Page 843 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 823

Section 7.9 LU Factorization
⎛ ⎞ ⎛ ⎞
2 4 −6 1 0 0
1. U = ⎝ 0 −14 25 ⎠ and L = ⎝ 4 1 0 ⎠
0 0 136/7 −2 −6/7 1
⎛ ⎞ ⎛ ⎞
−2 1 12 1 0 0
3. U = ⎝ 0 −5 13 ⎠ and L = ⎝ −1 1 0 ⎠
0 0 119/5 −1 −3/5 1
⎛ ⎞ ⎛ ⎞
1 4 2 −1 4 1 0 0 0
⎜0 −5 2 0 0 ⎟ ⎜ 1 1 0 0⎟
0 0 88/5 4 6 ⎠ and L = ⎝ −2 −14/5 1 0 ⎠
5. U = ⎝
0 0 0 195/22 −691/44 4 14/5 −63/88 1
7. First obtain
⎛ ⎞ ⎛ ⎞
4 4 2 1 0 0
U = ⎝ 0 −2 5/2 ⎠ and L = ⎝ 1/4 1 0 ⎠ .
0 0 21/4 1/4 −3/2 1
Solve
⎛ ⎞ ⎛ ⎞
1 1
LY = ⎝ 0 ⎠ to obtain Y = ⎝ −1/4 ⎠ .
1 3/8
Solve UX = Y to obtain
⎛ ⎞
0
X = ⎝ 3/14 ⎠ .
1/4
9.
⎛ ⎞ ⎛ ⎞
−1 1 1 6 1 0 0
U = ⎝ 0 3 2 16 ⎠ and L = ⎝ −2 1 0 ⎠ .
0 0 17/3 52/3 −1 −1/3 1
⎛ ⎞ ⎛ ⎞
⎛ ⎞ 1 0
2
⎠ .
⎜ 28/3 ⎟ ⎜−5/3⎟
26/3 ⎠ + ⎝ −1/3
Solve LY = B and then UX = Y to obtain Y = ⎝ 5 ⎠ and X = α ⎝
29/3
−17/6 2/3
11.
6 1 −1 3 1 0 0 0
⎛ ⎞ ⎛ ⎞
⎜0 4/3 5/3 3 ⎟ ⎜ 2/3 1 0 0⎟
0 0 13/4 13/4 ⎠ and L = ⎝ −2/3 5/4 1 0 ⎠ .
U = ⎝
0 0 0 5 1/3 −1 4/13 1
4 −263/130
⎛ ⎞ ⎛ ⎞
⎠ .
⎜ 28/3 ⎟ ⎜ 537/65 ⎟
−7 ⎠ and X = ⎝ −233/65
Solve LY = B and then UX = Y to obtain Y = ⎝
43/13 93/65
Section 7.10 Matrices and Linear Transformations
1. T is linear and T(1,0,0) = (3,1,0), T(0,1,0) = (0,−1,0), and T(0,0,1) = (0,0,2), so
⎛ ⎞
3 0 0
A T = ⎝ 1 −1 0 ⎠ .
0 0 2
T is onto and one-to-one because A T has rank 3 one-to-one. The null space of T has dimension
m − rankA T = 3 − 3 = 0.
3. Nonlinear (because of the 2xy term)

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 17:50 THM/NEIL Page-823 27410_25_Ans_p801-866