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.

3.\(\left( {\begin{aligned}{{}}1\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 2}\\{ - 4}\\8\end{aligned}} \right),\left( {\begin{aligned}{{}}2\\{ - 1}\\{11}\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\{15}\\{ - 9}\end{aligned}} \right)\)

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 \({c_1},{c_2},...,{c_p}\) not all zero, and the sum must be zero \({c_1} + {c_2} + ... + {c_p} = 0\), and \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

Let \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}1\\2\\{ - 1}\end{aligned}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{{}}{ - 2}\\{ - 4}\\8\end{aligned}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{{}}2\\{ - 1}\\{11}\end{aligned}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{aligned}{{}}0\\{15}\\9\end{aligned}} \right)\).

Compute \({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1}\), \({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}\), and \({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}\) as shown below

\({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}{ - 3}\\{ - 6}\\9\end{aligned}} \right)\), \({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}1\\{ - 3}\\{12}\end{aligned}} \right)\), \({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}{ - 1}\\{13}\\{ - 8}\end{aligned}} \right)\)

Write the augmented matrix as shown below:

\(\left( {\begin{aligned}{{}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}}&{{{\bf{v}}_4}}\end{aligned}} \right) \sim \left( {\begin{aligned}{{}}1&{ - 2}&2&0\\2&{ - 4}&{ - 1}&{15}\\{ - 1}&8&{11}&9\end{aligned}} \right)\)

At row 2, multiply row 1 by 2 and subtract it from row 2.

\( \sim \left( {\begin{aligned}{{}}1&{ - 2}&2&0\\0&0&{ - 5}&{15}\\{ - 1}&8&{11}&{ - 9}\end{aligned}} \right)\)

At row 3, add row 1 and row 3

\( \sim \left( {\begin{aligned}{{}}1&{ - 2}&2&0\\0&0&{ - 5}&{15}\\0&6&{13}&{ - 9}\end{aligned}} \right)\)

Interchange row 2 and row 3

\( \sim \left( {\begin{aligned}{{}}1&{ - 2}&2&0\\0&6&{13}&{ - 9}\\0&0&{ - 5}&{15}\end{aligned}} \right)\)

At row 2, multiply row 2 by \(\frac{1}{6}\)

\( \sim \left( {\begin{aligned}{{}}1&{ - 2}&2&0\\0&6&{\frac{{13}}{6}}&{ - \frac{3}{2}}\\0&0&{ - 5}&{15}\end{aligned}} \right)\)

At row 1, multiply row 2 by 2 and add it to row 1

\( \sim \left( {\begin{aligned}{{}}1&0&{\frac{{19}}{3}}&{ - 3}\\0&6&{\frac{{13}}{6}}&{ - \frac{3}{2}}\\0&0&{ - 5}&{15}\end{aligned}} \right)\)

At row 3, multiply row 3 by \(\frac{1}{{ - 5}}\)

\( \sim \left( {\begin{aligned}{{}}1&0&{\frac{{19}}{3}}&{ - 3}\\0&6&{\frac{{13}}{6}}&{ - \frac{3}{2}}\\0&0&1&{ - 3}\end{aligned}} \right)\)

At row 1, multiply row 3 by \(\frac{{19}}{3}\) and subtract it from row 1. At row 2, multiply row 3 by \(\frac{{13}}{6}\) and remove it from row 2

\( \sim \left( {\begin{aligned}{{}}1&0&0&{16}\\0&1&0&5\\0&0&1&{ - 3}\end{aligned}} \right)\)

The columns are linearly independent because every column is a pivot column.

The solution of the matrix is \({{\mathop{\rm x}\nolimits} _1} = 16,{{\mathop{\rm x}\nolimits} _2} = 5,{{\mathop{\rm x}\nolimits} _3} = - 3\).

Theorem 5states that an indexed set \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p}} \right\}\) in \({\mathbb{R}^n}\), with \(p \ge 2\), the following statement is equivalent. This means that either all the statements are true or all the statements are false.

1. The set \(S\) isaffinely dependent.
2. Each of the points in \(S\)is an affine combination of the other points in \(S\).
3. In \({\mathbb{R}^n}\), the set \(\left\{ {{{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p} - {{\mathop{\rm v}\nolimits} _1}} \right\}\)is linearly dependent.
4. The set \(\left\{ {{{\bar v}_1},...,{{\bar v}_p}} \right\}\) of homogeneous forms in \({\mathbb{R}^{n + 1}}\) is linearly dependent.

If \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\) are points, then the set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is a basis for \({\mathbb{R}^3}\), and\({{\mathop{\rm v}\nolimits} _4} = 16{{\mathop{\rm v}\nolimits} _1} + 5{{\mathop{\rm v}\nolimits} _2} - 3{{\mathop{\rm v}\nolimits} _3}\). However, the weights in the linear combination do not sum to 1. The set \(S\) is affinely independent.

Thus, the set of points are affinely independent.