Question

Question involves spans & vectors, see attached picture.

Answer 1

I know I'm supposed to form a matrix, but I keep getting the answers wrong. Maybe I need to see one of these attempted by you.

Answer 2

What is the question exactly?

Answer 3

Let me get attach the question, I'm sorry I just noticed it was cut out from the pic I posted

Answer 4

Just attached the question

Answer 5

And what was the modified version of the algorithm mentioned?

Answer 6

I have no idea haha, all I know is to create a matrix using what's given

Answer 7

I'll show #11

Answer 8

lol I'm having trouble making a matrix on here

Answer 9

3 1 1 ; 3
5 1 2 ; 4
-2 4 -3 ; 5

Answer 10

how are you gonna get rid of the 5 , and -2, in the first column ?

Answer 11

working on it right now

Answer 12

1 5 -2 ; 8
0 7 -3 ; 11
0 14 -7 ; 21

Answer 13

so far

Answer 14

am I even doing the right thing lol

Answer 15

:(

Answer 16

v1 v2 v3
S = {(3,5,-2),(1,1,4),(1,2,-3)}

b = (3,4,5)

can b be formed by a linear combination of the vectors in S?

b = c1v1 + c2v2 + c3v3 ?? can we find coefficients that create b, from the vectors in S?

c1v1 = 3(c1), 5(c1), -2(c1)
c2v2 = 1(c2), 1(c2), 4(c2)
c3v3 = 1(c3), 2(c3), -3(c3)
---------------------------
b = 3 , 4 , 5


in other words:

3(c1) + 1(c2) + 1(c3) = 3
5(c1) + 1(c2) + 2(c3) = 4
-2(c1) + 4(c2) - 3(c3) = 5

this conforms to the matrix that you wrote up to start your solution, so yes you are on the right track

Answer 17

3 1 1 3
5 1 2 4
-2 4 -3 5

Answer 18

So do I go for reduced row echelon form?

Answer 19

the long version is to use elimination to solve the system of equation

3x + y + z = 3
5x + y + 2z = 4
-2x +4y - 3z = 5

Answer 20

apparently the answer is 1/2 v1 + 3/2 v2

Answer 21

I don't get how that's the answer

Answer 22

it is quite possible that the vectors in S are not independant ... that one or more can be written as a linear combination of the others, what is the determinant of S?

3 1 1
5 1 2
-2 4 -3


1 1/3 1/3
1 1/5 2/5
1 -2 3/2


1 1/3 1/3
0 -2/15 1/15
0 -7/3 7/6

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


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


1 0 1/2
0 1 -1/2
0 0 0
...................

what this tells us is that v3 is redundant as far as an efficient span goes, v3 can be formed as a linear combination of v1 and v2, thereby making any combination with v3 simplifies to v1 and v2 alone.

v3 = (v1-v2)/2

3/2 -1/2 = 1
5/2 -1/2 = 2
-2/2 -4/2 = -3

Answer 23

is b, a linear combination of v1 and v2? seeing that v3 is redundant (v3 is in the same plane as v1 and v2)

v1 v2 b
3 1 3
5 1 4
-2 4 5


rref{{3,1,3},{5,1,4},{-2,4,5}}

http://www.wolframalpha.com/input/?i=rref {{3%2C1%2C3}%2C{5%2C1%2C4}%2C{-2%2C4%2C5}}&dataset=

Answer 24

firefox doesnt like to copy paste links correctly .. but the wolf agrees with 1/2v1+ 3/2v2