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Chapter 8: Q16E (page 437)

Repeat Exercise 15 for points \({{\rm{v}}_1} = \left( \begin{aligned}{} - 1\\\,\,0\end{aligned} \right)\), \({{\rm{v}}_2} = \left( \begin{aligned}{}\,0\\\,3\end{aligned} \right)\), \({{\rm{v}}_3} = \left( \begin{aligned}{}\,3\\\,1\end{aligned} \right)\), \({{\rm{v}}_4} = \left( \begin{aligned}{}1\\ - \,1\end{aligned} \right)\), and \({\rm{p}} = \left( \begin{aligned}{}1\\2\end{aligned} \right)\) , given that \({\rm{p}}\, = \frac{1}{{121}}{{\rm{v}}_1} + \frac{{72}}{{121}}{{\rm{v}}_2} + \frac{{37}}{{121}}{{\rm{v}}_3} + \frac{1}{{11}}{{\rm{v}}_4}\)

and \({\rm{10}}{{\rm{v}}_1} - {\rm{6}}{{\rm{v}}_2} + 7{{\rm{v}}_3} - 11{{\rm{v}}_4} = 0\).

Short Answer

Expert verified

\({\bf{p}} = \frac{1}{6}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_4}\) and \({\rm{p}} = \frac{1}{{11}}{{\bf{v}}_1} + \frac{6}{{11}}{{\bf{v}}_2} + \frac{4}{{11}}{{\bf{v}}_3}\)

Step by step solution

01

Verify the expression for p

\(\begin{aligned}{}\frac{1}{{121}}{{\bf{v}}_1} + \frac{{72}}{{121}}{{\bf{v}}_2} + \frac{{37}}{{121}}{{\bf{v}}_3} + \frac{1}{{11}}{{\bf{v}}_4} &= \frac{1}{{121}}\left( {\begin{aligned}{{}}{ - 1}\\0\end{aligned}} \right) + \frac{{72}}{{121}}\left( {\begin{aligned}{{}}0\\3\end{aligned}} \right) + \frac{{37}}{{121}}\left( {\begin{aligned}{{}}3\\1\end{aligned}} \right) + \frac{1}{{11}}\left( {\begin{aligned}{{}}1\\{ - 1}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{ - \frac{1}{{121}} + 0 + \frac{{111}}{{121}} + \frac{1}{{11}}}\\{0 + \frac{{216}}{{121}} + \frac{{37}}{{121}} - \frac{1}{{11}}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}1\\2\end{aligned}} \right)\\ &= {\bf{p}}\end{aligned}\)

02

Verify the equation

Substitute value in the equation \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\).

\(\begin{aligned}{}10{{\bf{v}}_1} - 6{{\bf{v}}_2} + 7{{\bf{v}}_3} - 11{{\bf{v}}_4} &= 10\left( {\begin{aligned}{{}}{ - 1}\\0\end{aligned}} \right) - 6\left( {\begin{aligned}{{}}0\\3\end{aligned}} \right) + 7\left( {\begin{aligned}{{}}3\\1\end{aligned}} \right) - 11\left( {\begin{aligned}{{}}1\\{ - 1}\end{aligned}} \right)\\& = \left( {\begin{aligned}{{}}{ - 10 - 0 + 21 - 11}\\{0 - 18 + 7 + 11}\end{aligned}} \right)\\& = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\& = {\bf{0}}\end{aligned}\)

From two steps, the sum of the coefficients is 0, so the combination is an affine dependence relationship. Both the given equations are valid.

03

Consider the coefficients of \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{3}}}\)

From the equations \({\rm{p}}\, = \frac{1}{{121}}{{\rm{v}}_1} + \frac{{72}}{{121}}{{\rm{v}}_2} + \frac{{37}}{{121}}{{\rm{v}}_3} + \frac{1}{{11}}{{\rm{v}}_4}\)and\(10{{\rm{v}}_1} - 6{{\rm{v}}_2} + 7{{\rm{v}}_3} - 11{{\rm{v}}_4} = 0\), the coefficients \({{\bf{v}}_1}\) and \({{\bf{v}}_2}\) are \(\frac{1}{{1210}}\) and \(\frac{{37}}{{847}}\).

Subtract \(\frac{1}{{1210}}\) times \(10{{\rm{v}}_1} - 6{{\rm{v}}_2} + 7{{\rm{v}}_3} - 11{{\rm{v}}_4} = 0\) from \({\rm{p}}\, = \frac{1}{{121}}{{\rm{v}}_1} + \frac{{72}}{{121}}{{\rm{v}}_2} + \frac{{37}}{{121}}{{\rm{v}}_3} + \frac{1}{{11}}{{\rm{v}}_4}\).

\(\begin{aligned}{}{\rm{p}} &= \frac{1}{{121}}{{\bf{v}}_1} + \frac{{72}}{{121}}{{\bf{v}}_2} + \frac{{37}}{{121}}{{\bf{v}}_3} + \frac{1}{{11}}{{\bf{v}}_4} - \frac{1}{{1210}}\left( {10{{\bf{v}}_1} - 6{{\bf{v}}_2} + 7{{\bf{v}}_3} - 11{{\bf{v}}_4}} \right)\\ &= \left( {\frac{1}{{121}} - \frac{1}{{121}}} \right){{\bf{v}}_1} + \left( {\frac{{72}}{{121}} + \frac{6}{{1210}}} \right){{\bf{v}}_2} + \left( {\frac{{37}}{{121}} - \frac{7}{{1210}}} \right){{\bf{v}}_3} + \left( {\frac{1}{{11}} + \frac{{11}}{{1210}}} \right){{\bf{v}}_4}\\ &= \left( 0 \right){{\bf{v}}_1} + \left( {\frac{3}{5}} \right){{\bf{v}}_2} + \left( {\frac{3}{{10}}} \right){{\bf{v}}_3} + \left( {\frac{1}{{10}}} \right){{\bf{v}}_4}\\ &= \frac{3}{5}{{\bf{v}}_2} + \frac{3}{{10}}{{\bf{v}}_3} + \frac{1}{{10}}{{\bf{v}}_4}\end{aligned}\)

Similarly, add \(\frac{1}{{1210}}\) times \(10{{\rm{v}}_1} - 6{{\rm{v}}_2} + 7{{\rm{v}}_3} - 11{{\rm{v}}_4} = 0\) from \({\rm{p}}\, = \frac{1}{{121}}{{\rm{v}}_1} + \frac{{72}}{{121}}{{\rm{v}}_2} + \frac{{37}}{{121}}{{\rm{v}}_3} + \frac{1}{{11}}{{\rm{v}}_4}\).

\(\begin{aligned}{}{\rm{p}}& = \frac{1}{{121}}{{\bf{v}}_1} + \frac{{72}}{{121}}{{\bf{v}}_2} + \frac{{37}}{{121}}{{\bf{v}}_3} + \frac{1}{{11}}{{\bf{v}}_4} + \frac{1}{{1210}}\left( {10{{\bf{v}}_1} - 6{{\bf{v}}_2} + 7{{\bf{v}}_3} - 11{{\bf{v}}_4}} \right)\\ &= \left( {\frac{1}{{121}} + \frac{{10}}{{121}}} \right){{\bf{v}}_1} + \left( {\frac{{72}}{{121}} - \frac{6}{{121}}} \right){{\bf{v}}_2} + \left( {\frac{{37}}{{121}} + \frac{7}{{121}}} \right){{\bf{v}}_3} + \left( {\frac{1}{{11}} - \frac{{11}}{{121}}} \right){{\bf{v}}_4}\\ &= \frac{1}{{11}}{{\bf{v}}_1} + \frac{6}{{11}}{{\bf{v}}_2} + \frac{4}{{11}}{{\bf{v}}_3}\end{aligned}\)

So, the expressions for pare\({\rm{p}} = \frac{3}{5}{{\bf{v}}_2} + \frac{3}{{10}}{{\bf{v}}_3} + \frac{1}{{10}}{{\bf{v}}_4}\)or \({\rm{p}} = \frac{1}{{11}}{{\bf{v}}_1} + \frac{6}{{11}}{{\bf{v}}_2} + \frac{4}{{11}}{{\bf{v}}_3}\).

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Most popular questions from this chapter

In Exercises 9 and 10, mark each statement True or False. Justify each answer.

10.a. If \(\left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p}} \right\}\) is an affinely dependent set in \({\mathbb{R}^n}\), then the set \(\left\{ {{{\overline {\mathop{\rm v}\nolimits} }_1},...,{{\overline {\mathop{\rm v}\nolimits} }_p}} \right\}\) in \({\mathbb{R}^{n + 1}}\) of homogeneous forms may be linearly independent.

b. If \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\) and \({{\mathop{\rm v}\nolimits} _4}\) are in \({\mathbb{R}^3}\) and if the set \(\left\{ {{{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}} \right\}\) is linearly independent, then \(\left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}} \right\}\) is affinely independent.

c. Given \(S = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _k}} \right\}\) in \({\mathbb{R}^n}\), each \({\bf{p}}\) in\({\mathop{\rm aff}\nolimits} S\) has a unique representation as an affine combination of \({{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _k}\).

d. When color information is specified at each vertex \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\) of a triangle in \({\mathbb{R}^3}\), then the color may be interpolated at a point p in \({\mathop{\rm aff}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}} \right\}\) using the barycentric coordinates of p.

e. If T is a triangle in \({\mathbb{R}^2}\) and if a point p is on edge of the triangle, then the barycentric coordinates of p (for this triangle) are not all positive.

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 \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{aligned}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{aligned}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each of the given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{8}}\\{\bf{4}}\end{aligned}} \right)\)

b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{6}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right)\)

c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{ - {\bf{1}}}\\{ - {\bf{5}}}\end{aligned}} \right)\)

Question: In Exercise 6, determine whether or not each set is compact and whether or not it is convex.

6. Use the sets from Exercise 4.

Question: In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{array}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{array}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each is given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{ - {\bf{19}}}\\{\bf{5}}\end{array}} \right)\) b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{{\bf{1}}.{\bf{5}}}\\{ - {\bf{1}}.{\bf{3}}}\\{ - .{\bf{5}}}\end{array}} \right)\) c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{4}}}\\{\bf{0}}\end{array}} \right)\)
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