Question

In Linear algebra:
The Question gives a system of linear equations:
x+ 2y+ 3z+ t = 0
2x+ 4y+ 7z+ 4t = 0
3x+ 6y+ 10z+ 5t = 0

How do you find out the dimension and basis of the solution space?

Thanks in advance!

Answer 1

Welcome to Open Study.

Answer 2

Hi!

Answer 3

any idea on how to solve?

Answer 4

Are you in Uni ?

Answer 5

Yes.

Answer 6

Hi there, you could use gaussian elimination

Answer 7

Find out how many pivot rows you have, that is equal to it's dimension

Answer 8

And you can use each independent outcome row as a vector element of an arbitrary basis for the solution set

Answer 9

OK thanks a lot, I will try and do it now

Answer 10

when I do the gaussian elimination, I get:

1 2 3 1 0
0 0 -1 -2
0 0 0 0

However, this isn't correct is it?

Answer 11

Something like that, you have 2 pivot rows there

Answer 12

1 is a pivot row, what is the other? thanks for all this help btw
0
0
0

Answer 13

1 2 3 1 is a pivot row
0 0 1 2 is a pivot row too

Answer 14

To be a pivot, all the elements in the row before the pivot should be 0, the pivot itself should be 1, and everything above the pivot should be 0. But you can always get that 0 by substracting the 2nd pivot row from the first

Answer 15

OK thanks now that we've got the pivot rows, you then said I can use each independent outcome row as a vector element of an arbitrary basis for the solution set to find the basis

I'm a bit confused on what this means?

Answer 16

Let me see, I don't know the english terms well

Answer 17

Ok thanks :)- just basically want to know what comes next to find the basis

Answer 18

Hmm I'm starting to think that I am wrong but we are certainly heading the right way let me recheck something

Answer 19

hmm well you can solve for x and z now right?

Answer 20

can you?

Answer 21

What you can do now is solver for x, y, z, t. You will have 2 'free' variables meaning that the dimension of the solution set is R^4 - 2 = 2. You can thus treat those free variables as scalars and the vectors as a basis

Answer 22

dim(R^4) - 2*

Answer 23

Though I can not rigorously prove it, I really need to recheck my LA, but I'm sure the answer would be right

Answer 24

the answer that my lecturer got is: (x, y, z, t) - y(-2,1,0,0) + t(-2,1,0,0) and (5,0,-2,1)

but I am unsure as to how he got this as the basis.

Answer 25

*** sorry the answer is (x, y, z, t) = y(-2,1,0,0) +t(5,0,-2,1)

and the basis is therefore (-2,1,0,0) and (5,0-2,1)

Answer 26

however, I am not sure how he got that..

Answer 27

I have the same answer

Answer 28

So the method is right :)

Answer 29

Just solve like I told you and if you have questions I'll be here but I need to brb 5 mins

Answer 30

OK thanks!

Answer 31

OK so the method is to use gaussian elimination and I get:

1 2 3 1 0
0 0 -1 -2 0
0 0 0 0 0

then from this I can see that the pivot row is 2 = dim(2)

after this, how do I solve the above matrix? in order to get the answer that my lecturer gave?

Answer 32

\[ M*(x y z t)^T = (0 0 0 0)^T \]

Answer 33

so now you can solve for x y z t

Answer 34

but you only have two pivot rows which means only two variables can be found, while the other are 'free variables', eg x_i \in R

Answer 35

how does M∗(xyzt)T=(0000)T help you to solve for xyzt?

Answer 36

multiply by M-1 or just use the systems of equations approach

Answer 37

M^-1 *

Answer 38

I solved it by just using systems of equations

Answer 39

so when you solve by systems of equations you get?

x+2y+3z+t = 0
and -z-2t=0

Answer 40

and then x=-2t

Answer 41

sorry z=-2t*

Answer 42

yep then what is x?

Answer 43

not sure- how would i get x and y?

Answer 44

x+2y+3z+t = 0
x = 2y -3(-2t) + t

Answer 45

y and t are free variables, meaning, they can be any element in \mathbb{R}

Answer 46

So now show me your solution set for x y z t

Answer 47

so id they are free variables then we can put t as say 0

therefore, x=2y?

Answer 48

you could, but then you would get a specific solution, not the solution set

Answer 49

What you need here is the solution set

Answer 50

so
x = -2y + 5t
y = free variable in R (or whatever field you're working in)
z = -2t
t = free variable in R (or whatever field you're working in)

Answer 51

So the solution vector space = y*v1 + t*v2

Answer 52

Try to figure out what vectors v1 and v2 represent

Answer 53

hmm, they represent the basis but I'm not sure how you would put the solutions into it?

Answer 54

\[ y * v_1^T + t * v_2^T = ( -2y + 5t , y, -2t, t )^T\]

Answer 55

that's nearly the answer you should be able to work that out yourself on paper

Answer 56

Ok so the basis is basically 4 different equations from the matrix

Answer 57

all put into one solution using the formula of y*v1 + t*v2

Answer 58

Well yes that's how we did it. This is one way that works but there might be more ways, as is often the case in uni you just have to "work something out yourself" using what you know

Answer 59

(-2y +5t,y,-2t,t)

where did you get the second bit (,y,) from?

Answer 60

(x y z w) = (...)

Answer 61

y because y = y ^_^

Answer 62

oh yes of course haha

Answer 63

Thank you so much for your help- really appreciated!

Answer 64

yw