91Ó°ÊÓ

Write the augmented matrix by using the points as shown below:

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}&{{{\bf{p}}_1}}&{{{\bf{p}}_2}}&{{{\bf{p}}_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&2&3&4&{ - 5}&4\\0&1&0&{ - 1}&3&2\\3&6&{12}&{15}&{ - 8}&8\\{ - 2}&{ - 5}&{ - 6}&{ - 7}&6&5\end{array}} \right)\end{array}\)

The augmented matrix can be written as,

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}1&2&3&4&{ - 5}&4\\0&1&0&{ - 1}&3&6\\0&0&3&3&7&{ - 9}\\0&{ - 1}&0&1&{ - 4}&{ - 6}\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_3} \to {R_3} - 3{R_1}\\{R_4} \to {R_4} + 2{R_1}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&3&6&{ - 11}&{ - 11}\\0&1&0&{ - 1}&3&6\\0&0&3&3&7&{ - 9}\\0&0&0&0&{ - 1}&0\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_1} \to {R_1} - 2{R_2}\\{R_4} \to {R_4} + {R_2}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&3&{ - 18}&{ - 2}\\0&1&0&{ - 1}&0&6\\0&0&3&3&7&{ - 9}\\0&0&0&0&{ - 1}&0\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_1} \to {R_1} - {R_3}\\{R_2} \to {R_2} + 3{R_4}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&3&0&{ - 2}\\0&1&0&{ - 1}&0&6\\0&0&3&3&7&{ - 9}\\0&0&0&0&{ - 1}&0\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_1} \to {R_1} - 18{R_4}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&3&0&{ - 2}\\0&1&0&{ - 1}&0&6\\0&0&1&1&{\frac{7}{3}}&{ - 3}\\0&0&0&0&{ - 1}&0\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to \frac{1}{3}{R_3}} \right\}\end{array}\)

There are no zero rows in the augmented matrix. Hence the set is linearly independent.

Use the augmented matrix, \({{\bf{p}}_1}\) which can be expressed as shown below:

\({{\bf{p}}_1} = 3\left( {{{\bf{b}}_1}} \right) - 1\left( {{{\bf{b}}_2}} \right) + 1\left( {{{\bf{b}}_3}} \right)\)

The sum of coefficients is \(3 - 1 + 1 = 3 \ne 1\).

So, \({{\bf{p}}_1}\) is not an affine combination of point in S.

From the augmented matrix, it can be observed that, \({{\bf{p}}_2}\) can not be written as the linear combination of point of S.

\({{\bf{p}}_2}\)is not an affine combination of points in S.

Use the augmented matrix, \({{\bf{p}}_3}\) which can be expressed as shown below:

\({{\bf{p}}_3} = - 2\left( {{{\bf{b}}_1}} \right) + 6\left( {{{\bf{b}}_2}} \right) - 3\left( {{{\bf{b}}_3}} \right)\)

The sum of coefficients is \( - 2 + 6 - 3 = 1\).

So, \({{\bf{p}}_3}\) is an affine combination of point in S.

\({{\bf{p}}_3} = - 2\left( {{{\bf{b}}_1}} \right) + 6\left( {{{\bf{b}}_2}} \right) - 3\left( {{{\bf{b}}_3}} \right)\)

Hence, \({{\bf{p}}_3}\) is an affine combination of point in S.