> [!NOTE] Corollary > If $A\in \text{Mat}_{nn}$ is invertible then $A^{-1}$ is invertible with $(A^{-1})^{-1}=A$. *Proof*. $(A^{-1})A=A(A^{-1})=I_{n}$ and so $(A^{-1})^{-1}=A$ by uniqueness of inverses.