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Found in: Page 1

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 3 and 4, display the following vectors using arrowson an $$xy$$-graph: u, v, $$- {\bf{v}}$$, $$- 2{\bf{v}}$$, u + v , u - v , and u - 2v. Notice that is the vertex of a parallelogram whose other vertices are u, 0, and $$- {\bf{v}}$$.3. u and v as in Exercise 1

The graph of the vectors is shown below:

See the step by step solution

Step 1: Write vectors u, v, $$- {\bf{v}}$$, and$$- 2{\bf{v}}$$

From Exercise 1, the vectors are $$u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$$, and $$v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]$$.

Vector $$- v$$ can be written as $$\left( { - 1} \right)v$$.

Obtain the scalar multiple of vector $$v$$ by scalar $$\left( { - 1} \right)$$ to compute vector $$- v$$.

\begin{aligned}{c}\left( { - 1} \right)v &= \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{\left( { - 1} \right)\left( { - 3} \right)}\\{\left( { - 1} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\end{aligned}

Thus, $$- v = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]$$.

Vector $$- 2v$$ can be written as $$\left( { - 2} \right)v$$.

Obtain the scalar multiple of vector $$v$$ by scalar $$\left( { - 2} \right)$$ to compute vector $$- 2v$$.

\begin{aligned}{c}\left( { - 2} \right)v &= \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\end{aligned}

Thus, $$- 2v = \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]$$.

Step 2: Write vectors u + v , and u - v

Obtain the vector $$u + v$$ by using vectors $$u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$$, and $$v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]$$.

\begin{aligned}{c}u + v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + \left( { - 3} \right)}\\{2 + \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 - 3}\\{2 - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\end{aligned}

Thus, the vector is $$u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]$$.

Obtain the vector $$u - v$$ by using vectors $$u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$$, and $$v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]$$.

\begin{aligned}{c}u - v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 - \left( { - 3} \right)}\\{2 - \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + 3}\\{2 + 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\end{aligned}

Thus, the vector is $$u - v = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]$$.

Step 3: Compute vector $$u - 2v$$

Vector $$u - 2v$$ can be written as $$u + \left( { - 2} \right)v$$.

Obtain the scalar multiple of vector $$v$$ by scalar $$\left( { - 2} \right)$$, and then add the resultant vector with vector $$u$$.

\begin{aligned}{c}u + \left( { - 2} \right)v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + 6}\\{2 + 2}\end{array}} \right]\end{aligned}

Solve further to get:

$$u + \left( { - 2} \right)v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]$$

Thus, $$u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]$$.

Step 4: Display the vectors on a graph

The graph of vectors $$u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$$, $$v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]$$, $$- v = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]$$, $$- 2v = \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]$$, $$u - v = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]$$, $$u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]$$, and $$u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]$$ using arrows are shown below:

Thus, the graph is obtained.