Suppose the solution set of a certain system of linear equations can be described as \({x_1} = 5 + 4{x_3}\), \({x_2} = - 2 - 7{x_3}\), with \({x_3}\) free. Use vectors to describe this set as a line in \({\mathbb{R}^3}\).

If a line passes through a vector\({\bf{a}}\)and is parallel to vector b,then the parametric equation of the line is represented as\({\bf{x}} = {\bf{a}} + t{\bf{b}}\), where\(t\)is a parameter.

Here, \({\bf{x}}\) is represented as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\)

Use the entries \({x_1} = 5 + 4{x_3}\), \({x_2} = - 2 - 7{x_3}\) in vector \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\) as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{5 + 4{x_3}}\\{ - 2 - 7{x_3}}\\{{x_3}}\end{array}} \right]\)

Simplify the matrix form of equation \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{5 + 4{x_3}}\\{ - 2 - 7{x_3}}\\{{x_3}}\end{array}} \right)\).

\(\begin{array}{c}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{4{x_3}}\\{ - 7{x_3}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right]\end{array}\)

Compare the parametric equation \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right]\) with the general parametric equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\).

So, the vectors are \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right]\), where \({x_3}\) is the parameter.

The obtained vectors are \({\bf{a}} = \left( {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right)\)and\({\bf{b}} = \left( {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right)\).

The parametric equation \({\bf{x}} = \left( {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right)\) shows that the line passes through the vector\(\left( {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right)\)and is parallel to the vector\(\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right)\).

Thus, the solution set is the line passing through the vector \(\left( {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right)\), in the direction of \(\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right)\).