= Solution
A finite-dimensional <representation> of a <group> $G$ on a vector space $V$ is a <group homomorphism> $D:G\to\operatorname{GL}(V)$. Thus each $D(X)$ is invertible and
$$
D(XY)=D(X)D(Y),\qquad D(I)=I_V.
$$
Here the group identity and the identity linear map are different objects, represented by corresponding symbols. Applying the multiplication property to $XX^{-1}=X^{-1}X=I$ gives
$$
D(X)D(X^{-1})=I_V=D(X^{-1})D(X).
$$
By the defining property of a <matrix inverse>,
$$
\boxed{D(X^{-1})=D(X)^{-1}.}
$$
If the definition only states the multiplication law together with invertibility, $D(I)^2=D(I)$ already forces $D(I)=I_V$ after multiplication by its inverse.
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