applied linear algebra instructor’s solutions manual by peter j olver and chehrzad shakiban
applied linear algebra instructor’s solutions manual by peter j olver and chehrzad shakiban
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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+
Instructors’ Solutions Manual for Applied Linear Algebra by Peter J. Olver and Chehrzad Shakiban 2nd Edition A+
Exam (elaborations) TEST BANK FOR Applied Linear Algebra By Peter J. Olver and Chehrzad Shakiban (Instructor's Solution Manual)
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Applied Linear Algebra
Instructor’s Solutions Manual
by Peter J. Olver and Chehrzad Shakiban
Table of Contents
Chapter Page
1. Linear Algebraic Systems ............................................................ 1
2. Vector Spaces and Bases ............................................................. 46
3. Inner Products and Norms .......................................................... 78
4. Minimization and Least Squares Approximation .................. 114
5. Orthogonality .............................................................................. 131
6. Equilibrium ............................................................................... 174
7. Linearity ................................................................................... 193
8. Eigenvalues ................................................................................. 226
9. Linear Dynamical Systems ...................................................... 262
10. Iteration of Linear Systems ....................................................... 306
11. Boundary Value Problems in One Dimension.......................... 346
1
, Solutions — Chapter 1
1.1.1.
(a) Reduce the system to x − y = 7, 3 y = −4; then use Back Substitution to solve for
x = 173 , y = − 43.
(b) Reduce the system to 6 u + v = 5, − 5 v = 5 ; then use Back Substitution to solve for
2 2
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.
3
(d) Reduce the system to 2 u − v + 2 w = 2, − 2 v + 4 w = 2, − w = 0; then solve for
4
u = 13, v = − 3, w = 0.
(e) Reduce the system to 5 x1 + 3 x2 − x3 = 9, 1 x2 − 2 x3 = 2 , 2 x3 = −2; then solve for
5 5 5
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.(g
8
) Reduce the system to 3 x1 + x2 = 1, x + x3 = 23 , 21
3 2 8 3
x + x4 = 34 , 55 x = 5 ; then
21 4 7
3 2 2 3
solve for x1 = , x2 = , x3 = , x4 = .
11 11 11 11
1.1.2. Plugging in the given 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.
(b) We will get a diagonal system with the same solution.
(c) Start with the last equation and, assuming the coefficient of the last variable is /= 0, use
the operation to eliminate the last variable in all the preceding equations. Then, again
assuming 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 ).
Solving the reduced system by Forward Substitution reproduces the same solution (as
it must):
(a) The system reduces to 3 x = 17 , x + 2 y = 3.
2 2
(b) The reduced system is 152 u = 152, 3 u − 2 v = 5.
(c) The method doesn’t work since r doesn’t appear in the last equation.
(d) Reduce the system to 23 u = 12, 7 2u − v = 5 , 23 u − 2 w = −1.
(e) Reduce the system to3 2 x1 =3 8 , 4 x1 + 3 x2 = 4, x1 + x2 + x3 = −1.
(f )Doesn’t work since, after the first reduction, z doesn’t occur in the next to last
equation.
(g ) Reduce the system to 55 x1 = 5 , x2 + 21 x3 = 3 , x3 + 8 x4 = 2 , x3 + 3 x4 = 1.
21 7 8 4 3 3
, 1 1
1
0
2 3
1
1 2 3! 2 3 4
, (c) B4 7C
@
B A @ 5 6 A, (d) ( 1 2 3 4 ),
1.2.2. (a) 4 5 6C , (b) 1 4 5 8 9 3
7 8 9 7
0 1
1
B C
(e) @ 2 A, (f ) ( 1 ).
3
1.2.3. x = − 1 , y = 4
, z = −1, w = 2
.
3 3 3 3
1.2.4.
! ! !
−1 1 x 7
(a) A = , x= , b= ;
2! 1 y! 3!
1 6 u 5
(b) A = , x= , b= ;
3 −2 1
v 5
0 0 1 0 1
1 1 −1 p 0
B C B C B C
(c) A = @ 2 −1 3 A, x = @ q A, b= @ 3 A;
0 −1 −1 0 r 6
1 0 1 0 1
2 1 2 u 3
B C B C B C
(d) A = @ −1 3 3 A, x = @ v A, b = @ −2 A;
0 4 −3 0 w 7
1 0 1 0 1
5 3 −1 x1 9
B C
B C B C
(e) A = @3
2 −1 A, x = @ x2 A, b = @ 5 A;
0
1 1 2 1
x3
0 1 −1 0 1
1 0 1 −2 x −3
2 −1 2 −1 B yC 3C
(f ) A = B C, x = , b=B ;
0 −6 −4 2
@ A BwC 1C
@ zA 2 A
0 B
0
1 3 2 1
−1 x 1 @
1 0 1
3 1 0 0 1
B 2C B 1C
(g ) A = B 1
0 3
1 1
3 0
1 C, x = B x C , b = B
@ x3 A
C
@ 1 A.
@ A
0 0 1 3
x4 1
1.2.5.
(a) x − y = −1, 2 x + 3 y = −3. The solution is x = − 6 , y 5= − 1 . 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 = − 2 ,5 x3 = − 2 . 5
(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.
1.2.6. 0 1
01 0 0 01 0 0 0 0
B C
B0 1 0 0C
0 0 0 0 .
(a) I = B C, O=@ B
0 0 0C
@0 0 1 0A 0 A
0 0 0 1 0 0 0 0
(b) I + O = I , I O = O I = O. No, it does not. !
3 6 0
1.2.7. (a) undefined, (b) undefined, (c) , (d) undefined, (e) undefined,
0 1
−1 4 2 1
0
−
2
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