Problem:
Give a list of vectors that forms a basis for the null space of the following matrix:
$\left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 -2 & -7 & -6 & -6 \\
\end{array}
\right)$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Find the null space of the matrix }M: \\
 M=\left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 -2 & -7 & -6 & -6 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 -2 & -7 & -6 & -6 \\
\end{array}
\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
\end{array}
\right) \text{such }\text{that }M.v=0: \\
 \left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 -2 & -7 & -6 & -6 \\
\end{array}
\right).\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Reduce }\text{the }\text{matrix }\left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 -2 & -7 & -6 & -6 \\
\end{array}
\right) \text{to }\text{row }\text{echelon }\text{form}: \\
 \left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 -2 & -7 & -6 & -6 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }\frac{2}{3}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
 \left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 0 & -\frac{35}{3} & -\frac{38}{3} & -10 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Multiply }\text{row }2 \text{by }-\frac{3}{35}: \\
 \left(
\begin{array}{cccc}
 3 & -7 & -10 & -6 \\
 0 & 1 & \frac{38}{35} & \frac{6}{7} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }7\, \times \, \text{(row }2) \text{to }\text{row }1: \\
 \left(
\begin{array}{cccc}
 3 & 0 & -\frac{12}{5} & 0 \\
 0 & 1 & \frac{38}{35} & \frac{6}{7} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Divide }\text{row }1 \text{by }3: \\
 \left(
\begin{array}{cccc}
 1 & 0 & -\frac{4}{5} & 0 \\
 0 & 1 & \frac{38}{35} & \frac{6}{7} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
\end{array}
\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
\begin{array}{cccc}
 1 & 0 & -\frac{4}{5} & 0 \\
 0 & 1 & \frac{38}{35} & \frac{6}{7} \\
\end{array}
\right) \text{which }\text{have }\text{no }\text{pivot.} \\
 \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
\begin{array}{cccc}
 1 & 0 & -\frac{4}{5} & 0 \\
 0 & 1 & \frac{38}{35} & \frac{6}{7} \\
\end{array}
\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
\end{array}
\right): \\
 \left(
\begin{array}{cccc}
 1 & 0 & -\frac{4}{5} & 0 \\
 0 & 1 & \frac{38}{35} & \frac{6}{7} \\
\end{array}
\right).\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
\end{array}
\right)=\left(
\begin{array}{c}
 x_1-\frac{4 x_3}{5} \\
 x_2+\frac{38 x_3}{35}+\frac{6 x_4}{7} \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Solve }\text{the }\text{equations }\{
\begin{array}{l}
 x_1-\frac{4 x_3}{5}=0 \\
 x_2+\frac{38 x_3}{35}+\frac{6 x_4}{7}=0 \\
\end{array}
  \text{for }x_1 \text{and }x_2: \\
 \{
\begin{array}{l}
 x_1=\frac{4 x_3}{5} \\
 x_2=-\frac{38 x_3}{35}-\frac{6 x_4}{7} \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
 v=\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
\end{array}
\right)=\left(
\begin{array}{c}
 \frac{4 x_3}{5} \\
 -\frac{38 x_3}{35}-\frac{6 x_4}{7} \\
 x_3 \\
 x_4 \\
\end{array}
\right)=\left(
\begin{array}{c}
 \frac{4 x}{5} \\
 -\frac{6 y}{7}-\frac{38 x}{35} \\
 x \\
 y \\
\end{array}
\right)\text{ for }x,y\in \mathbb{R} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }35 x \text{and }7 y \text{respectively}: \\
 \left(
\begin{array}{c}
 \frac{4 x}{5} \\
 -\frac{6 y}{7}-\frac{38 x}{35} \\
 x \\
 y \\
\end{array}
\right)\, \rightarrow \, \left(
\begin{array}{c}
 \frac{4 (35 x)}{5} \\
 -\frac{6}{7} (7 y)-\frac{38 (35 x)}{35} \\
 35 x \\
 7 y \\
\end{array}
\right)=\left(
\begin{array}{c}
 28 x \\
 -6 y-38 x \\
 35 x \\
 7 y \\
\end{array}
\right)\text{ for }x,y\in \mathbb{R} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
\begin{array}{c}
 28 x \\
 -6 y-38 x \\
 35 x \\
 7 y \\
\end{array}
\right) \text{in }\text{set }\text{notation}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \{\, (28 x,-6 y-38 x,35 x,7 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/null_space_w_steps/2218.txt'}