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Instructors’ Solutions Manual for Applied Linear Algebra 2nd Edition By Peter J.Olver and Chehrzad Shakiban (All Chapters) Complete Guide A+
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Instructors’ Solutions Manual for Applied Linear Algebra 2nd Edition By Peter J.Olver and Chehrzad Shakiban (All Chapters) Complete Guide A+

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  • October 29, 2024
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  • solution
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  • applied
  • linear
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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


Current version (v. 2) posted July, 2019
v. 1 posted August, 2018




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,Table 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




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, Instructors’ 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) Reduce the system to 6 u + v = 5, − 52 v = 52 ; then use Back Substitution to solve
for u = 1, v = −1.
⋆ (c) Reduce the system to p + q − r = 0, −3 q + 5 r = 3, − r = 6; then solve for
p = 5, q = −11, r = −6.
(d) Reduce the system to 2 u − v + 2 w = 2, − 32 v + 4 w = 2, − w = 0; then solve for
u = 31 , v = − 43 , w = 0.
1 2 2
⋆ (e) Reduce the system to 5 x1 + 3 x2 − x3 = 9, 5 x2 − 5 x3 = 5, 2 x3 = −2; then solve for
x1 = 4, x2 = −4, x3 = −1.
(f ) Reduce the system to x + z − 2 w = − 3, − y + 3 w = 1, − 4 z − 16 w = − 4, 6 w = 6; then
solve for x = 2, y = 2, z = −3, w = 1.

⋆ 1.1.2. Plugging in the values of x, y and z gives a + 2 b − c = 3, a − 2 − c = 1, 1 + 2 b + c = 2.
Solving this system yields a = 4, b = 0, and c = 1.

♥ 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down.
Thus 2 x = −6 so x = −3. Plugging this into the second equation gives 12 + 3y = 3, and so
y = −3. Plugging the values of x and y in the third equation yields −3 + 4(−3) − z = 7, and
so z = −22.
⋆ (c) Start with the last equation and, assuming the coefficient of the last variable is 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) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f ). Solv-
ing the reduced system by Forward Substitution reproduces the same solution (as it must):
(a) The system reduces to 23 x = 17 15
2 , x + 2 y = 3. (b) The reduced system is 2 u = 2 ,
15

3 u − 2 v = 5. (d) Reduce the system to 32 u = 21 , 72 u − v = 52 , 3 u − 2 w = −1. (f ) Doesn’t
work since, after the first reduction, z doesn’t occur in the next to last equation.


 
0
 
1.2.1. (a) 3 × 4, (b) 7, (c) 6, (d) ( −2 0 1 2 ), (e)  2
 .
−6
     
1 2 3 ! 1 2 3 4 1

4
 1 2 3    
1.2.2. Examples: (a)  5 6, ⋆ (b) , (c) 4
 5 6 7, (e)  2 .
 
1 4 5
7 8 9 7 8 9 3 3
! ! !
6 1 u 5
1.2.4. (b) A = , x= , b= ;
3 −2 v 5




1 Solutions Manual




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, 2 Chapter 1: Instructors’ Solutions Manual
     
1 1 −1 p 0
     
⋆ (c) A=  2 −1 3 , x =  q , b =  3 
   
 ;
−1 −1 0 r 6
     
2 −1 2 u 2
     
(d) A = 
 −1 −1 3, x =  v , b =  1 ;
   
3 0 −2 w 1
     
5 3 −1 x 9
   1  
⋆ (e) A= 3 2 −1   x ,  5
 , x=  2 b=  ;
1 1 2 x3 −1
     
1 0 1 −2 x −3
     
2 −1 2 −1   y  −5 
(f ) A = 

, x =  , b = 
   
.

0 −6 −4 2  z  2
1 3 2 −1 w 1

1.2.5. (b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1.
(c) 3 x1 − x3 = 1, −2 x1 − x2 = 0, x1 + x2 − 3 x3 = 1.
The solution is x1 = 15 , x2 = − 25 , x3 = − 25 .
⋆ (d) x + y − z − w = 0, −x + z + 2 w = 4, x − y + z = 1, 2 y − z + w = 5.
The solution is x = 2, y = 1, z = 0, w = 3.



   
1 0 0 0 0 0 0 0 0 0
   
0 1 0 0 0 0 0 0 0 0
   

1.2.6. (a) I = 0

0 1 0 0,

O= 
0

0 0 0 0.


0 0 0 1 0

0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
(b) I + O = I , I O = O I = O. No, it does not.
!
3 6 0
1.2.7. (b) undefined, (c)
−1 4 2
, ⋆ (e) undefined,
   
1 11 9 9 −2 14
   
(f ) 3 −12 −12  ⋆ (h)  −8 −17 
 ,  6 .
7 8 8 12 −3 28

1.2.9. 1, 6, 11, 16.
 
  2 0 0 0
1 0 0  
  0 −2 0 0

1.2.10. (a)  0 0 0 ⋆ (b)  .
, 
0 0 3 0
0 0 −1
0 0 0 −3

1.2.11. (a) True, ⋆ (b) true.
! ! !
x y ax by ax ay
⋆ ♥ 1.2.12. (a) Let A =
z w
. Then A D =
az bw
=
bz bw
= D A, so if a 6= b these
!
a 0
are equal if and only if y = z = 0. (b) Every 2 × 2 matrix commutes with = a I.
0 a




Solutions Manual




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