1 -1

b:
0
0

Answer 13

Notice that by arranging linear combinations, F can be made to read
\(F:= \left[\begin{matrix}1\\1\\1\\1\\1\end{matrix}\right] +u\left[\begin{matrix}0\\0\\1\\1\\1\end{matrix}\right]+v\left[\begin{matrix}0\\1\\0\\0\\0\end{matrix}\right]+w\left[\begin{matrix}1\\0\\0\\0\\0\end{matrix}\right]\)
From where I had the idea that the last three variable share the same value.

Answer 14

@ganeshie8 I would like to know how you get that matrix, please

Answer 15

Does it work ?

Answer 16

I don't see the link between F and the matrix you gave me :(

Answer 17

Let's do this on a quick teamviewer meeting if you still have it

Answer 18

You take control, right?

Answer 19

No.. You may just start a meeting and send me the id...

Answer 20

My ID 927 189 615

Answer 21

That's for controlling
See if you can start a meeting

Answer 22

It says nothing :(

Answer 23

Meeting option will be there on right hand side of the teamviewer main window

Answer 24

From the right, it asks for partner ID

Answer 25

Right top side, I guess

Answer 26

http://cdn.free-power-point-templates.com/articles/wp-content/uploads/2012/05/meeting.jpg

Answer 27

This is what I have and it doesn't allow me to do anything

Answer 28

@ganeshie8 we work here, please.

Answer 29

I'm on mobile
Can't type much sry

Answer 30

I can't open that docx file

Answer 31

That's ok then. I show you my work, you check , ok?

Answer 32

Sure

Answer 33

May I know why you think the column vectors must have 5 components ?

Answer 34

because the way they defined F, any vector in F has 5 components

Answer 35

\(F:= \left[\begin{matrix}1\\1\\1\\1\\1\end{matrix}\right] +p\left[\begin{matrix}0\\0\\1\\1\\1\end{matrix}\right]+q\left[\begin{matrix}0\\1\\1\\1\\1\end{matrix}\right]+r\left[\begin{matrix}1\\1\\1\\1\\1\end{matrix}\right]\)

Answer 36

F is not a matrix

Answer 37

F is the set of vectors that are solutions to Ax = b

Answer 38

oh yeah.

Answer 39

Hey try starting a meeting if you can

Answer 40

Oh, I think I got your given matrix
Since the last 3 equation are the same, you have \(x_3=x_5\\x_4=x_5\)
that gives you the matrix above

Answer 41

:) ganeshie8 , every time I try to connect you on teamviewer, I am so embarrass when struggling with the process to access into it. It belittles me. I fell as if I am the most stupid people in the world hahaha....... everybody can use it but I. As a result, my face turns red now and I just want to give up no matter what it is.

Answer 42

Haha thats okay
Have u looked at the link I attached

Answer 43

yes, I did

Answer 44

You just have to click on Meeting,
then start meeting

Answer 45

Yes, I did also. It is ask for meeting ID

Answer 46

Youre on the right tab.

On the left side,
Click start instant maating

Answer 47

It is m, but when I try to click on "join meeting", it says "it is not a valid meeting-ID

Answer 48

You're trying to join a meeting.
You should start the meeting instead

Answer 49

AAAAAAAAAAHHH. I tried any button and got nothing. hahahaha.. ....
ganeshie8, please ....... as my friend, don't ask me to go there now. I promise. I will master it next week with ....another one, not you. hehehe..... then, I will contact you to show the resutl.

Answer 50

Try once
Its not rocket science. Not for you

Answer 51

Yes, sir. but later, not now. gtg.