Question: Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{3}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\)

\({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{\bf{5}}\\{\bf{3}}\end{array}} \right)\) b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{{\bf{10}}}\\{\bf{9}}\\{ - {\bf{13}}}\end{array}} \right)\) c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{2}}\\{\bf{8}}\\{\bf{5}}\end{array}} \right)\)

and \(S = \left\{ {{{\bf{v}}_1},\,\,{{\bf{v}}_2},\,{{\bf{v}}_3}} \right\}\). It can be shown that S is linearly independent.

a. Is \({{\bf{p}}_{\bf{1}}}\) is span S? Is \({{\bf{p}}_{\bf{1}}}\) is \({\bf{aff}}\,S\)?

b. Is \({{\bf{p}}_{\bf{2}}}\) is span S? Is \({{\bf{p}}_{\bf{2}}}\) is \({\bf{aff}}\,S\)?

c. Is \({{\bf{p}}_{\bf{3}}}\) is span S? Is \({{\bf{p}}_{\bf{3}}}\) is \({\bf{aff}}\,S\)?

The augmented matrix can be expressed as,

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}&{{{\bf{p}}_1}}&{{{\bf{p}}_2}}&{{{\bf{p}}_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&{ - 1}&2&{ - 3}&{10}&2\\3&0&1&5&9&8\\0&4&1&3&{ - 13}&5\end{array}} \right)\end{array}\)

The augmented matrix can be written as,

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&{ - 1}&2&{ - 3}&{10}&2\\0&{ - 6}&4&{ - 10}&{36}&{ - 4}\\0&4&1&3&{ - 13}&5\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to {R_3} - 3{R_1}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&{ - 1}&2&{ - 3}&{10}&2\\0&0&{ - 8}&8&{ - 24}&{ - 16}\\0&0&9&{ - 9}&{27}&{13}\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_3} \to {R_3} - 6{R_2}\\{R_4} \to {R_4} + 4{R_2}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&{ - 1}&2&{ - 3}&{10}&2\\0&0&1&{ - 1}&3&2\\0&0&9&{ - 9}&{27}&{13}\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_3} \to - \frac{1}{8}{R_3}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&1&{ - 2}&3&{ - 10}&{ - 2}\\0&0&1&{ - 1}&3&2\\0&0&0&0&0&{ - 5}\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_4} \to {R_4} - 9{R_3}\\{R_2} \to - {R_2}\end{array} \right\}\end{array}\)

Row reduce further,

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&1&{ - 2}&3&{ - 10}&{ - 2}\\0&0&1&{ - 1}&3&0\\0&0&0&0&0&1\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_3} \to - \frac{1}{5}{R_3}\\{R_3} \to {R_3} - 2{R_4}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&2&{ - 1}&5&{ - 9}&4\\0&1&0&1&{ - 4}&{ - 2}\\0&0&1&{ - 1}&3&0\\0&0&0&0&0&1\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_2} \to {R_2} + 2{R_3}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&2&0&4&{ - 6}&4\\0&1&0&1&{ - 4}&0\\0&0&1&{ - 1}&3&0\\0&0&0&0&0&1\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_2} \to {R_1} + 2{R_4}\\{R_1} \to {R_1} + {R_3}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&2&2&0\\0&1&0&1&{ - 4}&0\\0&0&1&{ - 1}&3&0\\0&0&0&0&0&1\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}\left\{ {{R_1} \to {R_1} - 4{R_4}} \right\}\\{R_1} \to {R_1} - 2{R_2}\end{array} \right\}\end{array}\)

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:

\({{\bf{p}}_1} = 2\left( {{{\bf{b}}_1}} \right) + 1\left( {{{\bf{b}}_2}} \right) - 1\left( {{{\bf{b}}_3}} \right)\)

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

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

Use the augmented matrix, \({{\bf{p}}_2}\) which can be expressed as shown below:

\({{\bf{p}}_2} = 2\left( {{{\bf{b}}_1}} \right) - 4\left( {{{\bf{b}}_2}} \right) + 3\left( {{{\bf{b}}_3}} \right)\)

The sum of coefficients is \(2 - 4 + 3 = 1\).

So, \({{\bf{p}}_2}\) is an affine combination of point in S.

\({{\bf{p}}_2} = 2\left( {{{\bf{b}}_1}} \right) - 4\left( {{{\bf{b}}_2}} \right) + 3\left( {{{\bf{b}}_3}} \right)\)

From the augmented matrix, it can be observed that, \({p_3}\) can not be written as the linear combination of point of S.

\({p_3}\)is not an affine combination of points in S.