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Linear Algebra - parameters

  1. Mar 20, 2007 #1

    daniel_i_l

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    Gold Member

    I was doing some problems on linear dependance and spanning that had parameters and i wasn't sure how to tell if i got the full answers. Here're two of them:

    1. The problem statement, all variables and given/known data
    Q1: Find all values of m so that the vectors:
    {(1-m,2,7),(0,-2-m,12),(0,0,-m)} are linearly dependant.

    Q2: Find all values of k so that the vectors:
    {(1,2,k),(0,1,k-1),(3,4,3)} span R^3.


    2. Relevant equations
    Q1: vectors are linearly dependent if and only if one of the vectors is a linear combination of the others

    Q2: 3 vectors span R^3 if and only if they're linearly independent.


    3. The attempt at a solution

    For Q1 i got: m=0,-2,1
    For Q2 i got: k != 1 - in other words, everything except 1.

    Are those answers right? How can i make sure that those are the only ones?
    Thanks.
     
  2. jcsd
  3. Mar 20, 2007 #2

    Dick

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    Homework Helper

    Are you using determinants?
     
  4. Mar 20, 2007 #3

    HallsofIvy

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    Use the definition of "linearly dependent", of course: there exist number A, B, C, not all 0, such that A(1-m, 2, 7)+ B(0,-2-m,12)+ C(0, 0, -m)= (0, 0, 0). That gives three equations: A(1-m)= 0, 2A+ B(-2-m)= 0, and 7A+ 12B- mC= 0. If, for example, m= 2 then the first equation is -2A= 0 so A= 0, the second equation is 2A- 4B= -4B= 0 so B= 0, and the third equation is 7A+ 12B- 2C= -2C= 0 so C= 0- ALL 0! For what values of m do you NOT get A= B= C= 0?

    Would it surprize you if I said "use the definition of "span"? A set of vectors spans R3 if some combination gives every vector in R3. That is, for any (x,y,z) there must be A, B, C, such that
    A(1, 2, k)+ B(0, 1, k-1)+ C(3, 4, 3)= (x,y,z). That gives the three equations
    A+ C= x, 2A+ B+ 3C= y, and kA+ (k-1)B+ 3C= z.

    From the first equation, A= x- C so the second equation becomes 2(x- C)+ B+ 3C= 2x+ B+ C= y. Then B= y- 2x- C. Putting A= x- C and B= y- 2x- C into the third equation, k(x- C)+ (k-1)(y- 2x- C)+ 3C= (4- 2k)+ ky- y- 3kx+2x= z or (4- 2k)C= z-ky+ y+3kx-2x. Obviously, if you can solve that for C, then you can find A and B. "Find all values of K" so that the given vectors span R3 is the same as finding all values of k so that you CAN solve that equation for C. It's simpler to first find the values of k for which you cannot solve that equation for C.


    2. Relevant equations
    Q1: vectors are linearly dependent if and only if one of the vectors is a linear combination of the others

    Q2: 3 vectors span R^3 if and only if they're linearly independent.


    3. The attempt at a solution

    For Q1 i got: m=0,-2,1
    For Q2 i got: k != 1 - in other words, everything except 1.

    Are those answers right? How can i make sure that those are the only ones?
    Thanks.[/QUOTE]

    Crud! I should have read the whole post before I wrote everything up! Yes, this is exactly correct. And you can be sure that those are the only ones because no other numbers make what I've said true.
     
  5. Mar 22, 2007 #4

    daniel_i_l

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    Gold Member

    Thanks for the help!
     
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