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Chapter 8: Q8.1-10E (page 437)

Question: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{4}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

Short Answer

Expert verified

The points are \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\) and \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\2\end{array}} \right)\).

Step by step solution

01

Substitute the given values in the equation of x

Substitute the given values in the equation \({\bf{x}} = {x_3}{\bf{u}} + {\bf{p}}\) as shown below:

\({\bf{x}} = {x_3}\left( {\begin{array}{*{20}{c}}5\\1\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\)

02

Find the value of \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\)

Substitute 0 for \({x_3}\) in the equation \({\bf{x}} = {x_3}\left( {\begin{array}{*{20}{c}}5\\1\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\) to find \({{\bf{v}}_1}\) as shown below:

\(\begin{array}{c}{{\bf{v}}_1} = 0\left( {\begin{array}{*{20}{c}}5\\1\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\end{array}\)

Substitute 1 for \({x_3}\) in the equation \({\bf{x}} = {x_3}\left( {\begin{array}{*{20}{c}}5\\1\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\) to find \({{\bf{v}}_2}\) as shown below:

\(\begin{array}{c}{{\bf{v}}_1} = 1\left( {\begin{array}{*{20}{c}}5\\1\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\2\end{array}} \right)\end{array}\)

So, one of the points are \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\4\end{array}} \right)\) and \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\2\end{array}} \right)\).

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Most popular questions from this chapter

Question: In Exercise 3, determine whether each set is open or closed or neither open nor closed.

3. a. \(\left\{ {\left( {x,y} \right):y > {\bf{0}}} \right\}\)

b. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} \le y \le {\bf{3}}} \right\}\)

c. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} < y < {\bf{3}}} \right\}\)

d. \(\left\{ {\left( {x,y} \right):xy = {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

e. \(\left\{ {\left( {x,y} \right):xy \ge {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\\{\bf{6}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{{\bf{17}}}\\{\bf{1}}\\{\bf{5}}\end{aligned}} \right)\)

In Exercises 21-26, prove the given statement about subsets A and B of \({\mathbb{R}^n}\), or provide the required example in \({\mathbb{R}^2}\). A proof for an exercise may use results from earlier exercises (as well as theorems already available in the text).

21. If \(A \subset B\), then B is affine, then \({\mathop{\rm aff}\nolimits} A \subset B\).

Question: 20. Let \(f:{\mathbb{R}^n} \to {\mathbb{R}^m}\) is a linear transformation, and let \(T\) be an affine subset of \({\mathbb{R}^{\bf{m}}}\), and let \(S = \left\{ {{\bf{x}} \in {\mathbb{R}^n}\,:\,f\left( {\bf{x}} \right) \in T} \right\}\). Show that \(S\) is an affine subset of \({\mathbb{R}^m}\).

Question: Let \({{\bf{a}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{5}}\end{array}} \right)\), \({{\bf{a}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{a}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{6}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{ - {\bf{2}}}\end{array}} \right)\),\({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{2}}\\{\bf{1}}\end{array}} \right)\) and \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\), and let \(A = \left\{ {{{\bf{a}}_{\bf{1}}},{{\bf{a}}_{\bf{2}}},{{\bf{a}}_{\bf{3}}}} \right\}\) and \(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\). Find a hyperplane H with normal n that separates A and B. Is there a hyperplane parallel to H that strictly separates A and B?
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