A 3-D ODE linear system solved by matrix exponential I am trying to solve the following linear system:
$$X'=\begin{pmatrix}0&1&0\\1&0&0\\1&1&1\end{pmatrix}X.$$
I'd like to use the exponential of the matrix, called $A$, so I've calculated
$$A^{2k}=\begin{pmatrix}1&0&0\\0&1&0\\2k&2k&1\end{pmatrix},\qquad A^{2k+1}=\begin{pmatrix}0&1&0\\1&0&0\\2k+1&2k+1&1\end{pmatrix}.$$
Then,
$$\exp(A)=\begin{pmatrix}e/2&e/2&0\\e/2&e/2&0\\e&e&e\end{pmatrix},$$
following that
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}e^t/2&e^t/2&0\\e^t/2&e^t/2&0\\e^t&e^t&e^t\end{pmatrix}\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}.$$
But so $z'=z$, and I should have $z'=x+y+z.$
What is the error?
Thank you so much!

There is some error in your computations. Note that if $v_1=(0,0,1)$, $v_2=\left(\frac12,\frac12,0\right)$, and $v_3=(1,0,0)$, then $A.v_1=v_1$, $A.v_2=v_2+v_1$, and $A.v_3=-v_3$. So, if$$P=\begin{bmatrix}0&\frac12&1\\0&\frac12&0\\1&0&0\end{bmatrix}$$(the columns of $P$ are $v_1$, $v_2$, and $v_3$), then$$P^{-1}AP=\begin{bmatrix}1&1&0\\0&1&0\\0&0&-1\end{bmatrix}.$$Therefore\begin{align}e^{At}&=P\exp\left(\begin{bmatrix}1&1&0\\0&1&0\\0&0&-1\end{bmatrix}t\right)P^{-1}\\&=P\begin{bmatrix}e^t&te^t&0\\0&e^t&0\\0&0&e^{-t}\end{bmatrix}P^{-1}\\&=\begin{bmatrix}\cosh(t)&\sinh(t)&0\\\sinh(t)&\cosh(t)&0\\te^t&te^t&e^t\end{bmatrix}.\end{align}Therefore, if$$\bigl(x(t),y(t),z(t)\bigr)=e^{At}(x_0,y_0,z_0),$$you have $x(t)=\cosh(t)x_0+\sinh(t)y_0$, $y(t)=\sinh(t)x_0+\cosh(t)y_0$, and $z(t)=te^tx_0+te^ty_0+e^tz_0$, and then $z'=x+y+z$.
