Step 1

Consider the matrix \(\begin{bmatrix}1 & 0 & 0 & -1 \\0 & 1 & 0 & -1\\ 0 & 0 & 1 & -1 \end{bmatrix}\)

Step 2

Here, \(\displaystyle{x}_{{{4}}}\) is a free variable and \(\displaystyle{x}_{{{1}}},{x}_{{{2}}},{x}_{{{3}}}\) are not. Column space of A contain all combination of vectors \(\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}\).

Therefore, it contains vector \(\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}\).

Thus, we consider the system \(\begin{bmatrix}1 & 0 & 0 & -1 \\0 & 1 & 0 & -1\\ 0 & 0 & 1 & -1 \end{bmatrix}=\begin{bmatrix}x_{1} \\x_{2}\\ x_{3} \end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}\)

Step 3

\(\displaystyle{x}_{{{1}}}-{x}_{{{4}}}={0}\Rightarrow{x}_{{{1}}}={x}_{{{4}}}\)

\(\displaystyle{x}_{{{2}}}-{x}_{{{4}}}={0}\Rightarrow{x}_{{{2}}}={x}_{{{4}}}\)

\(\displaystyle{x}_{{{3}}}-{x}_{{{4}}}={0}\Rightarrow{x}_{{{3}}}={x}_{{{4}}}\)

\(N(A)=x_{4} \begin{bmatrix}1\\1\\1\\1 \end{bmatrix}\)

So null space of A consists of all multiples of (1,1,1,1).

A matrix whose column space matrix contains (1,1,1) and whose null space is the line of multiples of (1,1,1,1) is \(A=\begin{bmatrix}1 & 0 & 0 & -1 \\0 & 1 & 0 & -1\\0 & 0 & 1 & -1 \end{bmatrix}\).