Proofread matrix multiply inverse of matrix are unique Prove $\textbf{A}^{-1}$ is unique for $\textbf{A}\textbf{A}^{-1} = \textbf{I}$
Assume 
$$\textbf{B} \neq \textbf{A}^{-1} \text{ and }\textbf{A}\textbf{B} = \textbf{I}$$
$$\textbf{A}\textbf{A}^{-1} = \textbf{I}$$
$$\Rightarrow \textbf{A}\textbf{A}^{-1} - \textbf{I} = \textbf{0}$$
$$\Rightarrow \textbf{A}\textbf{A}^{-1} - \textbf{A}\textbf{B} = \textbf{0}$$
$$\Rightarrow \textbf{A}(\textbf{A}^{-1} - \textbf{B})=\textbf{0}$$
$$\because (\textbf{A}^{-1}-\textbf{B}) \neq \textbf{0}$$
$$\therefore\text{matrix}\ (\textbf{A}^{-1}-\textbf{B})\ \text{contains at lease one column vector }\vec{v}_i \neq 0, \text{otherwise the matrix is zero matrix}$$
$$\text{let } \textbf{A}^{-1} - \textbf{B}=\lbrack\vec{v}_0, \vec{v}_1...\vec{v}_i...\vec{v}_n\rbrack$$
$$\textbf{A}\lbrack\vec{v}_0,\vec{v}_1...\vec{v}_i...\vec{v}_n\rbrack = \textbf{0}$$
$$\lbrack\textbf{A}\vec{v}_0,\textbf{A}\vec{v}_1...\textbf{A}\vec{v}_i...\textbf{A}\vec{v}_n\rbrack=\textbf{0}\tag{1}$$
$$\because \textbf{A}\text{ is invertible and }\vec{v}_i \neq \vec{0}$$
$$\therefore \textbf{A}\vec{v}_i \neq \vec{0}$$
$$\text{But }\vec{A}\vec{v}_i = \vec{0} \text{ from }\text{(1)} $$
$$\text{This contracts our assumption which is }\textbf{A}^{-1} \neq \textbf{B}$$
$$\Rightarrow \textbf{A}^{-1} = \textbf{B}$$
$$\textbf{Therefore }\textbf{A}^{-1}\text{ is unique}$$
 A: The justifcation does not hold:

"from $A(A^{-1}-B)=0$ and $A\neq 0$, it follows $A^{-1}-B=0$" 

A proper justification would be 

Since $A(A^{-1}-B)=0$ and $A$ is nonsingular, then $A^{-1}-B=0$.

This presupposes that you understand why $A$ is nonsingular and know that it makes a 1-1 mapping, though!
A: I'd like to give a counterexample, to give you a feel why the above is not precise enough.
You say:

Prove $A^{−1}$ is unique for $AA^{−1} = I$.

This reads:

We denote $A^{-1} := B$ for $B$ such that $AB = I$. Prove that such $B$ is unique.

Now, let
$$A := \begin{bmatrix} 1 & 0 \end{bmatrix}, \quad B := \begin{bmatrix} 1 \\ x \end{bmatrix}, \quad x \in \mathbb{R}.$$
Then
$$AB = \begin{bmatrix} 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix} = I,$$
for any $x \in \mathbb{R}$, so such $B$ is not unique.
You're missing the condition that $A$ is nonsingular, or that $A$ is square, or that $A^{-1}$ is both left and right inverse, or... But the way you put it, the statement is false and, hence, cannot be proven.
