Find 4x3 matrix B such that AB = I I got stuck with this problem.
\begin{matrix}
        1 & 1 & 1 & 1\\
        0 & 1 & 1 & 0\\
        0 & 0 & 1 & 1\\
        \end{matrix}
Consider the $3\times 4$ matrix $\bf A$ (above). Do the columns of $\bf A$ span $\mathbb R^3$? 
Prove your answer. Also, Find a $4\times 3$ matrix $\bf B$, such that $\bf AB = I_3$
--
I know that the columns of $\bf A$ span $\mathbb R^3$ as there more columns than rows. But I cannot understand how to find matrix $\bf B$ because I cannot implement "super-augmented" matrix and do Gauss-Jordan elimination. Looks like I need to do something with 4th column of $\bf A$ and 4th row of $\bf B$. What do you think?
Thanks!

B is a 4*3 matrix. we will find the columns of B using the fact that the columns of AB are linear combinations of columns of A with column elements of B.
$\\1\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}+0\begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix}+0\begin{bmatrix}
1\\ 
1\\ 
1
\end{bmatrix}+0\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}=\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}$ So the first column of B is given by $ \begin{bmatrix}
1\\ 
0\\ 
0\\ 
0
\end{bmatrix}$
 similarly $ \\-1\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}+1\begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix}+0\begin{bmatrix}
1\\ 
1\\ 
1
\end{bmatrix}+0\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}=\begin{bmatrix}
0\\ 
1\\ 
0
\end{bmatrix}$ so the second column of B is given by $\begin{bmatrix}
-1\\ 
1\\ 
0\\ 
0
\end{bmatrix}$
$\\0\begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}-1\begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix}+1\begin{bmatrix}
1\\ 
1\\ 
1
\end{bmatrix}+0\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}=\begin{bmatrix}
0\\ 
0\\ 
1
\end{bmatrix}$ So the third column of B is given by $\begin{bmatrix}
0\\ 
-1\\ 
1\\ 
0
\end{bmatrix}$
$B=\begin{bmatrix}
1 &-1  &0 \\ 
 0& 1 &-1 \\ 
 0& 0 &1 \\ 
 0& 0 & 0
\end{bmatrix}$.
Note that the matrix B is not unique. 
