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

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{2}}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{4}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{7}}\end{aligned}} \right)\), \({\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{aligned}} \right)\)

Write the translated points as shown below:

\({{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}{ - 3}\\0\end{aligned}} \right)\)

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

\({{\bf{v}}_4} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}2\\5\end{aligned}} \right)\)

\({\bf{y}} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}4\\1\end{aligned}} \right)\)

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

\(\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} &= {c_2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + {c_3}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + {c_4}\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right)\\\left( {\begin{aligned}{*{20}{c}}4\\1\end{aligned}} \right) &= {c_2}\left( {\begin{aligned}{*{20}{c}}{ - 3}\\0\end{aligned}} \right) + {c_3}\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right) + {c_4}\left( {\begin{aligned}{*{20}{c}}2\\5\end{aligned}} \right)\end{aligned}\)

The augmented matrixcan be written as shown below:

\(M = \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 1}&2&4\\0&2&5&1\end{aligned}} \right)\)

Row reduce the augmented matrix as shown below:

\(\begin{aligned}{c}M &= \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 1}&2&4\\0&2&5&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}1&{ - 1}&2&4\\0&2&5&1\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{C_1} \to \frac{{{C_1}}}{3}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}1&{\frac{1}{3}}&{ - \frac{2}{3}}&{ - \frac{4}{3}}\\0&1&{ - \frac{5}{2}}&{\frac{1}{2}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to - \frac{1}{3}{R_1},\,{R_2} \to \frac{1}{2}{R_2}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}1&0&{ - \frac{9}{6}}&{ - \frac{9}{6}}\\0&1&{\frac{5}{2}}&{\frac{1}{2}}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - \frac{1}{3}{R_2}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}1&0&{ - \frac{3}{2}}&{ - \frac{3}{2}}\\0&1&{\frac{5}{2}}&{\frac{1}{2}}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{c}{c_2} - \frac{3}{4}{c_4} &= - \frac{3}{2}\\{c_2} &= \frac{3}{2}{c_4} - \frac{3}{2}\end{aligned}\)

Let \({c_4} = 0\). Thus the value is shown below:

\({c_2} = - \frac{3}{2}\)and \({c_3} = \frac{1}{2}\)

Substitute the values in the equation of translated point as shown below:

\(\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} &= - \frac{3}{2}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{1}{2}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right) + 0\left( {{{\bf{v}}_4} - {{\bf{v}}_1}} \right)\\{\bf{y}} &= {{\bf{v}}_1} - \frac{3}{2}{{\bf{v}}_2} + \frac{3}{2}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_3} - \frac{1}{2}{{\bf{v}}_1}\\{\bf{y}} &= 2{{\bf{v}}_1} - \frac{3}{2}{{\bf{v}}_2} + \frac{1}{2}{{\bf{v}}_3}\end{aligned}\)

So, the vector \({\bf{y}}\) is \(2{{\bf{v}}_1} - \frac{3}{2}{{\bf{v}}_2} + \frac{1}{2}{{\bf{v}}_3}\).