# How can I find the Moore-Penrose pseudoinverse of a matrix?

I am trying to find Moore-Penrose pseudoinverse of the following matrix:

$$\begin{bmatrix} 1 & 0 & c\\ 0 & 1 & c \end{bmatrix}$$

where $$c \in \Bbb R$$. I found the following:

$$\begin{bmatrix} 1-c &-c \\ -c & 1-c \\ 1 &1 \end{bmatrix}$$

But this is not something that I am looking for. I am wondering how standard solver solve this problem, for example, for $$c=4$$. All online solvers give me this:

$$\begin{bmatrix} 0.5152 & -0.4848\\ -0.4848 & 0.5152\\ 0.1212 & 0.1212 \end{bmatrix}$$

Does any body have any idea? Also, do you know a solver that accepts letter as input to calculate the Moore-Penrose pseudoinverse?

The matrix that you were given has $$3$$ columns and $$2$$ rows and therefore its Moore-Penrose inverse shall have $$2$$ columns and $$3$$ rows. Actually, it is this matrix:$$\begin{bmatrix}\frac{c^2+1}{2 c^2+1} & -\frac{c^2}{2 c^2+1} \\ -\frac{c^2}{2 c^2+1} & \frac{c^2+1}{2 c^2+1} \\ \frac{c}{2 c^2+1} & \frac{c}{2 c^2+1}\end{bmatrix}.$$This follows from the fact that, if $$A$$ is your matrix, then it has full rank and therefore its Moore-Penrose inverse is $$A^t(AA^t)^{-1}$$.