• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q4E

Expert-verified
Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

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

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

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 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 \(\frac{1}{5}{{\bf{v}}_1} - \frac{2}{5}{{\bf{v}}_2} + \frac{6}{5}{{\bf{v}}_3}\).

Most popular questions for Math Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.