Let $\mathcal{L}^{−1}\{⋅\}(z)$ be an inverse Laplace transform. Let $A, B$ be square matrixes, $I$ an identity matrix, and $\hat{\Phi}_z = \mathcal{L}\{\Phi(t)\}(z)$.
I have: $\hat{\Phi}_z = A[Iz - B]^{-1}$.
$\mathcal{L}^{-1}\{[Iz - B]^{-1}\}(z) = e^{Bt}$ by the properties of matrix exponential and by convolution theorem: $\mathcal{L}\{(f\star g)(t)\}(z) = F(z)G(z)$, what result can I get for $\Phi(t)$?
Is it $\Phi(t) = Ae^{Bt}$?
We have that $$\hat{\Phi}_z = \mathcal{L}\{\Phi(t)\}(z) = A[Iz - B]^{-1}$$ But $\mathcal{L}^{-1}\{[Iz - B]^{-1}\}(z) = e^{Bt}$ or $[Iz - B]^{-1}= \mathcal{L}\{ e^{Bt} \}$ so $$\mathcal{L}\{\Phi(t)\}(z) = A\mathcal{L}\{ e^{Bt} \}$$ But since $A$ is independent of $z$, we have that $$\mathcal{L}\{\Phi(t)\}(z) = \mathcal{L}\{ Ae^{Bt} \}$$ Applying inverse-Laplace on both sides $$\Phi(t) = Ae^{Bt} $$