So, let us suppose that $A$ is a square matrix, and that $B$ is a matrix such that $BA=I$. You want to show that $B$ is the unique left inverse of $A$ (that is).
Note that a system $A\mathbf{x}=\mathbf{b}$ has at most one solution, namely $B\mathbf{b}$: if $A\mathbf{x}=\mathbf{b}$, then
$$\mathbf{x} = I\mathbf{x} = BA\mathbf{x} = B\mathbf{b}.$$
If $CA=I$, then again a system $A\mathbf{x}=\mathbf{b}$ has at most one solution, namely $C\mathbf{b}$. Thus, $B\mathbf{b}=C\mathbf{b}$ for any $\mathbf{b}$ for which the system has a solution.
If we can show that $A\mathbf{x}=\mathbf{e}_i$ has a solution for each $i$, where $\mathbf{e}_i$ is the $i$th standard basis vector ($1$ in the $i$th entry, $0$s elsewhere) this will show that $B=C$, since they have the same columns.
Because $A\mathbf{x}=\mathbf{0}$ has a solution, that solution must be $B\mathbf{0}=\mathbf{0}$. That means that the reduced row-echelon form of $A$ is $I$. Because the reduced row-echelon form of $A$ is $I$, performing row reduction on the augmented coefficient matrix $[A|\mathbf{e}_i]$ yields the matrix $[I|\mathbf{y}]$ for some $\mathbf{y}$, with $\mathbf{y}$ being the solution to $A\mathbf{x}=\mathbf{e}_i$. Since this vector is equal to both $\mathbf{b}_i=B\mathbf{e}_i$ (the $i$th column of $B$) and to $\mathbf{c}_i=C\mathbf{e}_i$, as noted above, then the $i$th columns of $B$ and $C$ are equal; thus, $B=C$, and the matrix has a unique left inverse.
Now, let us suppose that $A$ is a square matrix and has a right inverse, $AB=I$. We want to show that $B$ is the unique right inverse of $A$. Taking transposes, we get $I = I^T = (AB)^T = B^TA^T$. By what was proven above, $B^T$ is the unique left inverse of $A^T$. If $AC=I$, then $C^TA^T=I^T = I$, so $C^T=B^T$, hence $C=B$. Thus, $B$ is the unique right inverse of $A$.
The determinant is given by
$$\det A =
\begin{vmatrix}
1 & 2 & -1 \\
\color{red}2 & \color{red}0 & \color{red}2\\
-1 & 2 & k
\end {vmatrix} =
-\color{red} 2\begin{vmatrix}2 & -1 \\ 2 & k \end {vmatrix}
+\color{red} 0\begin{vmatrix}1 & -1 \\ -1 & k \end {vmatrix}
-\color{red} 2\begin{vmatrix}1 & 2 \\ -1 & 2 \end {vmatrix}
\\= -2(2k+2) - 2(2 + 2) = -4k - 12$$
Now $A$ is not invertible $\Leftrightarrow \det A = 0$. Then $k = -3$.
Best Answer
Let $A$ be a full rank $m\times n$ matrix. By full rank we mean $\DeclareMathOperator{rank}{rank}\rank(A)=\min\{m,n\}$.
If $m<n$, then $A$ has a right inverse given by $$ A^{-1}_{\text{right}}=A^\top(AA^\top)^{-1} $$
If $m>n$, then $A$ has a left inverse given by $$ A^{-1}_{\text{left}}=(A^\top A)^{-1} A^\top $$
Of course, a right inverse is useful for solving an equation of the form $XA=Y$ and a left inverse is useful for solving an equation of the form $AX=Y$.