Problem:
Give a list of vectors that forms a basis for the null space of the following matrix:
$\left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 -9 & -8 & -8 \\
 -5 & -9 & 9 \\
\end{array}
\right)$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Find the null space of the matrix }M: \\
 M=\left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 -9 & -8 & -8 \\
 -5 & -9 & 9 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 -9 & -8 & -8 \\
 -5 & -9 & 9 \\
\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 \\
\end{array}
\right) \text{such }\text{that }M.v=0: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 -9 & -8 & -8 \\
 -5 & -9 & 9 \\
\end{array}
\right).\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Reduce }\text{the }\text{matrix }\left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 -9 & -8 & -8 \\
 -5 & -9 & 9 \\
\end{array}
\right) \text{to }\text{row }\text{echelon }\text{form}: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 -9 & -8 & -8 \\
 -5 & -9 & 9 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{9}{10}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 0 & -\frac{71}{10} & \frac{1}{10} \\
 -5 & -9 & 9 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{1}{2}\, \times \, \text{(row }1) \text{from }\text{row }3: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 0 & -\frac{71}{10} & \frac{1}{10} \\
 0 & -\frac{17}{2} & \frac{27}{2} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Swap }\text{row }2 \text{with }\text{row }3: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 0 & -\frac{17}{2} & \frac{27}{2} \\
 0 & -\frac{71}{10} & \frac{1}{10} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{71}{85}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 0 & -\frac{17}{2} & \frac{27}{2} \\
 0 & 0 & -\frac{190}{17} \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Multiply }\text{row }3 \text{by }-\frac{17}{190}: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 0 & -\frac{17}{2} & \frac{27}{2} \\
 0 & 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Subtract }\frac{27}{2}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & -9 \\
 0 & -\frac{17}{2} & 0 \\
 0 & 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }9\, \times \, \text{(row }3) \text{to }\text{row }1: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & 0 \\
 0 & -\frac{17}{2} & 0 \\
 0 & 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Multiply }\text{row }2 \text{by }-\frac{2}{17}: \\
 \left(
\begin{array}{ccc}
 -10 & -1 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Add }\text{row }2 \text{to }\text{row }1: \\
 \left(
\begin{array}{ccc}
 -10 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Divide }\text{row }1 \text{by }-10: \\
 \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\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 \\
\end{array}
\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right) \text{which }\text{have }\text{no }\text{pivot.} \\
 \text{There }\text{are }\text{no }\text{free }\text{variables} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{only }\text{value }\text{of }v=\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
\end{array}
\right) \text{that }\text{would }\text{make }\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right).\left(
\begin{array}{c}
 x_1 \\
 x_2 \\
 x_3 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right) \text{is }\text{the }\text{zero }\text{vector }\left(
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right): \\
 \left(
\begin{array}{c}
 0 \\
 0 \\
 0 \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \{\, (0,0,0)\, \} \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/null_space_w_steps/1522.txt'}