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Use cross product formula in R^4 to obtain orthogonal vector

  1. Sep 18, 2011 #1

    sharks

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    Use cross product formula in R^4 to obtain a vector that is orthogonal to rows of A

    Please help with first part and check if i answered the questions correctly.

    The matrix A =
    1 4 -1 2
    0 1 0 -1
    2 9 -2 2


    1. Use cross product formula in R^4 to obtain a vector that is orthogonal to the rows of A.

    I multiply the matrix A by a 4x1 matrix X and equate to 0.

    Matrix X =
    x1
    x2
    x3
    x4

    Then i get:
    x1+ 4x2 - 3x3 + 2x4 =0
    x2 + x4 = 0
    2x1 + 9x2 - 2x3 + 2x4 =0


    x1 = -4x2 + 3x3 - 2x4
    x2 = - x4
    x3 = x3
    2x4 = -2x1- 9x2 + 2x3

    And then... i'm stuck.

    2. How is this vector related to the null space of A?
    My answer: That vector is perpendicular to the null space of A.
    Is this correct and is there another way to put it?
     
    Last edited: Sep 18, 2011
  2. jcsd
  3. Sep 18, 2011 #2

    D H

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    Your first two equations are inconsistent with the matrix you posted.

    The title of the thread doesn't make much sense. There is no cross product formula in R^4. Do you mean the formula for the product of a matrix and a vector? (Which you did use, if erroneously.)

    You will end up with three equations in four unknowns. The second row of A lets you quickly eliminate one of x1, x2, x3, and x4. That will leave you with two equations in three unknowns. See if you can get to, and possibly beyond, this point.
     
  4. Sep 18, 2011 #3

    sharks

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    OK, i will get to it for the first part. But is the second part correct?
     
  5. Sep 18, 2011 #4

    D H

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    What second part?
     
  6. Sep 18, 2011 #5

    sharks

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    This part.
     
  7. Sep 18, 2011 #6

    D H

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    The vector is in the null space of A, not perpendicular to it.
     
  8. Sep 18, 2011 #7

    sharks

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    I'm sorry but i haven't understood anything from your replies. It's obvious that you haven't even taken the time to read my question properly. You missed my second question despite using bold for the questions, and you haven't been helpful either with the first part of the question. The question is as it is. The title is based on part of the first question. The entire question itself (as well as my attempts) is in the first post.
    Not being ungrateful here, but your replies have made this whole troubleshooting process even more confusing, if that's even possible.

    I have fixed the title to make it match the first question.
     
  9. Sep 18, 2011 #8

    D H

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    What is unclear about say that you did the matrix multiplication wrong? I gave you a hint for how to go further after you get that multiplication correct. I'll help you more if you get stuck.

    What is unclear about saying that you were wrong in saying that "That vector is perpendicular to the null space of A"? For crying out loud, I out and out gave you the answer to that part of the question.
     
  10. Sep 18, 2011 #9

    vela

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    For part 1, I think you're actually being asked to calculate
    \begin{vmatrix}
    \hat{x} & \hat{y} & \hat{z} & \hat{w} \\
    1 &4 &-1 &2 \\
    0 &1 &0 &-1 \\
    2 &9 &-2 & 2
    \end{vmatrix}
     
  11. Sep 18, 2011 #10

    D H

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    I took the first part as meaning finding a 4-vector such that it is orthogonal to each of the three rows of the given 3x4 matrix. I take the title of the thread as meaning that any 4-vector that satisfies these conditions will suffice; all that is needed is one (as opposed to a a general expression that describes all such 4-vectors).

    Since the OP hasn't come back in a while, the problem with the first part is that the matrix multiplication does not jibe with the matrix. Given

    [tex]A=\begin{bmatrix}
    1 & 4 & -1 & \phantom{-}2 \\
    0 & 1 & \phantom{-}0 & -1 \\
    2 & 9 & -2 & \phantom{-}2\end{bmatrix}[/tex]

    and [itex]x=[x_1\,\,x_2\,\,x_3\,\,x_4]^T[/itex], the matrix multiplication [itex]Ax[/itex] yields

    [tex]
    \begin{matrix}
    x_1 &+& 4x_2 &-& x_3 &+& 2x_4 &=& 0 \\
    &&\phantom{1}x_2 && &-& \phantom{1}x_4 &=& 0 \\
    2x_1 &+& 9x_2 &-& 2x_3 &+& 2x_4 &=& 0
    \end{matrix}[/tex]

    The OP came up with
    This corresponds to a matrix

    [tex]A=\begin{bmatrix}
    1 & 4 & -3 & 2 \\
    0 & 1 & \phantom{-}0 & 1 \\
    2 & 9 & -2 & 2\end{bmatrix}[/tex]

    So which is correct, the matrix as shown in the OP, or the matrix as inferred from the OP's matrix multiplication?
     
  12. Sep 18, 2011 #11

    vela

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    I don't know either. The OP needs to clarify which matrix is correct and exactly what the problem statement was. I just inferred from the mention of the "cross product" that the OP was meant to use the generalization to R4 involving the determinant.
     
  13. Sep 19, 2011 #12

    sharks

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    The matrix multiplication that i did is wrong. By the way, how do i type the matrices in here as neatly as you did?
     
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