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Chapter 1: Q13E (page 1)

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 \({\mathbb{R}^3}\).

Short Answer

Expert verified

The solution set consists of a line that runs through the vector \(\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right]\) and is parallel to the vector \(\left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right]\).

Step by step solution

01

Write the general parametric equation of the line

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:

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

02

Convert the matrix equation into the parametric form

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

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

03

Obtain vectors a and b

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.

04

Describe the obtained vectors

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 \(\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\1\end{array}} \right)\).

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Most popular questions from this chapter

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

Suppose \(a,b,c,\) and \(d\) are constants such that \(a\) is not zero and the system below is consistent for all possible values of \(f\) and \(g\). What can you say about the numbers \(a,b,c,\) and \(d\)? Justify your answer.

28. \(\begin{array}{l}a{x_1} + b{x_2} = f\\c{x_1} + d{x_2} = g\end{array}\)

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. (Hint: Given u, v in \({\mathbb{R}^n}\), let \({\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\). Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \(T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \). Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).)

In (a) and (b), suppose the vectors are linearly independent. What can you say about the numbers \(a,....,f\) ? Justify your answers. (Hint: Use a theorem for (b).)

  1. \(\left( {\begin{aligned}{*{20}{c}}a\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\d\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\end{aligned}} \right)\)
  2. \(\left( {\begin{aligned}{*{20}{c}}a\\1\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\1\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\\1\end{aligned}} \right)\)

In Exercises 31, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

31. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\4&{ - 1}&3&{ - 6}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\0&7&{ - 1}&{ - 6}\end{array}} \right]\)
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