how I can find Moore-Penrose Pseudoinverse of a matrix

Question

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

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

where $c$ is a real number in 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 solver 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 accept letter as input to calculate Moore-Penrose Pseudoinverse ?

Answer 1

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 $A^t(AA^t)^{-1}$.