Question

In Exercises 1-4, write y as an affine combination of the other point listed, if possible.

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\\{ - {\bf{2}}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\\{\bf{6}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{{\bf{17}}}\\{\bf{1}}\\{\bf{5}}\end{aligned}} \right)\)

Short answer

The affine combination is \({\bf{y}} = - 3{{\bf{v}}_1} + 2{{\bf{v}}_2} + 2{{\bf{v}}_3}\).

Step-by-step solution

Find the translated point

Write the translated points as shown below:

\({{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}3\\3\\{ - 3}\end{aligned}} \right)\)

\({{\bf{v}}_3} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}7\\{ - 3}\\5\end{aligned}} \right)\)

\({\bf{y}} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}{20}\\0\\4\end{aligned}} \right)\)

Write the equation by using the translated matrix as shown below:

\(\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} = {c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\\left( {\begin{aligned}{*{20}{c}}{20}\\0\\4\end{aligned}} \right) = {c_2}\left( {\begin{aligned}{*{20}{c}}3\\3\\{ - 3}\end{aligned}} \right) + {c_3}\left( {\begin{aligned}{*{20}{c}}7\\{ - 3}\\5\end{aligned}} \right)\end{aligned}\)

Write the augmented matrix

The augmented matrix can be written as shown below:

\(M = \left( {\begin{aligned}{*{20}{c}}3&7&{20}\\3&{ - 3}&0\\{ - 3}&5&4\end{aligned}} \right)\)

Row reduce the augmented matrix as shown below:

\(\begin{aligned}{c}M = \left( {\begin{aligned}{*{20}{c}}3&7&{20}\\3&{ - 3}&0\\{ - 3}&5&4\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&7&{20}\\0&{ - 10}&{ - 20}\\0&{12}&{24}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_2} \to {R_2} - {R_1}\\{R_3} \to {R_3} + {R_1}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&7&{20}\\0&1&2\\0&1&2\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_2} \to - \frac{1}{{10}}{R_2}\\{R_3} \to \frac{1}{{12}}{R_3}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&0&6\\0&1&2\\0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_3} \to {R_3} - {R_2}\\{R_1} \to {R_1} - 7{R_2}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&2\\0&1&2\\0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to \frac{1}{3}{R_1}} \right)\end{aligned}\)

Write the system of equations

From the augmented matrix, the system of equation is shown below:

\({c_2} = 2\)

And,

\({c_3} = 2\)

Substitute the value in the equation of translated points as shown below:

\(\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} = 2\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + 2\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\{\bf{y}} = {{\bf{v}}_1} + 2{{\bf{v}}_2} - 2{{\bf{v}}_1} + 2{{\bf{v}}_3} - 2{{\bf{v}}_1}\\{\bf{y}} = - 3{{\bf{v}}_1} + 2{{\bf{v}}_2} + 2{{\bf{v}}_3}\end{aligned}\)

So, the vector \({\bf{y}}\) is \( - 3{{\bf{v}}_1} + 2{{\bf{v}}_2} + 2{{\bf{v}}_3}\).