Question 2.2.9: Find the general solution of the homogeneous system 2x1 + x2...

We row reduce the coefficient matrix of the system to RREF:

\left [ \begin{matrix} 2 & 1 & 0 \\ 1 & 1 & -1 \\ 0 & -1 & 2 \end{matrix} \right ] \begin{matrix} R_{1} – R_{2} \\ \\ \\ \end{matrix} \thicksim \left [ \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & -1 \\ 0 & -1 & 2 \end{matrix} \right ] \begin{matrix} \\ R_{2} – R_{1} \\ \\ \end{matrix} \thicksim \\ \left [ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & -1 & 2 \end{matrix} \right ] \begin{matrix} \\ \\R_{3} + R_{2} \end{matrix} \thicksim \left [ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{matrix} \right ]

This corresponds to the homogeneous system

x_{1} \ \ \ \ \ \ \ \ + x_{3} = 0 \\  \ \ \ \ \ x_{2}  –  2x_{3} =0

Hence, x_{3} is a free variable, so we let x_{3} = t \in \mathbb{R}. Then . x_{1} = -x_{3} = -t, x_{2} = 2x_{3} = 2t, and the general solution is

\left [ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix} \right ] = \left [ \begin{matrix} -t \\ 2t \\ t \end{matrix} \right ] = t\left [ \begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right ] , \ \ \ \ \ t \in \mathbb{R}