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Find elements of a matrix such that its determinant is zero

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Please see the attached file if my inline insertion does not work.

    Screenshot from 2015-04-06 17:51:06.jpeg

    2. Relevant equations

    ##det(A)=det(A^T)##



    3. The attempt at a solution
    Since a matrix has a determinant of zero only when it's columns are linearly dependent, we look for a set of points [x1 x2] such that the first column of the matrix is linearly dependent. This is about as far as I got and I was wondering if this is the right approach or if there's a much more simpler way to approach this problem.
     
  2. jcsd
  3. Apr 6, 2015 #2

    BvU

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    Well, linear dependency means what in this case ? That you can write column 1 as a linear combination of columns 2,3,4. What do you get ?
     
  4. Apr 6, 2015 #3
    Yes, it means that. [x1 x2] can be any scalar multiple of [a1 a2], [b1 b2], and/or [c1 c2].

    Could I say that [x1 x2] is spanned by the vectors [a1 a2], [b1 b2], and[c1 c2]?
     
  5. Apr 6, 2015 #4

    BvU

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    That's 2 equations with 3 unknowns. You can pick the first row or the last row for the third unknown.
     
  6. Apr 6, 2015 #5
    I'm sorry I do not understand. What are you referring to as 2 equations and 3 unknowns?
     
  7. Apr 6, 2015 #6

    BvU

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    any ?

    Column one has four elements. you write down something for two of them only!

    When you work out what the three coefficients are

    $$\begin{pmatrix}x_1\\x_2\end{pmatrix} = u \begin{pmatrix}a_1\\a_2\end{pmatrix} + v \begin{pmatrix}b_1\\b_2\end{pmatrix} + w \begin{pmatrix}c_1\\c_2\end{pmatrix}$$

    2 eqns, 3 unknowns. Not good enough...
     
  8. Apr 6, 2015 #7
    Oh! I see what you are saying.
    We need one additional equation and that can be obtained by formulating an equation using the 1st row (i.e., 1=u+v+w) or the last column (i.e., ##x_1^2+x_2^2 = u(a_1^2+a_2^2)+v(b_1^2+b_2^2)+w(c_1^2+c_2^2)##)
     
  9. Apr 6, 2015 #8

    BvU

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    Yep. There are two more equations, so one of them to solve, the other to find a relationship between x1 and x2.
     
  10. Apr 6, 2015 #9

    Dick

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    You might also try the approach of trying to guess what the answer is. If you expand that determinant you are going to get a quadratic form in ##x_1## and ##x_2##. The solution to that is a conic section. Can you deduce enough facts about that conic section to guess what it is without cranking through a lot of algebra? Knowing that might guide you with the algebra. If you deduce enough you might be able to show it without the algebra.
     
  11. Apr 7, 2015 #10

    Dick

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    If you people have given up on this you should come back. There is actually a very neat way to solve this one.
     
  12. Apr 7, 2015 #11

    Mark44

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    It makes no sense to talk about a single column being dependent (excepting the possibility that all of the elements of the column are zero).
     
  13. Apr 11, 2015 #12
    Hello all, sorry for the late reply. I was caught up in other courses and neglected this one. I am doing this now, first using the 3 equation, 3 unknowns approach.

    I haven't attempted to expand, as I was intimated by a 4x4 matrix of all non-zeros.
     
  14. Apr 11, 2015 #13
    Is the neat way by expanding out the determinant?
     
  15. Apr 11, 2015 #14
    I took the determinant of this matrix using matlab.
    Here is what I got:

    a1^2*b1*c2 - a1^2*b1*x2 - a1^2*b2*c1 + a1^2*b2*x1 + a1^2*c1*x2 - a1^2*c2*x1 - a1*b1^2*c2 + a1*b1^2*x2 - a1*b2^2*c2 + a1*b2^2*x2 + a1*b2*c1^2 + a1*b2*c2^2 - a1*b2*x1^2 - a1*b2*x2^2 - a1*c1^2*x2 - a1*c2^2*x2 + a1*c2*x1^2 + a1*c2*x2^2 + a2^2*b1*c2 - a2^2*b1*x2 - a2^2*b2*c1 + a2^2*b2*x1 + a2^2*c1*x2 - a2^2*c2*x1 + a2*b1^2*c1 - a2*b1^2*x1 - a2*b1*c1^2 - a2*b1*c2^2 + a2*b1*x1^2 + a2*b1*x2^2 + a2*b2^2*c1 - a2*b2^2*x1 + a2*c1^2*x1 - a2*c1*x1^2 - a2*c1*x2^2 + a2*c2^2*x1 - b1^2*c1*x2 + b1^2*c2*x1 + b1*c1^2*x2 + b1*c2^2*x2 - b1*c2*x1^2 - b1*c2*x2^2 - b2^2*c1*x2 + b2^2*c2*x1 - b2*c1^2*x1 + b2*c1*x1^2 + b2*c1*x2^2 - b2*c2^2*x1


    OUCH!!!! There must be an easier way to do this.
     
  16. Apr 11, 2015 #15

    Dick

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    Yes, there is a much easier way. Your expansion is quadratic in ##x_1## and ##x_2##. Where it's zero is a conic section as I said before. I.e. it's an ellipse, hyperbola, etc. That narrows it down a bit. Now go back to thinking about the determinant. Can you name three points on the conic section by thinking about the determinant? Can you narrow down the form of the conic by thinking more about the coefficients of the quadratic terms?
     
  17. Apr 11, 2015 #16
    I am struggling to view this in terms of geometry. I'm not a very visual person when it comes to math.

    Can you explain what you mean by "your expansion is quadratic in ##x_1## and ##x_1##"?
     
  18. Apr 11, 2015 #17

    Dick

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    I mean that it is like ##x^2+y^2=1## is a circle. And ##x^2-y^2=1## is a hyperbola. Like conic sections. You must have done these at some point. Try to visualize the graph in ##x_1## and ##x_2## like it's a graph in ##x## and ##y##. All the terms have degree 2 or less. It must be a conic section, yes?
     
  19. Apr 11, 2015 #18
    Yes, quite awhile ago. I am familiar with conics as I do a numerical work with hyperbolic PDEs, but I am having a hard time tying geometrical concepts with this matrix.

    So the 4 terms in the last row all represent a conic section?
     
  20. Apr 12, 2015 #19

    Dick

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    ALL of terms put together represent a conic section. It has the form ##Ax_1^2+Bx_1x_2+Cx_2^2+Dx_1+Ex_2+F=0##, yes? Where the constants are combinations of a's b's and c's? What is B? That's a start.
     
    Last edited: Apr 12, 2015
  21. Apr 12, 2015 #20
    Oh I see.
    I'm guessing B=0? But I am unsure why.

    If that's right, then we would only have linear terms.
     
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