Question

Suppose the solution set of a certain system of linear equations can be described as\({x_1} = 3{x_4}\),\({x_2} = 8 + {x_4}\),\({x_3} = 2 - 5{x_4}\)with\({x_4}\)free. Use vectors to describe this set as a line in\({\mathbb{R}^4}\).

Short answer

The solution set is the line passing through the vector \(\left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right]\) and parallel to the vector \(\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\).

Step-by-step solution

Write the general parametric equation of the line

If a line passes through 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:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\)

Convert the matrix equation into the parametric form

Use the entries \({x_1} = 3{x_4}\),\({x_2} = 8 + {x_4}\),\({x_3} = 2 - 5{x_4}\) in vector \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\) as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{3{x_4}}\\{8 + {x_4}}\\{2 - 5{x_4}}\\{{x_4}}\end{array}} \right]\)

Simplify the matrix form of equation \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{3{x_4}}\\{8 + {x_4}}\\{2 - 5{x_4}}\\{{x_4}}\end{array}} \right]\).

\[\begin{array}{c}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{3{x_4}}\\{{x_4}}\\{ - 5{x_4}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\end{array}\]

Obtain vectors a and b

Compare the parametric equation \[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\] with the general parametric equation \({\bf{x}} = {\bf{a}} + t{\bf{b}}\).

So, the vectors are\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\), where \({x_4}\) is the parameter.

Describe the obtained vectors

The obtained vectors are \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\).

The parametric equation \[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\] shows that the line passes through the vector\(\left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right]\)and is parallel to the vector\(\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\).

Thus, the solution set is a line passing through the vector \(\left[ {\begin{array}{*{20}{c}}0\\8\\2\\0\end{array}} \right]\), in the direction of \(\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\).