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Found in: Page 437

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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}}\\{\bf{0}}\end{aligned}} \right), {{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{6}}}\\{\bf{7}}\end{aligned}} \right), {{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}\\{\bf{3}}\\{\bf{1}}\end{aligned}} \right), {\bf{y}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{ - {\bf{4}}}\end{aligned}} \right)

The affine combination is $${\bf{y}} = \frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}$$.

See the step by step solution

## Step 1: Find the translated point

Write the translated points as shown below:

{{\bf{v}}_2} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}1\\8\\{ - 7}\end{aligned}} \right)

{{\bf{v}}_3} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}3\\1\\1\end{aligned}} \right)

{\bf{y}} - {{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\2\\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}}{ - 4}\\2\\4\end{aligned}} \right) = {c_2}\left( {\begin{aligned}{*{20}{c}}1\\{ - 8}\\7\end{aligned}} \right) + {c_3}\left( {\begin{aligned}{*{20}{c}}3\\1\\1\end{aligned}} \right)\end{aligned}

## Step 2: Write the augmented matrix

The augmented matrix can be written as shown below:

M = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\{ - 8}&1&2\\7&1&{ - 4}\end{aligned}} \right)

Row reduce the augmented matrix as shown below:

\begin{aligned}{c}M = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\{ - 8}&1&2\\7&1&{ - 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\0&{25}&{ - 30}\\0&{ - 20}&{24}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_2} \to {R_2} + 8{R_1}\\{R_3} \to {R_3} - 7{R_1}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\0&5&{ - 6}\\0&{ - 10}&{12}\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_2} \to \frac{1}{5}{R_2}\\{R_3} \to \frac{1}{2}{R_3}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&3&{ - 4}\\0&1&{ - \frac{6}{5}}\\0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{aligned}{l}{R_3} \to {R_3} + 2{R_2}\\{R_2} \to \frac{1}{5}{R_2}\end{aligned} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&{ - \frac{2}{5}}\\0&1&{\frac{6}{5}}\\0&0&0\end{aligned}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to {R_1} - 3{R_2}} \right)\end{aligned}

## Step 3: Write the system of equations

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

$${c_2} = - \frac{2}{5}$$

And,

$${c_3} = \frac{6}{5}$$

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

\begin{aligned}{c}{\bf{y}} - {{\bf{v}}_1} = - \frac{2}{5}\left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) + \frac{6}{5}\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\\{\bf{y}} = {{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{2}{5}{{\bf{v}}_1} + \frac{6}{5}{{\bf{v}}_3} - \frac{6}{5}{{\bf{v}}_1}\\{\bf{y}} = \frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}\end{aligned}

So, the vector $${\bf{y}}$$ is $$\frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}$$.