How to pass from Laplace transform ${\hat{\mathrm{\Psi}}}_{z}$ to ${\mathrm{\Psi}}_{z}$?

Let's define the Laplace Transform

$${\hat{\mathrm{\Psi}}}_{z}={\int}_{\mathbb{R}}{e}^{-izt}{\mathrm{\Psi}}_{t}\phantom{\rule{thinmathspace}{0ex}}dt$$

and the Laplace Transform

$${\hat{\mathrm{\Phi}}}_{z}={\int}_{\mathbb{R}}{e}^{-izt}{\mathrm{\Phi}}_{z}\phantom{\rule{thinmathspace}{0ex}}dt$$

If I know that

$${\hat{\mathrm{\Psi}}}_{z}={\int}_{\mathbb{R}}{e}^{-izt}{\mathrm{\Psi}}_{t}\phantom{\rule{thinmathspace}{0ex}}dt={\hat{\mathrm{\Phi}}}_{z}(I-{\hat{\mathrm{\Phi}}}_{z}{)}^{-1}$$

where I is the identity matrix - how can I get ${\mathrm{\Psi}}_{z}?$?

Let's define the Laplace Transform

$${\hat{\mathrm{\Psi}}}_{z}={\int}_{\mathbb{R}}{e}^{-izt}{\mathrm{\Psi}}_{t}\phantom{\rule{thinmathspace}{0ex}}dt$$

and the Laplace Transform

$${\hat{\mathrm{\Phi}}}_{z}={\int}_{\mathbb{R}}{e}^{-izt}{\mathrm{\Phi}}_{z}\phantom{\rule{thinmathspace}{0ex}}dt$$

If I know that

$${\hat{\mathrm{\Psi}}}_{z}={\int}_{\mathbb{R}}{e}^{-izt}{\mathrm{\Psi}}_{t}\phantom{\rule{thinmathspace}{0ex}}dt={\hat{\mathrm{\Phi}}}_{z}(I-{\hat{\mathrm{\Phi}}}_{z}{)}^{-1}$$

where I is the identity matrix - how can I get ${\mathrm{\Psi}}_{z}?$?