The ilaplace(M) calling sequence computes the element-wise inverse Laplace transform of M.  The result, R, is formed as R[i,j] = ilaplace(M[i,j], y, x).

ilaplace(L) is the inverse Laplace transform of the scalar L with default independent variable s.  If L is not a function of s, then L is  assumed to be a function of the independent variable returned by findsym(L,1).The default return is a function of t.

If L = L(t), then ilaplace returns a function of x.

By definition,

ilaplace(L,y) makes F a function of the variable y instead of the default t.

ilaplace(L,y,x) takes L to be a function of x instead of the default t. The integration is then with respect to y.

$\mathrm{with}\left(\mathrm{MTM}\right)\:$

$\mathrm{ilaplace}\left(\frac{1}{s^2+4}\right)$

$\frac{\sin \left(2t\right)}{2}$

$\mathrm{ilaplace}\left(\frac{1}{t^2+4}\right)$

$\frac{\sin \left(2x\right)}{2}$

$\mathrm{ilaplace}\left(\frac{1}{s^2+4}\,w\right)$

$\frac{\sin \left(2w\right)}{2}$

$\mathrm{ilaplace}\left(\frac{z}{s^2+4}\,z\,q\right)$

$\frac{\mathrm{Dirac}\left(1\,q\right)}{s^2+4}$

$M≔\mathrm{Matrix}\left(\left[\frac{1}{s^2+4}\,\frac{z}{s^2+4}\right]\right)\:$

$\mathrm{ilaplace}\left(M\right)$

$\left[\begin{array}{cc}\frac{\sin \left(2t\right)}{2}& \frac{z\sin \left(2t\right)}{2}\end{array}\right]$