Finding right inverse matrix Given a $3\times 4$ matrix $A$ such as 
$$
        \begin{pmatrix}
        1 & 1 & 1 & 1 \\
        0 & 1 & 1 & 0 \\
        0 & 0 & 1 & 1 \\
        \end{pmatrix}
,$$ 
find a matrix $B_{4\times 3}$  such that
$$AB =
        \begin{pmatrix}
        1 & 0 & 0  \\
        0 & 1 & 0  \\
        0 & 0 & 1  \\
        \end{pmatrix}
$$
Apart from simply multiplying $A$ with $B$ and generating a $12$ variable system of equations, is there any simpler way of finding $B$ ?

Your matrix 
$$A=\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}$$ 
has rank=$3$ and you can see that the square matrix $AA^T$ is invertible. Now note that $AA^T(AA^T)^{-1}=I$ so the matrix $B=A^T(AA^T)^{-1}$ is a right inverse of $A$ (but it is not the unique).
in this case we have:
$$
AA^T=\begin{bmatrix}4&2&2\\2&2&1\\2&1&2\end{bmatrix}
$$
$$
(AA^T)^{-1}=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\-2&4&0\\-2&0&4\end{bmatrix}
$$
$$
B=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\1&2&-2\\-1&2&2\\1&-2&2\end{bmatrix}
$$
