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

I am trying to solve this differential equation system:

$\displaystyle \begin{cases}x'_1=x_3 \\ x'_2=2x_1-x_3 \\ x'_3=-x_1 \end{cases}$

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:

$\displaystyle \begin{pmatrix}0 & 0 & 1 \\ 2 & 0 & -1 \\ -1 & 0 & 0 \end{pmatrix}$

This doesn't make any sense though since the determinant here is 0, which it shouldn't be. So what am I doing wrong? :(

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 $\displaystyle x_i$ 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 $\displaystyle x_1''= x_3'= -x_1$ so that $\displaystyle x_1''+ x_1= 0$. That equation has general solution $\displaystyle x_1(t)= Acos(t)+ Bsin(t)$. Then $\displaystyle x_3= x_1'$ so that $\displaystyle x_3= -Asin(t)+ Bcos(t)$. Finally, $\displaystyle x_2'= 2x_1- x_3= (2A- B)cos(t)+ (2B+A)sin(t)$ so that $\displaystyle x_2= (2A- B)sin(t)- (2B+A)cos(t)+ C$ just as I said.

Re: Why can I not get this matrix to make any sense?

Re: Why can I not get this matrix to make any sense?

Quote:

Originally Posted by

**HallsofIvy** 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 $\displaystyle x_i$ 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 $\displaystyle x_1''= x_3'= -x_1$ so that $\displaystyle x_1''+ x_1= 0$. That equation has general solution $\displaystyle x_1(t)= Acos(t)+ Bsin(t)$. Then $\displaystyle x_3= x_1'$ so that $\displaystyle x_3= -Asin(t)+ Bcos(t)$. Finally, $\displaystyle x_2'= 2x_1- x_3= (2A- B)cos(t)+ (2B+A)sin(t)$ so that $\displaystyle x_2= (2A- B)sin(t)- (2B+A)cos(t)+ C$ 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:

$\displaystyle x(t)=c_1 (0,1,0)+c_2e^{it}(1,-1-2i,i)+c_3e^{-it}(1,-1+2i,-i)$

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:

$\displaystyle x(t)=e^{\lambda t} v$

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.