##### Example2.2.5

Describe the solution space of the equation

\begin{equation*} \left[\begin{array}{rrr} 2 \amp 0 \amp 2 \\ 4 \amp -1 \amp 6 \\ 1 \amp 3 \amp -5 \\ \end{array}\right] \xvec = \left[\begin{array}{r} 0 \\ -5 \\ 15 \end{array}\right] \end{equation*}

By Proposition 4, the solution space to this equation is the same as the equation

\begin{equation*} x_1\left[\begin{array}{r}2\\4\\1\end{array}\right] + x_2\left[\begin{array}{r}0\\-1\\3\end{array}\right]+ x_3\left[\begin{array}{r}2\\6\\-5\end{array}\right]= \left[\begin{array}{r}0\\-5\\15\end{array}\right] \text{,} \end{equation*}

which is the same as the linear system corresponding to

\begin{equation*} \left[\begin{array}{rrr|r} 2 \amp 0 \amp 2 \amp 0 \\ 4 \amp -1 \amp 6 \amp -5 \\ 1 \amp 3 \amp -5 \amp 15 \\ \end{array} \right] \text{.} \end{equation*}

We will study the solutions to this linear system by finding the reduced row echelon form of the augmented matrix:

\begin{equation*} \left[\begin{array}{rrr|r} 2 \amp 0 \amp 2 \amp 0 \\ 4 \amp -1 \amp 6 \amp -5 \\ 1 \amp 3 \amp -5 \amp 15 \\ \end{array} \right] \sim \left[\begin{array}{rrr|r} 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp -2 \amp 5 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \text{.} \end{equation*}

This gives us the system of equations

\begin{equation*} \begin{alignedat}{4} x_1 \amp \amp \amp {}+{} \amp x_3 \amp {}={} \amp 0 \\ \amp \amp x_2 \amp {}-{} \amp 2x_3 \amp {}={} \amp 5 \\ \end{alignedat} \text{.} \end{equation*}

The variable $$x_3$$ is free so we may write the solution space parametrically as

\begin{equation*} \begin{aligned} x_1 \amp {}={} -x_3 \\ x_2 \amp {}={} 5+2x_3 \\ \end{aligned} \text{.} \end{equation*}

Since we originally asked to describe the solutions to the equation $$A\xvec = \bvec\text{,}$$ we will express the solution in terms of the vector $$\xvec\text{:}$$

\begin{equation*} \xvec =\left[ \begin{array}{r} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{r} -x_3 \\ 5 + 2x_3 \\ x_3 \end{array} \right] =\left[\begin{array}{r}0\\5\\0\end{array}\right] +x_3\left[\begin{array}{r}-1\\2\\1\end{array}\right] \end{equation*}

This shows that the solutions $$\xvec$$ may be written in the form $$\vvec + x_3\wvec\text{,}$$ for appropriate vectors $$\vvec$$ and $$\wvec\text{.}$$ Geometrically, the solution space is a line in $$\real^3$$ through $$\vvec$$ moving parallel to $$\wvec\text{.}$$

in-context