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Chapter 8: Q8.1-9E (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 \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{0}}\end{array}} \right)\). 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 \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right)\) and \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right)\).

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:

\({\bf{x}} = {x_3}\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right)\)

02

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

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

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

Substitute 1 for \({x_3}\) in the equation \({\bf{x}} = {x_3}\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right)\) to find \({{\bf{v}}_2}\) as shown below:

\(\begin{array}{c}{{\bf{v}}_2} = 1\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right)\end{array}\)

So, \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right)\) and \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right)\).

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Most popular questions from this chapter

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).

25. \({\mathop{\rm aff}\nolimits} \left( {A \cap B} \right) \subset \left( {{\mathop{\rm aff}\nolimits} A \cap {\mathop{\rm aff}\nolimits} B} \right)\)

Question:28. Give an example of a compact set\(A\)and a closed set\(B\)in\({\mathbb{R}^2}\)such that\(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) = \emptyset \)but\(A\)and\(B\)cannot be strictly separated by a hyperplane.

In Exercises 7 and 8, find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.

8. \(\left( {\begin{array}{{}}0\\1\\{ - 2}\\1\end{array}} \right),\left( {\begin{array}{{}}1\\1\\0\\2\end{array}} \right),\left( {\begin{array}{{}}1\\4\\{ - 6}\\5\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}{ - 1}\\1\\{ - 4}\\0\end{array}} \right)\)

Question: In Exercise 4, determine whether each set is open or closed or neither open nor closed.

4. a. \(\left\{ {\left( {x,y} \right):{x^{\bf{2}}} + {y^{\bf{2}}} = {\bf{1}}} \right\}\)

b. \(\left\{ {\left( {x,y} \right):{x^{\bf{2}}} + {y^{\bf{2}}} > {\bf{1}}} \right\}\)

c. \(\left\{ {\left( {x,y} \right):{x^{\bf{2}}} + {y^{\bf{2}}} \le {\bf{1}}\,\,\,and\,\,y > {\bf{0}}} \right\}\)

d. \(\left\{ {\left( {x,y} \right):y \ge {x^{\bf{2}}}} \right\}\)

e. \(\left\{ {\left( {x,y} \right):y < {x^{\bf{2}}}} \right\}\)

TrueType fonts, created by Apple Computer and Adobe Systems, use quadratic Bezier curves, while PostScript fonts, created by Microsoft, use cubic Bezier curves. The cubic curves provide more flexibility for typeface design, but it is important to Microsoft that every typeface using quadratic curves can be transformed into one that used cubic curves. Suppose that \({\mathop{\rm w}\nolimits} \left( t \right)\) is a quadratic curve, with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\).

  1. Find control points \({{\mathop{\rm r}\nolimits} _0},{{\mathop{\rm r}\nolimits} _1},{{\mathop{\rm r}\nolimits} _2},\), and \({{\mathop{\rm r}\nolimits} _3}\) such that the cubic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with these control points has the property that \({\mathop{\rm x}\nolimits} \left( t \right)\) and \({\mathop{\rm w}\nolimits} \left( t \right)\) have the same initial and terminal points and the same tangent vectors at \(t = 0\)and\(t = 1\). (See Exercise 16.)
  1. Show that if \({\mathop{\rm x}\nolimits} \left( t \right)\) is constructed as in part (a), then \({\mathop{\rm x}\nolimits} \left( t \right) = {\mathop{\rm w}\nolimits} \left( t \right)\) for \(0 \le t \le 1\).
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