Exercises 7-10 use the terminology from section 8.2.

Repeat Exercise 7 for \(T = \left\{ {\left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{0}}\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{5}}\end{array}} \right],\,\left[ {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{1}}\end{array}} \right]} \right\}\) and \({{\bf{p}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\end{array}} \right]\), \({{\bf{p}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{1}}\end{array}} \right]\), \({{\bf{p}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\frac{{\bf{1}}}{{\bf{3}}}}\end{array}} \right]\), and \({{\bf{p}}_{\bf{4}}} = \left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\end{array}} \right]\).

Use the vectors of T and \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), \({{\bf{p}}_3}\), and \({{\bf{p}}_4}\) the augmented matrixcan be expressed as shown below:

\(\left[ {\begin{array}{*{20}{c}}{\widetilde {{{\bf{v}}_1}}}&{\widetilde {{{\bf{v}}_2}}}&{\widetilde {{{\bf{v}}_3}}}&{\widetilde {{{\bf{p}}_1}}}&{\widetilde {{{\bf{p}}_2}}}&{\widetilde {{{\bf{p}}_3}}}&{\widetilde {{{\bf{p}}_4}}}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\2&0&{ - 1}&2&1&1&1\\0&5&1&1&1&{\frac{1}{3}}&0\end{array}} \right]\)

\[\begin{array}{c}M = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\0&{ - 2}&{ - 3}&0&{ - 1}&{ - 1}&{ - 1}\\0&5&1&1&1&{\frac{1}{3}}&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} - 2{R_1}} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\0&1&{\frac{3}{2}}&0&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\0&5&1&1&1&{\frac{1}{3}}&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to - \frac{1}{2}{R_2}} \right)\end{array}\]

Row reduce further,

\[\begin{array}{c}M = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\0&1&{\frac{3}{2}}&0&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\0&0&{ - \frac{{13}}{2}}&1&{ - \frac{3}{2}}&{ - \frac{{13}}{6}}&{ - \frac{5}{2}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_3} \to {R_3} - 5{R_2}} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\0&1&{\frac{3}{2}}&0&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\0&0&1&{ - \frac{2}{{13}}}&{\frac{3}{{13}}}&{\frac{1}{3}}&{\frac{5}{{13}}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_3} \to - \frac{2}{{13}}{R_3}} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&1&0&{\frac{{15}}{{13}}}&{\frac{{10}}{{13}}}&{\frac{2}{3}}&{\frac{8}{{13}}}\\0&1&0&{\frac{3}{{13}}}&{\frac{2}{{13}}}&0&{ - \frac{1}{{13}}}\\0&0&1&{ - \frac{2}{{13}}}&{\frac{3}{{13}}}&{\frac{1}{3}}&{\frac{5}{{13}}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}{R_1} \to {R_1} - {R_3}\\{R_2} \to {R_2} - \frac{3}{2}{R_3}\end{array} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&{\frac{{12}}{{13}}}&{\frac{8}{{13}}}&{\frac{2}{3}}&{\frac{9}{{13}}}\\0&1&0&{\frac{3}{{13}}}&{\frac{2}{{13}}}&0&{ - \frac{1}{{13}}}\\0&0&1&{ - \frac{2}{{13}}}&{\frac{3}{{13}}}&{\frac{1}{3}}&{\frac{5}{{13}}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - {R_2}} \right)\end{array}\]

In the row reduced matrix column 4, column 5, column 6, and column 7 represents the barycentric coordinates of \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), \({{\bf{p}}_3}\), and \({{\bf{p}}_4}\) respectively.

So, for \({{\bf{p}}_1}\), \({{\bf{p}}_2}\), \({{\bf{p}}_3}\), and \({{\bf{p}}_4}\) the barycentric coordinate is\(\left( {\frac{{12}}{{13}},\,\frac{3}{{13}}, - \frac{2}{{13}}} \right)\), \(\left( {\frac{8}{{13}},\frac{2}{{13}},\frac{3}{{13}}} \right)\), \(\left( {\frac{2}{3},0,\frac{1}{3}} \right)\), and \(\left( {\frac{9}{{13}}, - \frac{1}{{13}},\frac{5}{{13}}} \right)\) respectively.

As barycentric coordinateof \({{\bf{p}}_1}\) and \({{\bf{p}}_4}\) has one negative value, therefore \({{\bf{p}}_3}\) and \({{\bf{p}}_4}\) are outside conv T.

As all coordinates for \({{\bf{p}}_2}\) are positive, therefore, it will be inside conv T. For \({{\bf{p}}_3}\), its second coordinate is zero, therefore \({{\bf{p}}_3}\) is on the edge of conv T.