/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q23Q A Givens rotation is a linear tr... [FREE SOLUTION] | 91影视

91影视

A Givens rotation is a linear transformation from \({\mathbb{R}^{\bf{n}}}\) to \({\mathbb{R}^{\bf{n}}}\) used in computer programs to create a zero entry in a vector (usually a column of matrix). The standard matrix of a given rotations in \({\mathbb{R}^{\bf{2}}}\) has the form

\(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\), \({a^2} + {b^2} = 1\)

Find \(a\) and \(b\) such that \(\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right)\) is rotated into \(\left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\).

Short Answer

Expert verified

\(a = \frac{4}{5}\) and \(b = - \frac{3}{5}\)

Step by step solution

01

Find the equation for rotation

As the rotation matrix is \(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right)\) is rotated through \(\left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\),

\(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\).

The above equation implies that \(4a - 3b = 5\) and \(3a + 4b = 0\).

02

Form the augmented matrix for \(a\) and \(b\)

The augmented matrix for the equation of \(a\) and \(b\) is \(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\3&4&0\end{aligned}} \right)\).

03

Solve the augmented matrix using row operations

At row two, multiply row two by 4 and row one by 3 and subtract row one from row two, i.e., \({R_2} \to 4{R_2} - 3{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\{3 \times 4 - 4 \times 3}&{4 \times 4 - 3\left( { - 3} \right)}&{0 - 5 \times 3}\end{aligned}} \right)\)

After performing the row operations, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\0&{25}&{ - 15}\end{aligned}} \right)\).

Divide row two by 25, i.e., \({R_2} \to \frac{{{R_2}}}{{25}}\).

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 3}&5\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\)

04

Solve the augmented matrix using row operations

At row one, multiply row two by 3 and add it to row one, i.e., \({R_1} \to {R_1} + 3{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{4 + 0}&{ - 3 + 3 \times 1}&{5 - 3 \times \frac{3}{5}}\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}4&0&{\frac{{16}}{5}}\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\).

Divide row one by \(4\), i.e., \({R_1} \to \frac{{{R_1}}}{4}\).

\(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{4}{5}}\\0&1&{ - \frac{3}{5}}\end{aligned}} \right)\)

So, \(a = \frac{4}{5}\) and \(b = - \frac{3}{5}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercise 23 and 24, make each statement True or False. Justify each answer.

23.

a. Another notation for the vector \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}{ - 4}&3\end{array}} \right]\).

b. The points in the plane corresponding to \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{ - 5}\\2\end{array}} \right]\) lie on a line through the origin.

c. An example of a linear combination of vectors \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) is the vector \(\frac{1}{2}{{\mathop{\rm v}\nolimits} _1}\).

d. The solution set of the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}&b\end{array}} \right]\) is the same as the solution set of the equation\({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + {x_3}{a_3} = b\).

e. The set Span \(\left\{ {u,v} \right\}\) is always visualized as a plane through the origin.


Consider two vectors v1 andv2in R3 that are not parallel.

Which vectors inlocalid="1668167992227" 3are linear combinations ofv1andv2? Describe the set of these vectors geometrically. Include a sketch in your answer.

In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

16. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\)

In (a) and (b), suppose the vectors are linearly independent. What can you say about the numbers \(a,....,f\) ? Justify your answers. (Hint: Use a theorem for (b).)

  1. \(\left( {\begin{aligned}{*{20}{c}}a\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\d\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\end{aligned}} \right)\)
  2. \(\left( {\begin{aligned}{*{20}{c}}a\\1\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\1\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\\1\end{aligned}} \right)\)

Question: There exists a 2x2 matrix such thatA[12]=[34].

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.