Question

Let\({v_1} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{5}}\end{array}} \right]\),\({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{3}}\end{array}} \right]\),\({p_1} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{array}} \right]\),\({p_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\end{array}} \right]\),\({p_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\end{array}} \right]\),\({p_{\bf{4}}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{0}}\end{array}} \right]\),\({p_{\bf{5}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{array}} \right]\),\({p_{\bf{6}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{array}} \right]\),\({p_{\bf{7}}} = \left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{4}}\end{array}} \right]\)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\({p_1}\),\({p_{\bf{2}}}\), and\({p_{\bf{3}}}\)with respect to S.
3. On graph paper, sketch the triangle\(T\)with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\), extend the sides as in Figure 8, and plot the points\({p_{\bf{4}}}\),\({p_{\bf{5}}}\),\({p_{\bf{6}}}\), and\({p_{\bf{7}}}\). Without calculating the actual values, determine the signs of the barycentric coordinates of points\({p_{\bf{4}}}\),\({p_{\bf{5}}}\),\({p_{\bf{6}}}\), and\({p_{\bf{7}}}\).

Short answer

1. The set is affinely independent.
2. The barycentric coordinates of points are\({{\bf{p}}_1} \leftrightarrow \left( { - \frac{2}{7},\frac{5}{7},\frac{4}{7}} \right)\),\({{\bf{p}}_2} \leftrightarrow \left( {\frac{2}{7}, - \frac{5}{7},\frac{{10}}{7}} \right)\), and \({{\bf{p}}_3} \leftrightarrow \left( {\frac{2}{7},\frac{2}{7},\frac{3}{7}} \right)\).
3. The signs of the barycentric points are \({{\bf{p}}_4} \leftrightarrow \left( { + , - , - } \right)\), \({{\bf{p}}_5} \leftrightarrow \left( { + , + , - } \right)\), \({{\bf{p}}_6} \leftrightarrow \left( { + , + , + } \right)\) and \({{\bf{p}}_7} \leftrightarrow \left( { - ,0, + } \right)\).

Step-by-step solution

State the condition for affinely dependence

The set is said to be affinely dependent if the set\(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},...,{{\bf{v}}_p}} \right\}\)in the dimension\({\mathbb{R}^n}\)exists such that for non-zero scalars\({c_1},{c_2},...,{c_p}\), the sum of scalars is zero i.e.\({c_1} + {c_2} + ... + {c_p} = 0\), and\({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

(a)

Show affinely independence

Consider the set of vectors\(S = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},{{\bf{v}}_3}} \right\}\), where\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\),\({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}1\\5\end{array}} \right]\), and\({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}4\\3\end{array}} \right]\).

Let the newpoints be \({{\bf{v}}_3} - {{\bf{v}}_1}\) and \({{\bf{v}}_2} - {{\bf{v}}_1}\), be obtained by eliminating the first point.

\(\begin{array}{c}{{\bf{v}}_3} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}4\\3\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}4\\2\end{array}} \right]\end{array}\)

And

\(\begin{array}{c}{{\bf{v}}_2} - {{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}1\\5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\4\end{array}} \right]\end{array}\)

It is observed that \({{\bf{v}}_2} - {{\bf{v}}_1}\) and \({{\bf{v}}_3} - {{\bf{v}}_1}\) are not proportional or multiples of each other. They are linearly independent. Also, the sum of weights cannot be 0.

Thus, the indexed set \(S\)is affinely independent.

(b)

State the barycentric coordinates of points

Obtain the barycentric coordinates of points using the vectors\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\),\({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}1\\5\end{array}} \right]\),\({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}4\\3\end{array}} \right]\), and the points\({{\bf{p}}_1} = \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right]\),\({{\bf{p}}_2} = \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\), and\({{\bf{p}}_3} = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\).

Construct an augmented matrix using the above vectors and points.

\(M = \left[ {\begin{array}{*{20}{c}}0&1&4&3&5&2\\1&5&3&5&1&3\\1&1&1&1&1&1\end{array}} \right]\)

Obtain the row-reduced echelon form as shown below:

Interchange rows 1 and 3.

\(M = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1\\1&5&3&5&1&3\\0&1&4&3&5&2\end{array}} \right]\)

Add\( - 1\)times row 1 to row 2 to get row 2.

\(M = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1\\0&4&2&4&0&2\\0&1&4&3&5&2\end{array}} \right]\)

Add\( - 3\)times row 3 to row 2 to get row 2.

\(M = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1\\0&1&{ - 10}&{ - 5}&{ - 15}&{ - 4}\\0&1&4&3&5&2\end{array}} \right]\)

Add\( - 1\)times row 2 to row 1 to get row 1. Then, add\( - 1\)times row 2 to row 3 to get row 3.

\(M = \left[ {\begin{array}{*{20}{c}}1&0&{11}&6&{16}&5\\0&1&{ - 10}&{ - 5}&{ - 15}&{ - 4}\\0&0&{14}&8&{20}&6\end{array}} \right]\)

Multiply row 3 by\(\frac{1}{{14}}\)to get row 3.

\(M = \left[ {\begin{array}{*{20}{c}}1&0&{11}&6&{16}&5\\0&1&{ - 10}&{ - 5}&{ - 15}&{ - 4}\\0&0&1&{8/14}&{20/14}&{6/14}\end{array}} \right]\)

Add\( - 11\)times row 3 to row 1 to get row 1. Then, add 10 times row 3 to row 2 to get row 2.

\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 4/14}&{4/14}&{4/14}\\0&1&0&{10/14}&{ - 10/14}&{4/14}\\0&0&1&{8/14}&{20/14}&{6/14}\end{array}} \right]\)

Thus, the barycentric coordinates of points are\({{\bf{p}}_1} \leftrightarrow \left( { - \frac{2}{7},\frac{5}{7},\frac{4}{7}} \right)\),\({{\bf{p}}_2} \leftrightarrow \left( {\frac{2}{7}, - \frac{5}{7},\frac{{10}}{7}} \right)\), and \({{\bf{p}}_3} \leftrightarrow \left( {\frac{2}{7},\frac{2}{7},\frac{3}{7}} \right)\).

(c)

Find the signs of the barycentric coordinates

Sketch the triangle using the vertices \({v_1} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\end{array}} \right]\), \({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{5}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{3}}\end{array}} \right]\). Also, extend the sides and plot the points \({p_1} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{array}} \right]\), \({p_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{1}}\end{array}} \right]\), \({p_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\end{array}} \right]\), \({p_{\bf{4}}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{0}}\end{array}} \right]\), \({p_{\bf{5}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{array}} \right]\), \({p_{\bf{6}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{array}} \right]\), \({p_{\bf{7}}} = \left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{4}}\end{array}} \right]\) as shown:

Obtain the signs of the coordinates\({p_{\bf{4}}}\), \({{\bf{p}}_5}\),\({{\bf{p}}_6}\), and\({{\bf{p}}_7}\)from the figure.

Point\({p_{\bf{4}}}\)has the coordinate signs \(\left( { + , - , - } \right)\), point\({{\bf{p}}_5}\) has the coordinate signs \(\left( { + , + , - } \right)\), point\({{\bf{p}}_6}\) has the coordinate signs \(\left( { + , + , + } \right)\), and point\({{\bf{p}}_7}\) has the coordinate signs \(\left( { - ,0, + } \right)\),

Therefore, the signs of the barycentric points are \({{\bf{p}}_4} \leftrightarrow \left( { + , - , - } \right)\), \({{\bf{p}}_5} \leftrightarrow \left( { + , + , - } \right)\), \({{\bf{p}}_6} \leftrightarrow \left( { + , + , + } \right)\) and \({{\bf{p}}_7} \leftrightarrow \left( { - ,0, + } \ri