If I have this:
Suppose I have a matrix:
If I have this:
$$\dfrac{d}{dt} \begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
and when $t=0$ $$\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$$
And additionaly \begin{align}e^{tA}&=M^{-1}\begin{bmatrix}1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix}M\\&=\begin{bmatrix}\frac{1}{2} \left(-t^2+t+2\right) & \frac{1}{16} (t-4) t & -\frac{1}{32} (t-2) t \\ 2 t (t+3) & 1-\frac{t^2}{4} & \frac{1}{8} t (t+2) \\ 4 t (3 t+1) & \frac{1}{2} (8-3 t) t & \frac{1}{4} \left(3 t^2-2 t+4\right)\end{bmatrix}\end{align}
Then how am I supposed to solve this problem when $t=0$.
Do I simply substitute $t=0$ into $\exp\left[tA \right]$ as this appears too obvious?
I also apologise for the mess with matrices, I am relatively new to this forum.
Thank you for any help.
If you have the dynamical system
$$\boldsymbol{\dot{x}}(t) = \boldsymbol{Ax} \qquad ,\quad \boldsymbol{x}(t=t_0)=\boldsymbol{x}_0.$$
Then the general solution is given by
$$\boldsymbol{x}(t) = \exp\left[(t-t_0)\boldsymbol{A} \right]\boldsymbol{x}_0.$$