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Linear Algebra - parameters

  • Thread starter daniel_i_l
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  • #1
daniel_i_l
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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:

Homework Statement


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.


Homework 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.


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.
 

Answers and Replies

  • #2
Dick
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Are you using determinants?
 
  • #3
HallsofIvy
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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:

Homework Statement


Q1: Find all values of m so that the vectors:
{(1-m,2,7),(0,-2-m,12),(0,0,-m)} are linearly dependant.
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?

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


Homework 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.


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.
 
  • #4
daniel_i_l
Gold Member
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Thanks for the help!
 

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