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Q8E

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Linear Algebra and its Applications
Found in: Page 331
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

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Short Answer

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

The length of x is \(\left\| {\mathop{\rm x}\nolimits} \right\| = 7\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: The length of a vector

The nonnegative scalar \(\left\| {\mathop{\rm v}\nolimits} \right\|\) is defined as length (or norm) as shown below:

\(\left\| {\mathop{\rm v}\nolimits} \right\| = \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } = \sqrt {v_1^2 + v_2^2 + \cdots + v_n^2} ,\) and \({\left\| {\mathop{\rm v}\nolimits} \right\|^2} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} \)

Step 2: Compute

\(\left\| {\mathop{\rm x}\nolimits} \right\|\)

It is given that \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\).

Compute the length of \({\mathop{\rm x}\nolimits} \) as shown below:

\(\begin{aligned}{c}\left\| {\mathop{\rm x}\nolimits} \right\| &= \sqrt {{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} } \\ &= \sqrt {{6^2} + {{\left( { - 2} \right)}^2} + {3^2}} \\ &= \sqrt {36 + 4 + 9} \\ &= \sqrt {49} \\ &= 7\end{aligned}\)

Thus, the length of x is \(\left\| {\mathop{\rm x}\nolimits} \right\| = 7\).

Most popular questions for Math Textbooks

Find a \(QR\) factorization of the matrix in Exercise 11.

Let \(X\) be the design matrix used to find the least square line of fit data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Use a theorem in Section 6.5 to show that the normal equations have a unique solution if and only if the data include at least two data points with different \(x\)-coordinates.

In Exercises 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]

In Exercises 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

11. \[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

In Exercises 9-12, find a unit vector in the direction of the given vector.

9. \(\left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\)
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