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zonazoo Group Title

For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).

  • one year ago
  • one year ago

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  1. vf321 Group Title
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    In \(ax + by + cz = d\), if x and y are any number in R(1), then z is constrained to \[z=\frac{d-ax-by}{c}\] The subspace S = { any (x, y, z)} is a plane which does not pass through (0,0,0) and is therefore not a subspace.

    • one year ago
  2. vf321 Group Title
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    Actually, I failed to address all the trivial cases. If a = b = c = 0, there is no solution. The empty set is not a subspace. If one or two of a,b, or c equals 0, then you can rotate your axes such that the one that is not 0 is the constant for z, in which case the theorem holds.

    • one year ago
  3. zonazoo Group Title
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    so your saying that if atleast a, b, or c is 0 then the others can be non zero numbers and be a subspace of R(3)?

    • one year ago
  4. vf321 Group Title
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    No, I'm saying that if d = 0 then the solution set (x, y, z) is a subspace of R(3) UNLESS all three are equal to 0.

    • one year ago
  5. vf321 Group Title
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    ll three a, b, c****

    • one year ago
  6. zonazoo Group Title
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    and that would be the same thing if it was a, b, c to the equation ax+by=c form a subspace of R(2)... then c=0 in this case.

    • one year ago
  7. zonazoo Group Title
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    also, if d=0, then as long as atleast one of a, b, or c is non zero then it IS a subspace right?

    • one year ago
  8. helder_edwin Group Title
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    actually the set u r given is a subspace if and only if d=0.

    • one year ago
  9. zonazoo Group Title
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    but why? and does it matter what a, b, and c are?

    • one year ago
  10. helder_edwin Group Title
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    no.

    • one year ago
  11. helder_edwin Group Title
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    u have \[ \large D=\{(x,y,z):ax+by+cz=d\} \]

    • one year ago
  12. helder_edwin Group Title
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    if d=0 then u have \[ \large D=\{(x,y,z):ax+by+cz=0\}. \] So let \(u=(x_0,y_0,z_0)\), \(v=(x_1,y_1,z_1)\) both in \(D\) and \(\alpha\in\mathbb{R}\) so (i) \(u+v=(x_0+x_1,y_0+y_1,z_0+z_1)\) so \[ \large a(x_0+x_1)+b(y_0+y_1)+c(z_0+z_1)= \] \[ \large =(ax_0+by_0+cz_0)+(ax_1+by_1+cz_1)=0+0=0 \] (ii) \(\alpha v=(\alpha x_1,\alpha y_1,\alpha z_1)\) so \[ \large a(\alpha x_1)+b(\alpha y_1)+c(\alpha z_1)=\alpha(ax_1+by_1+cz_1= \alpha0=0 \] this proves that D is a subspace of R^3.

    • one year ago
  13. helder_edwin Group Title
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    For the converse. Let's suppose that \[ \large D=\{(x,y,z):ax+by+cz=d\} \] is a subspace of R^3. then the zero vector MUST be in D so \[ \large 0=a0+b0+c0=d \]

    • one year ago
  14. zonazoo Group Title
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    I do understand that for sure. but how did you know to make d=0, or how do you know it doesn't work for any non zero number?

    • one year ago
  15. zonazoo Group Title
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    ohhh okay.

    • one year ago
  16. helder_edwin Group Title
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    therefore D is a subspace if and only if d=0

    • one year ago
  17. zonazoo Group Title
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    so that same idea applies if the question is ax+by=c, for scalars a, b, c ... subspace of R(2) right

    • one year ago
  18. helder_edwin Group Title
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    yes

    • one year ago
  19. zonazoo Group Title
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    It seems like the book is trying to trick me... why give one question stating for ax+by=c, and then the question right after that be ax+by+cz=d, when c has to be 0 for the first and d has to be 0 for the second. Its the same question isnt it, or is there more to it that I am missing.

    • one year ago
  20. helder_edwin Group Title
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    yes it is. the book is trying to stress that for a subset of a given to be a subspace (i.e., a space on its own) the zero vector (which is unique) must also belong to the subset in question.

    • one year ago
  21. helder_edwin Group Title
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    that's what u r supposed to get from this two exercises.

    • one year ago
  22. helder_edwin Group Title
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    *given space*

    • one year ago
  23. helder_edwin Group Title
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    i hope i was helpful gotta go

    • one year ago
  24. zonazoo Group Title
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    it most definitely was... i hope to see you around again, you have been very helpful. thank you.

    • one year ago
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