ISBN 9783319910406 - Manual for Applied Linear Algebra by Peter J. O...

Instructors’ Solutions Manual for Applied Linear Algebra by Peter J. Olver and Chehrzad Shakiban Second Edition Undergraduate Texts in Mathematics

ISBN 978–3–319–91040–6

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Stuvia.com - The Marketplace to Buy and Sell your Study MaterialTable of Contents Chapter 1. Linear Algebraic Systems . . . . . . . . . . . . . . . . . 1 Chapter 2. Vector Spaces and Bases . . . . . . . . . . . . . . . . 22 Chapter 3. Inner Products and Norms . . . . . . . . . . . . . . . 40 Chapter 4. Orthogonality . . . . . . . . . . . . . . . . . . . . . 59 Chapter 5. Minimization and Least Squares . . . . . . . . . . . . . 77 Chapter 6. Equilibrium . . . . . . . . . . . . . . . . . . . . . 94 Chapter 7. Linearity . . . . . . . . . . . . . . . . . . . . . . . 105 Chapter 8. Eigenvalues and Singular Values . . . . . . . . . . . . . 124 Chapter 9. Iteration . . . . . . . . . . . . . . . . . . . . . . . 150 Chapter 10. Dynamics . . . . . . . . . . . . . . . . . . . . . . 176 Stuvia.com - The Marketplace to Buy and Sell your Study Material

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Stuvia.com - The Marketplace to Buy and Sell your Study MaterialInstructors’ Solutions Manual for

Chapter 1: Linear Algebraic Systems

Note:Solutions marked with a⋆do not appear in the Students’ Solutions Manual.

1.1.1. (b) u+v= 5,− 5 2 v= 5 2 ; then use Back Substitution to solve foru= 1, v=−1.⋆(c) p+q−r= 0,−3q+ 5r= 3,−r= 6; then solve for p= 5, q=−11, r=−6.(d) Reduce the system to 2u−v+ 2w= 2,− 3 2 v+ 4w= 2,−w= 0; then solve for u= 1 3 , v=− 4 3 , w= 0.⋆(e) x 1

• 3x

2 −x 3 = 9, 1 5 x 2 − 2 5 x 3 = 2 5 ,2x 3 =−2 ; then solve for x 1 = 4, x 2 =−4, x 3

=−1.

(f) Reduce the system tox+z−2w=−3,−y+ 3w= 1,−4z−16w=−4,6w= 6; then solve forx= 2, y= 2, z=−3, w= 1.⋆1.1.2. Plugging in the values ofx , yandzgivesa+ 2b−c= 3, a−2−c= 1,1 + 2b+c= 2.Solving this system yieldsa= 4, b= 0, andc= 1.♥1.1.3. (a) With Forward Substitution, we just start with the top equation and work down.Thus 2x=−6 sox=−3. Plugging this into the second equation gives 12 + 3y= 3, and so y=−3. Plugging the values ofxandyin the third equation yields−3 + 4(−3)−z= 7, and soz=−22.⋆(c) 6= 0, use the operation to eliminate the last variable in all the preceding equations. Then, again as- suming the coefficient of the next-to-last variable is non-zero, eliminate it from all but the last two equations, and so on.⋆(d)

• and (f). Solv-

ing the reduced system by Forward Substitution reproduces the same solution (as it must): (a) The system reduces to 3 2 x= 1 7 2 , x+ 2y= 3.(b) The reduced system is 15 2 u= 1 5 2 , 3u−2v=

• (d) Reduce the system to

3 2 u= 1 2 , 7 2 u−v= 5 2 ,3u−2w=−1 . (f) Doesn’t work since, after the first reduction,zdoesn’t occur in the next to last equation.

1.2.1. (a) ×4, (b) 7, (c) 6, (d) (−2 0 1 2 ), (e)    

2 −6     .

1.2.2. Examples: (a)

   

• 2 3
• 5 6
• 8 9

    ,⋆(b)

• 2 3
• 4 5

!,(c)    

1 2 3 4

4 5 6 7

7 8 9 3

    ,(e)     1 2 3     .

1.2.4. (b)A=

• 1

3−2 !,x=

u v !,b=

5 5 !; 1 Solutions Manual Stuvia.com - The Marketplace to Buy and Sell your Study Material

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Stuvia.com - The Marketplace to Buy and Sell your Study Material2Chapter 1: Instructors’ Solutions Manual ⋆(c)A=     1 1−1

2−1 3

−1−1 0

    ,x=     p q r     ,b=    

3 6     ; (d)A=    

2−1 2

−1−1 3

3 0 −2

    ,x=     u v w     ,b=     2 1 1     ; ⋆(e)A=     5 3−1

3 2−1

• 1 2

    ,x=     x 1 x 2 x 3     ,b=     9 5 −1     ; (f)A=      

1 0 1 −2

2−1 2 −1

0−6−4 2

1 3 2 −1

      ,x=       x y z w       ,b=       −3 −5 2 1       .

1.2.5. (b)u+w=−1, u+v=−1, v+w= 2.The solution isu=−2, v= 1, w= 1.(c) 3x 1 −x 3 = 1,−2x 1 −x 2 = 0, x 1 +x 2 −3x 3 = 1.The solution isx 1 = 1 5 , x 2 =− 2 5 , x 3 =− 2 5 .⋆(d)x+y−z−w= 0,−x+z+ 2w= 4, x−y+z= 1,2y−z+w= 5.The solution isx= 2, y= 1, z= 0, w= 3.

1.2.6. (a) I =         

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

         ,O =         

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

         .(b) I + O = I,I O = O I = O. No, it does not.

1.2.7. (b) undefined, (c)

• 6 0

−1 4 2

!,⋆(e)

(f)    

1 11 9

3−12−12

• 8 8

    ,⋆(h)     9−2 14

−8 6 −17

12−3 28

    .

1.2.9. 1,6,11,16.

1.2.10. (a)    

• 0 0
• 0 0

0 0−1

    ,⋆(b)       2

• 0 0

0−2 0 0

0 0 3 0

0 0 0 −3

      .

1.2.11. (a) True,⋆(b)

⋆♥1.2.12. (a) LetA=

x y z w !. ThenAD=

a x b y a z b w !=

a x a y b z b w !=D A, so ifa6 =bthese are equal if and only ify=z= 0. (b) Every 2×2 matrix commutes with

a0 0a !=aI .

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