Linear algebra with applications 5th edition solutions

See our solution for Question 3E from Chapter 1.9 from Lay's Linear Algebra and Its Applications, 5th Edition.

Step 1
Let the Vectors are:\[{{\bf{e}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right];\,\,\,{{\bf{e}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\]Given Transformation (see Figure 1 in the section 9.1 of textbook):\[T:{{\rm{R}}^2} \to {{\rm{R}}^2}::T\left( {{{\bf{e}}_1}} \right) = \left[ {\begin{array}{*{20}{c}}{\cos \phi }\\{\sin \phi }\end{array}} \right],\,\,:T\left( {{{\bf{e}}_2}} \right) = \left[ {\begin{array}{*{20}{c}}{ - \sin \phi }\\{\cos \phi }\end{array}} \right]\,\]We have to find the transformation matrix for the above transformation.

Step 2: The Rotation Angle
\[\begin{array}{l}\phi = \dfrac{{3\pi }}{2}\\\\\phi = \dfrac{{3\pi }}{2}{\kern 1pt} \dfrac{{180^\circ }}{\pi }\\\\\phi = 270^\circ \end{array}\]

Step 3: The Transformation Matrix
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{\cos \phi }&{ - \sin \phi }\\{\sin \phi }&{\cos \phi }\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{\cos 270^\circ }&{ - \sin 270^\circ }\\{\sin 270^\circ }&{\cos 270^\circ }\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\end{array}\]

ANSWER

\[A = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\]

Book Details

Offering the most geometric presentation available, Linear Algebra with Applications, Fifth Edition emphasizes linear transformations as a unifying theme. This elegant textbook combines a user-friendly presentation with straightforward, lucid language to clarify and organize the techniques and applications of linear algebra. Exercises and examples make up the heart of the text, with abstract exposition kept to a minimum. Exercise sets are broad and varied and reflect the author's creativity and passion for this course. This revision reflects careful review and appropriate edits throughout, while preserving the order of topics of the previous edition.

TRUE OR FALSE? 19 Determine whether the statements that follow are true or false, and justify your...The matrix [5665] represents a rotation combinedwith a scaling.If v1,v2,...,vn and w1,w2,...,wm are any twobases of a subspace Vof 10 , then n must equal m.The polynomials of degree less than 7 form a seven dimensional subspace of the linear space of all...If T is a linear transformation from n to n suchthat T(e1),T(e2),...,T(en) are all unit vectors,...Given information: An statement is given as - If B is obtained be multiplying a column of A by 9 ,...Given information : If 0 is an Eigen value of a matrix A . Since determinant of a matrix is the...Given: The singular values of any diagonal matrix D are the absolute values of the diagonal entries...

Copyright © 2016 Pearson Education, Inc. 1-1

1.1 SOLUTIONS

Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand

for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.

1. 12

12

57

275

xx

xx





157

275









Replace R2 by R2 + (2)R1 and obtain: 12

2

57

39

xx

x



157

039







Scale R2 by 1/3: 12

2

57

3

xx

x



157

013







Replace R1 by R1 + (–5)R2: 1

2

8

3

x

x



108

013









The solution is (x1, x2) = (–8, 3), or simply (–8, 3).

2. 12

12

244

5711

xx

xx





244

5711









Scale R1 by 1/2 and obtain: 12

12

22

5711

xx

xx





122

5711









Replace R2 by R2 + (–5)R1: 12

2

22

321

xx

x



 122

0321











Scale R2 by –1/3: 12

2

22

7

xx

x



 122

017











Replace R1 by R1 + (–2)R2: 1

2

12

7

x

x



 1012

017









The solution is (x1, x2) = (12, –7), or simply (12, –7).