Let the matrix is "A=\\begin{bmatrix}\n1 & 0 & 0 & 1\\\\\n0 & 1 & 0 & 1 \\\\\n0 & 0 & 1 & 1\n\\end{bmatrix}"
Basis for row space of "A=\\begin{Bmatrix}\n[1 & 0 & 0 & 1], [0 & 1 & 0 & 1],[0 & 0 & 1 & 1]\n\\end{Bmatrix}"
In the 1, 2 and 3, it contains pivot hence,
"=\\{\\begin{bmatrix}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n0 & 0 & 1 \n\\end{bmatrix}\\}"
If "Ax=0"
"\\begin{bmatrix}\nx_1\\\\\nx_2 \\\\\nx_3\\\\\nx_4\n\\end{bmatrix}" "=\\begin{bmatrix}\n-1\\\\\n-1 \\\\\n-1\\\\\n1\n\\end{bmatrix}"
Hence, it is much better for the null space "=\\begin{bmatrix}\n-1\\\\\n-1 \\\\\n-1\\\\\n1\n\\end{bmatrix}"
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