Proof of Rules for Matrix Inverses

Each result cited is universally quantified over the data in its own statement. The rules of Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes are used with $m=n=p=q$. By Inverse Matrix and Invertible Real Square Matrix §invertible, an invertible $X\in\mathcal{M}_{n}(\mathbb{R})$ has exactly one inverse $X^{-1}\in\mathcal{M}_{n}(\mathbb{R})$, and $XX^{-1}=I_{n}=X^{-1}X$ by Inverse Matrix and Invertible Real Square Matrix §inverse; so if some $Y$ is an inverse of $X$, then $X$ is invertible and $X^{-1}=Y$.

identity. $I_{n}I_{n}=I_{n}$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity, so $I_{n}$ is an inverse of $I_{n}$; hence $I_{n}$ is invertible and $I_{n}^{-1}=I_{n}$.

one-sided. If $AC=I_{n}$, then $A$ is invertible by the equivalence of Criteria for Invertibility of a Square Real Matrix §right-inverse and Criteria for Invertibility of a Square Real Matrix §invertible, with $B=C$; from $AC=I_{n}$ and $A^{-1}A=I_{n}$, Uniqueness of the Matrix Inverse §unique (with $C$ and $A^{-1}$ in place of $B$ and $C$) gives $C=A^{-1}$. If $CA=I_{n}$, then $A$ is invertible by the equivalence of Criteria for Invertibility of a Square Real Matrix §left-inverse and Criteria for Invertibility of a Square Real Matrix §invertible; from $AA^{-1}=I_{n}$ and $CA=I_{n}$, Uniqueness of the Matrix Inverse §unique (with $A^{-1}$ in place of $B$) gives $A^{-1}=C$.

inverse. If $A$ is invertible, then $A^{-1}A=I_{n}$ and $AA^{-1}=I_{n}$, so $A$ is an inverse of $A^{-1}$; hence $A^{-1}$ is invertible and $(A^{-1})^{-1}=A$.

product. Let $A$ and $B$ be invertible. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §associative and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity,

$$(AB)(B^{-1}A^{-1})=A\bigl(B(B^{-1}A^{-1})\bigr)=A\bigl((BB^{-1})A^{-1}\bigr)=A(I_{n}A^{-1})=AA^{-1}=I_{n},$$

and likewise $(B^{-1}A^{-1})(AB)=B^{-1}\bigl(A^{-1}(AB)\bigr)=B^{-1}\bigl((A^{-1}A)B\bigr)=B^{-1}(I_{n}B)=B^{-1}B=I_{n}$. So $B^{-1}A^{-1}$ is an inverse of $AB$; hence $AB$ is invertible and $(AB)^{-1}=B^{-1}A^{-1}$.

transpose. Let $A$ be invertible. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §transpose, $A^{\top}(A^{-1})^{\top}=(A^{-1}A)^{\top}=I_{n}^{\top}=I_{n}$ and $(A^{-1})^{\top}A^{\top}=(AA^{-1})^{\top}=I_{n}^{\top}=I_{n}$. So $(A^{-1})^{\top}$ is an inverse of $A^{\top}$; hence $A^{\top}$ is invertible and $(A^{\top})^{-1}=(A^{-1})^{\top}$.

solution. Let $A$ be invertible and $b\in\mathbb{R}^{n}$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $A(A^{-1}b)=(AA^{-1})b=I_{n}b=b$. Conversely, if $x\in\mathbb{R}^{n}$ and $Ax=b$, then, again by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $x=I_{n}x=(A^{-1}A)x=A^{-1}(Ax)=A^{-1}b$. So $A^{-1}b$ is the only $x\in\mathbb{R}^{n}$ with $Ax=b$.