SOLUTIONS MANUAL
,
, Solutions Manual for Applied Linear
Algebra 2ndEd Undergraduate Texts in
Mathematics
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
, 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, — 5 v = 5 ; then use Back Substitution to solve
for u = 1, v = —1. 2 2
⋆
(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, 2— 3 v + 4 w = 2, — w = 0; then solve for
u = 1 , v = — 4 , w = 0.
3 3
⋆ (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.
⋆ 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 /=
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 3 x = 17 , x + 2 y = 3. (b) The reduced system is 15 u = 15 ,
2 2 2 2
3 u — 2 v = 5. (d) Reduce the system2 to 3 u2 = 21 , 7 u — v2 = 5 , 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
. 1 2 3 . .
1.2.2. Examples: .4 5 6 ,., ⋆ (b) 1 4 5
, (c) .4 5 6 .,
7, (e) .2 .,.
(a) 7 8 9 3
7 8 9 3
! ! !
6 1 u 5
1.2.4. (b) A = , x= , b= ;
3 —2 v 5
1 ⃝c 2019 Peter J. Olver and Chehrzad Shakiban
,
, Solutions Manual for Applied Linear
Algebra 2ndEd Undergraduate Texts in
Mathematics
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
, 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, — 5 v = 5 ; then use Back Substitution to solve
for u = 1, v = —1. 2 2
⋆
(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, 2— 3 v + 4 w = 2, — w = 0; then solve for
u = 1 , v = — 4 , w = 0.
3 3
⋆ (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.
⋆ 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 /=
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 3 x = 17 , x + 2 y = 3. (b) The reduced system is 15 u = 15 ,
2 2 2 2
3 u — 2 v = 5. (d) Reduce the system2 to 3 u2 = 21 , 7 u — v2 = 5 , 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
. 1 2 3 . .
1.2.2. Examples: .4 5 6 ,., ⋆ (b) 1 4 5
, (c) .4 5 6 .,
7, (e) .2 .,.
(a) 7 8 9 3
7 8 9 3
! ! !
6 1 u 5
1.2.4. (b) A = , x= , b= ;
3 —2 v 5
1 ⃝c 2019 Peter J. Olver and Chehrzad Shakiban
