find two different generalized inverse of the given matrix 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.

The matrix and its Moore-Penrose pseudoinverse are
$$
\mathbf{A} = 
\left[
\begin{array}{rrrr}
 1 & 0 & -1 & 2 \\
 2 & 0 & -2 & 4 \\
 -1 & 1 & 1 & 3 \\
 -2 & 2 & 2 & 6 \\
\end{array}
\right],
\qquad
\mathbf{A}^{\dagger} = \frac{1}{40}
\left[
\begin{array}{rrrr}
 8 & 16 & -5 & -10 \\
 -2 & -4 & 3 & 6 \\
 -8 & -16 & 5 & 10 \\
 6 & 12 & 5 & 10 \\
\end{array}
\right].
$$
The pseudoinverse satisfies all four requirements: 


*

*$\mathbf{A} \, \mathbf{A}^{\dagger} \mathbf{A} = \mathbf{A}$

*$\mathbf{A}^{\dagger} \mathbf{A} \, \mathbf{A} = \mathbf{A}^{\dagger}$

*$\left( \mathbf{A} \, \mathbf{A}^{\dagger} \right)^{*} = \mathbf{A} \, \mathbf{A}^{\dagger}$

*$\left( \mathbf{A}^{\dagger} \mathbf{A} \right)^{*} = \mathbf{A}^{\dagger} \mathbf{A}$


The inverses presented in the question are problematic:
$$
\mathbf{G}_{1} = \left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 \frac{1}{2} & 0 & 0 & 0 \\
 \frac{1}{2} & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right), \qquad
%
 \mathbf{A} \, \mathbf{G}_{1} \mathbf{A} = 
\left(
\begin{array}{rrrr}
 -\frac{5}{2} & 0 & \frac{5}{2} & -5 \\
 -5 & 0 & 5 & -10 \\
 3 & 0 & -3 & 6 \\
 6 & 0 & -6 & 12 \\
\end{array}
\right)
\ne \mathbf{A}
$$
$$
\mathbf{G}_{2} = \left(
\begin{array}{cccr}
 0 & 0 & 0 & 0 \\
 0 & 0 & \frac{3}{10} & -\frac{2}{5} \\
 0 & 0 & \frac{1}{10} & \frac{1}{10} \\
 0 & 0 & 0 & 0 \\
\end{array}
\right), \qquad
%
 \mathbf{A} \, \mathbf{G}_{2} \mathbf{A} = 
\left(
\begin{array}{rrrr}
 \frac{3}{10} & -\frac{3}{10} & -\frac{3}{10} &
   -\frac{9}{10} \\
 \frac{3}{5} & -\frac{3}{5} & -\frac{3}{5} & -\frac{9}{5}
   \\
 \frac{1}{5} & -\frac{1}{5} & -\frac{1}{5} & -\frac{3}{5}
   \\
 \frac{2}{5} & -\frac{2}{5} & -\frac{2}{5} & -\frac{6}{5}
\end{array}
\right)
\ne \mathbf{A}
$$
