Proofread matrix multiply inverse of matrix are unique

Proofread matrix multiply inverse of matrix are unique

Prove $\textbf{A}^{-1}$ is unique for $\textbf{A}\textbf{A}^{-1} =
\textbf{I}$
Assume $$\textbf{B} \neq \textbf{A}^{-1} \text{ and }\textbf{A}\textbf{B}
= \textbf{I}$$ $$\textbf{A}\textbf{A}^{-1} = \textbf{I}$$ $$\Rightarrow
\textbf{A}\textbf{A}^{-1} - \textbf{I} = \textbf{0}$$ $$\Rightarrow
\textbf{A}\textbf{A}^{-1} - \textbf{A}\textbf{B} = \textbf{0}$$
$$\Rightarrow \textbf{A}(\textbf{A}^{-1} - \textbf{B})=\textbf{0}$$
$$\because (\textbf{A}^{-1}-\textbf{B}) \neq \textbf{0}$$
$$\therefore\text{matrix}\ (\textbf{A}^{-1}-\textbf{B})\ \text{contains at
lease one column vector }\vec{v}_i \neq 0, \text{otherwise the matrix is
zero matrix}$$ $$\text{let } \textbf{A}^{-1} -
\textbf{B}=\lbrack\vec{v}_0, \vec{v}_1...\vec{v}_i...\vec{v}_n\rbrack$$
$$\textbf{A}\lbrack\vec{v}_0,\vec{v}_1...\vec{v}_i...\vec{v}_n\rbrack =
\textbf{0}$$
$$\lbrack\textbf{A}\vec{v}_0,\textbf{A}\vec{v}_1...\textbf{A}\vec{v}_i...\textbf{A}\vec{v}_n\rbrack=\textbf{0}\tag{1}$$
$$\because \textbf{A}\text{ is invertible and }\vec{v}_i \neq \vec{0}$$
$$\therefore \textbf{A}\vec{v}_i \neq \vec{0}$$ $$\text{But
}\vec{A}\vec{v}_i = \vec{0} \text{ from }\text{(1)} $$ $$\text{This
contracts our assumption which is }\textbf{A}^{-1} \neq \textbf{B}$$
$$\Rightarrow \textbf{A}^{-1} = \textbf{B}$$ $$\textbf{Therefore
}\textbf{A}^{-1}\text{ is unique}$$