Question

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

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{2}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}\\{\bf{7}}\end{aligned}} \right)\)

Short answer

The affine combination is \({\bf{y}} = - 5{{\bf{v}}_1} + 2{{\bf{v}}_2} + 4{{\bf{v}}_3}\).

Step-by-step solution

Find the translated point

Write the translated points as shown below:

\({{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\1\end{aligned}} \right)\)

\({{\bf{v}}_3} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}2\\1\end{aligned}} \right)\)

\({\bf{y}} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\)

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

\(\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} = {c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\\left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right) = {c_2}\left( {\begin{aligned}{*{20}{c}}{ - 2}\\1\end{aligned}} \right) + {c_3}\left( {\begin{aligned}{*{20}{c}}2\\1\end{aligned}} \right)\end{aligned}\)

Write the augmented matrix

The augmented matrix can be written as shown below:

\(M = \left( {\begin{aligned}{*{20}{c}}{ - 2}&2&4\\1&1&6\end{aligned}} \right)\)

Row reduce the augmented matrix as shown below:

\(\begin{aligned}{c}M = \left( {\begin{aligned}{*{20}{c}}{ - 2}&2&4\\1&1&6\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&{ - 1}&{ - 2}\\1&1&6\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to - \frac{1}{2}{R_1}} \right)\,\\ = \left( {\begin{aligned}{*{20}{c}}1&{ - 1}&{ - 2}\\0&2&8\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_2} - {R_1}} \right)\,\\ = \left( {\begin{aligned}{*{20}{c}}1&{ - 1}&{ - 2}\\0&1&4\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to \frac{1}{2}{R_2}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&2\\0&1&4\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} + {R_2}} \right)\end{aligned}\)

Write the system of equations

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

\({c_2} = 2\)

And,

\({c_3} = 4\)

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

\(\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} &= 2\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + 4\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\{\bf{y}} &= {{\bf{v}}_1} + 2{{\bf{v}}_2} - 2{{\bf{v}}_1} + 4{{\bf{v}}_3} - 4{{\bf{v}}_1}\\{\bf{y}} &= - 5{{\bf{v}}_1} + 2{{\bf{v}}_2} + 4{{\bf{v}}_3}\end{aligned}\)

So, the vector \({\bf{y}}\) is \( - 5{{\bf{v}}_1} + 2{{\bf{v}}_2} + 4{{\bf{v}}_3}\).