Question involves spans & vectors, see attached picture.

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Question involves spans & vectors, see attached picture.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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.
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Let me get attach the question, I'm sorry I just noticed it was cut out from the pic I posted
Just attached the question
And what was the modified version of the algorithm mentioned?
I have no idea haha, all I know is to create a matrix using what's given
I'll show #11
lol I'm having trouble making a matrix on here
3 1 1 ; 3 5 1 2 ; 4 -2 4 -3 ; 5
how are you gonna get rid of the 5 , and -2, in the first column ?
working on it right now
1 5 -2 ; 8 0 7 -3 ; 11 0 14 -7 ; 21
so far
am I even doing the right thing lol
:(
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
3 1 1 3 5 1 2 4 -2 4 -3 5
So do I go for reduced row echelon form?
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
apparently the answer is 1/2 v1 + 3/2 v2
I don't get how that's the answer
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
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=
firefox doesnt like to copy paste links correctly .. but the wolf agrees with 1/2v1+ 3/2v2

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