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Chapter 8: Q19E (page 437)

Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d, so that \(\left( {f:d} \right)\) the hyperplane H described in the exercise.

Let H be the column space of the matrix \(B = \left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{ - {\bf{4}}}&{\bf{2}}\\{\bf{7}}&{ - {\bf{6}}}\end{array}} \right)\). That is, \(H = {\bf{Col}}\,B\).(Hint: How is \({\bf{Col}}\,B\) related to Nul \({B^T}\)? See section 6.1)

Short Answer

Expert verified

The linear functional is \(f\left( {{x_1},{x_2},{x_3}} \right) = - 11{x_1} + 4{x_2} + {x_3}\) and \(d = 0\).

Step by step solution

01

Write the matrix equation

The matrix equation can be written as follows:

\(\begin{array}{c}{B^T}{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}1&{ - 4}&7\\0&2&{ - 6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\)

02

Write the equation using matrix multiplication

The matrix equation can be simplified as shown below:

\({x_1} + 4{x_2} - 7{x_3} = 0\)

And,

\(\begin{array}{c}2{x_2} - 6{x_3} = 0\\{x_2} = 3{x_3}\end{array}\)

Simplifying the above equations:

\(\begin{array}{c}{x_1} + 4\left( {3{x_3}} \right) - 7{x_3} = 0\\{x_1} = - 5{x_3}\end{array}\)

03

Write the general solution

The general solution is:

\(\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 5{x_3}}\\{3{x_3}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\1\end{array}} \right){x_3}\end{array}\)

So, \(f\left( {{x_1},{x_2},{x_3}} \right) = - 5{x_1} + 3{x_2} + {x_3}\) and \(d = 0\).

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Most popular questions from this chapter

Question: 29. Prove that the open ball \(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x:}}\left\| {{\rm{x - p}}} \right\| < \delta } \right\}\)is a convex set. (Hint: Use the Triangle Inequality).

Explain why any set of five or more points in \({\mathbb{R}^3}\) must be affinely dependent.

Let\({v_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{array}} \right]\),\({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{0}}\end{array}} \right]\), and let\(S = \left\{ {{v_1},{v_2},{v_3}} \right\}\).

  1. Show that the set is affinely independent.
  2. Find the barycentric coordinates of\({p_1} = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\),\({p_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{array}} \right]\),\({p_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{\bf{1}}\end{array}} \right]\),\({p_{\bf{4}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\end{array}} \right]\), and\({p_{\bf{5}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{1}}\end{array}} \right]\), with respect to S.
  3. Let\(T\)be the triangle with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\). When the sides of\(T\)are extended, the lines divide\({\mathbb{R}^{\bf{2}}}\)into seven regions. See Figure 8. Note the signs of the barycentric coordinates of the points in each region. For example,\({{\bf{p}}_{\bf{5}}}\)is inside the triangle\(T\)and all its barycentric coordinates are positive. Point\({{\bf{p}}_{\bf{1}}}\)has coordinates\(\left( { - , + , + } \right)\). Its third coordinate is positive because\({{\bf{p}}_{\bf{1}}}\)is on the\({{\bf{v}}_{\bf{3}}}\)side of the line through\({{\bf{v}}_{\bf{1}}}\)and\({{\bf{v}}_{\bf{2}}}\). Its first coordinate is negative because\({{\bf{p}}_{\bf{1}}}\)is opposite the\({{\bf{v}}_{\bf{1}}}\)side of the line through\({{\bf{v}}_{\bf{2}}}\)and\({{\bf{v}}_{\bf{3}}}\). Point\({{\bf{p}}_{\bf{2}}}\)is on the\({{\bf{v}}_{\bf{2}}}{{\bf{v}}_{\bf{3}}}\)edge of\(T\). Its coordinates are\(\left( {0, + , + } \right)\). Without calculating the actual values, determine the signs of the barycentric coordinates of points\({{\bf{p}}_{\bf{6}}}\),\({{\bf{p}}_{\bf{7}}}\), and\({{\bf{p}}_{\bf{8}}}\)as shown in Figure 8.

In Exercises 21–24, a, b, and c are non-collinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \) and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

22. Let p be a point on the line through a and b. Show that\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

In Exercises 1-6, determine if the set of points is affinely dependent. (See Practice Problem 2.) If so, construct an affine dependence relation for the points.

5.\(\left( {\begin{aligned}{{}}1\\0\\{ - 2}\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 1}\\5\\1\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\5\\{ - 3}\end{aligned}} \right)\)
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