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Math Help - Why can I not get this matrix to make any sense?

  1. #1
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    Why can I not get this matrix to make any sense?

    I am trying to solve this differential equation system:
    \begin{cases}x'_1=x_3 \\ x'_2=2x_1-x_3 \\ x'_3=-x_1 \end{cases}

    I'm trying to write this into a matrix in the form of:

    x' = A x

    Where "A" is my matrix for this equation system. But when I try to create the matrix out of my own logic thinking, I get this:
    \begin{pmatrix}0 & 0 & 1 \\ 2 & 0 & -1 \\ -1 & 0 & 0 \end{pmatrix}

    This doesn't make any sense though since the determinant here is 0, which it shouldn't be. So what am I doing wrong?
    Last edited by MathIsOhSoHard; September 15th 2012 at 01:41 PM.
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    Re: Why can I not get this matrix to make any sense?

    Why do you say that does not make sense? Why can the determinant not be 0? That matrix has eigenvalues 0 (because the determinant is 0), i and -i. That means each x_i will be of the form A+ B cos(x)+ C sin(x).

    But, personally, I wouldn't use a matrix solution. Differentiating the first equation again, we have x_1''= x_3'= -x_1 so that x_1''+ x_1= 0. That equation has general solution x_1(t)= Acos(t)+ Bsin(t). Then x_3= x_1' so that x_3= -Asin(t)+ Bcos(t). Finally, x_2'= 2x_1- x_3= (2A- B)cos(t)+ (2B+A)sin(t) so that x_2= (2A- B)sin(t)- (2B+A)cos(t)+ C just as I said.
    Last edited by HallsofIvy; September 15th 2012 at 02:01 PM.
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    Re: Why can I not get this matrix to make any sense?

    I was rather slow there.
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    Re: Why can I not get this matrix to make any sense?

    Quote Originally Posted by HallsofIvy View Post
    Why do you say that does not make sense? Why can the determinant not be 0? That matrix has eigenvalues 0 (because the determinant is 0), i and -i. That means each x_i will be of the form A+ B cos(x)+ C sin(x).

    But, personally, I wouldn't use a matrix solution. Differentiating the first equation again, we have x_1''= x_3'= -x_1 so that x_1''+ x_1= 0. That equation has general solution x_1(t)= Acos(t)+ Bsin(t). Then x_3= x_1' so that x_3= -Asin(t)+ Bcos(t). Finally, x_2'= 2x_1- x_3= (2A- B)cos(t)+ (2B+A)sin(t) so that x_2= (2A- B)sin(t)- (2B+A)cos(t)+ C just as I said.
    How did you conclude that the matrix has the eigenvalues i and -i?

    My assignment says I need to find the complete complex solution for the differential equation system. And that the facit is this:
    x(t)=c_1 (0,1,0)+c_2e^{it}(1,-1-2i,i)+c_3e^{-it}(1,-1+2i,-i)

    By looking at the facit, it says that the eigenvalues are 0, i and -i but our textbook didn't mention anything about a matrix with eigenvalues of 0 so I have no idea how to conclude this.

    We haven't been taught any other way of solving this. According to my textbook, once I have the eigenvalues and eigenvectors, I can insert them into this solution:
    x(t)=e^{\lambda t} v
    where "lambda" is my eigenvalue and "v" is my eigenvectors. Since I have a system of 3 equations, I should end up with 3 solutions.

    But nobody ever mentioned to me that a matrix with determinant 0 also has the eigenvalues i and -i?
    I understand that -i is the conjugated value of i.
    Last edited by MathIsOhSoHard; September 15th 2012 at 02:16 PM.
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