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Chapter 6: Q8E (page 331)

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

Short Answer

Expert verified

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

Step by step solution

01

The length of a vector

The nonnegative scalar \(\left\| {\mathop{\rm v}\nolimits} \right\|\) is defined aslength (ornorm) 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 \({\left\| {\mathop{\rm v}\nolimits} \right\|^2} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} \)

02

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 \(\left\| {\mathop{\rm x}\nolimits} \right\| = 7\).

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Most popular questions from this chapter

Compute the quantities in Exercises 1-8 using the vectors

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

7. \(\left\| {\mathop{\rm w}\nolimits} \right\|\)

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

3. \(\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right)\)

Given \(A = QR\) as in Theorem 12, describe how to find an orthogonal\(m \times m\)(square) matrix \({Q_1}\) and an invertible \(n \times n\) upper triangular matrix \(R\) such that

\(A = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\)

The MATLAB qr command supplies this 鈥渇ull鈥� QR factorization

when rank \(A = n\).

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

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)
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