Differential equation with matrix

Question

How can we find $x′ = Ax$ for the matrix

\begin{pmatrix}3 &1 &0\\ 0& 3 &0\\ 0& 0& 2\end{pmatrix}

Answer 1

Solving $x' = ax$ for real constant $a$ is $x(t) = e^{ta}x(0)$. By analogy in the multi-dimensional world, $x'=Ax$ is solved by $\vec{x} = e^{tA} \vec{x}_0$.

Exponentiating $A$ typically requires diagonalizing it, as if $A = VDV^T$ then $e^{tA} = Ve^{tD}V^T$ and exponents of diagonal matrices just exponentiate the diagonal...