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

Expert verified

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

01

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:

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

02

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

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

03

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.

04

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 \(\left[ {\begin{array}{*{20}{c}}3\\1\\{ - 5}\\1\end{array}} \right]\).

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Most popular questions from this chapter

Consider a dynamical systemwith two components. The accompanying sketch shows the initial state vectorx0and two eigen vectorsυ1andυ2of A (with eigen values λ1andλ2respectively). For the given values ofλ1andλ2, draw a rough trajectory. Consider the future and the past of the system.

λ1=0.9,λ2=0.9

Let \({{\bf{a}}_1}\) \({{\bf{a}}_2}\), and b be the vectors in \({\mathbb{R}^{\bf{2}}}\) shown in the figure, and let \(A = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}\end{aligned}} \right)\). Does the equation \(A{\bf{x}} = {\bf{b}}\) have a solution? If so, is the solution unique? Explain.

In Exercise 2, compute \(u + v\) and \(u - 2v\).

2. \(u = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\).

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

Let \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\8\end{array}} \right],\) and \({\rm{y = }}\left[ {\begin{array}{*{20}{c}}h\\{ - 5}\\{ - 3}\end{array}} \right]\). For what values(s) of \(h\) is \(y\) in the plane generated by \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\)

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