The Inverse command, when applied to an object, represents the object's (noncommutative) multiplicative inverse; that is, Inverse(Z) * Z = Z * Inverse(Z) = 1, where * herein represents the Physics[*] product, whose commutativity depends on the operands (see also type, commutative).

The %Inverse command is the inert form of Inverse; that is, it represents the same mathematical operation while displaying the operation unevaluated. To evaluate the operation, use the value command.

The results returned by Inverse are constructed as follows:

- If $f$ is of commutative type, then return $\frac{1}{f}$.

- If $f$ is a matrix, then return its inverse.

- If $f$ is equal to Inverse(g) for some $g$, then return $g$.

- If $f$ is a noncommutative product, then distribute:$\mathrm{Inverse}\left(A*B\right)\to \mathrm{Inverse}\left(B\right)*\mathrm{Inverse}\left(A\right)$.

- If $f$ is a * (commutative) product, then distribute:$\mathrm{Inverse}\left(A*B\right)\to \mathrm{Inverse}\left(A\right)*\mathrm{Inverse}\left(B\right)$.

- Otherwise, return the unevaluated expression $'\mathrm{Inverse}'\left(f\right)$.

All noncommutative products introduced by Inverse have their operands sorted and normalized automatically by the Physics[*] operator. This ensures that the basic simplifications and identities for these products are taken into account in the returned results.

A `print/Inverse` procedure makes the display of this function appear as a power, as in

Inverse(Q);

$Q^{\mathrm{-1}}$

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

$\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$

$\left[\mathrm{mathematicalnotation}=\mathrm{true}\right]$

$\mathrm{Setup}\left(\mathrm{anticommutativeprefix}=Q\,\mathrm{noncommutativeprefix}=Z\right)$

$\left[\mathrm{anticommutativeprefix}=\left\{Q\right\}\,\mathrm{noncommutativeprefix}=\left\{Z\right\}\right]$

$\mathrm{Inverse}\left(\mathrm{Z1}\right)\mathrm{Z1}$

$1$

$\left[a\,\mathrm{Inverse}\left(Q\right)\,\mathrm{Q1}\mathrm{Z2}\,AB\,a\mathrm{Q1}\mathrm{Q2}\right]$

$\left[a\,Q^{-1}\,\mathrm{Q1}\mathrm{Z2}\,AB\,a\mathrm{Q1}\mathrm{Q2}\right]$

$\mathrm{map}\left(\mathrm{Inverse}\,\right)$

$\left[\frac{1}{a}\,Q\,{\mathrm{Z2}}^{-1}{\mathrm{Q1}}^{-1}\,\frac{1}{AB}\,-\frac{{\mathrm{Q1}}^{-1}{\mathrm{Q2}}^{-1}}{a}\right]$

$\mathrm{map}\left(\mathrm{Inverse}\,\right)$

$\left[a\,Q^{-1}\,\mathrm{Q1}\mathrm{Z2}\,AB\,a\mathrm{Q1}\mathrm{Q2}\right]$