Definition:
For a given matrix $A_{m\times n}$, a matrix $G_{n\times m}$ is said to be a generalized inverse of $A$, if it satisfies $$AGA=A.$$
Question:
Find two different generalized inverse of the given matrix
$$\begin{pmatrix} 1 & 0 &-1 & 2\\2 & 0 &-2 & 4 \\-1 & 1 & 1 & 3\\ -2 & 2 & 2 & 6 \end{pmatrix}$$
Work done:
Since the echelon form of the matrix is, $$ \left(\begin{array}{rrrr} 1 & 0 & -1 & 2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$ rank is 2.
since there are two distinct $2\times 2$ minors,
one of the generalized inverse is, $$\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ \frac 1 2 &0 & 0 & 0 \\ \frac 1 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$ and the other one is,
$$\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 &0 & \frac 3 {10} & -\frac 4{10} \\ 0& 0 & \frac 1 {10} & \frac 2 {10} \\ 0 & 0 & 0 & 0 \end{array}\right)$$
Luckily we get two different solutions,
But if the question is to find 5 different generalized inverses, How to do that?
As we know there are plenty of generalized inverses are there for this given matrix, different possible ways are welcome.
Thanks in advance.
If $AGA=A$, then $A(G+uv^T)A=A$ if $u\in\ker A$ or $v\in\ker A^T$. Note that when $A$ is not a nonsingular matrix (this includes the case where $A$ is not square), at least one of $A$ or $A^T$ has a nonzero nullspace. Therefore, if you can find one generalised inverse of $A$, you can find infinitely many others if the field is infinite.
By the way, the two matrices that you claim to be generalised inverses of your example $A$ do not seem to be correct.