Question

Solve x'=(1, 0, -1, 1, 2, 1, 2, 2, 3)x, x(0)=(2, 0, 1). (1, 0, -1 on the left, 1, 2, 1 in the middle, 2, 2, 3 on the right. I was trying to find the roots, but I got -x(x^2+11)+6(x^2+1) after factoring by grouping. can someone check my work?)

Answer 1

\[x'=\begin{pmatrix}1&0&-1\\1&2&1\\2&2&3\end{pmatrix}x\]
\[\begin{vmatrix}1-\lambda&0&-1\\1&2-\lambda&1\\2&2&3-\lambda\end{vmatrix}\\
~~~~~~=(1-\lambda)\begin{vmatrix}2-\lambda&1\\2&3-\lambda\end{vmatrix}+(-1)(0)\begin{vmatrix}1&1\\2&3-\lambda\end{vmatrix}+(-1)\begin{vmatrix}1&2-\lambda\\2&2\end{vmatrix}\\
~~~~~~=(1-\lambda)\bigg((2-\lambda)(3-\lambda)-2\bigg)-\bigg(2-2(2-\lambda)\bigg)\\
~~~~~~=-(\lambda-1)(\lambda^2-5\lambda+4)-2(\lambda-1)\\
~~~~~~=-(\lambda-1)\bigg(\lambda^2-5\lambda +4+2\bigg)\\
~~~~~~=-(\lambda-1)(\lambda^2-5\lambda+6)\\
~~~~~~=-(\lambda-1)(\lambda-3)(\lambda-2)
\]

Answer 2

@Loser66

Answer 3

hey, girl, use the draw box to draw out your problem, I don't get the matrix and the initial value of x(0)

Answer 4

this what I get. I assume the matrix like this. If it is not, you just imitate the steps. ok?

Answer 5

Is the latex not showing up? See attached.

Answer 6

@SithsAndGiggles It showed. However, she explained that the vectors are [yours] transpose, not that. She said: 1, 0, -1 on the left. I thought it should be \[\left[\begin{matrix}1\\0\\-1\end{matrix}\right]\text{on the left side, like this}\]. To me, the order of the vectors is important since we find the eigenvectors base on that matrix.
@Idealist Please, practice latex. Nothing is easy at the beginning but it is NOT impossible.

Answer 7

Ah, right. I think I had even asked Idealist to list the matrix elements one row at a time... Make latex typing more convenient.