Elementary Linear Algebra - Solutions to Problems (Matthews).pdf

文档介绍：SOLUTIONS TO PROBLEMS
ELEMENTARY
LINEAR ALGEBRA
K. R. MATTHEWS
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF QUEENSLAND
First Printing, 1991
CONTENTS
PROBLEMS ............................................ 1
PROBLEMS ............................................ 12
PROBLEMS ............................................ 18
PROBLEMS ............................................ 32
PROBLEMS ............................................ 45
PROBLEMS ............................................ 58
PROBLEMS ............................................ 69
PROBLEMS ............................................ 83
PROBLEMS ............................................ 91
i
SECTION 1:6
0 0 0 2 4 0 1 2 0
2. (i) R R R 1 R ;
2 4 0 1 ↔ 2 0 0 0 1 → 2 1 0 0 0
· ¸ · ¸ · ¸
0 1 3 1 2 4 1 0 2
(ii) R1 R2 R1 R1 2R2 −;
1 2 4 ↔ 0 1 3 →− 0 1 3
· ¸ · ¸ · ¸
1 1 1 1 1 0
R R R
(iii) 1 1 0 2 → 2 − 1 0 0 1
 R3 R3 R1 −
1 0 0 →− 0 1 1
−−

R1 R1 + R3 1 0 0 1 0 0
→ R2 R2 + R3
R3 R3 0 1 1 → 0 1 0 ;
→− R3 R3 
R2 R3 0 0 1 →− 0 0 1
↔−
2 0 0  1 0 0 
R3 R3 + 2R1
(iv) 0 0 0 → 1 0 0 0 .
 R1 R1 
4 0 0 → 2 0 0 0
−
 1 1 1 2  1 1 1 2
R R 2R
3. (a) 2 3 1 8 2 → 2 − 1 0 1 3 4
− R3 R3 R1 −
1 1 1 8 →− 0 2 2 10
−−−−−−
 1 0  4 2  1 0 4 2 
R1 R1 R2 − 1
→− 0 1 3 4 R3 −8 R3 0 1 3 4
R3 R3 + 2R2 −→−
→ 0 0 8 2 0 0 1 1
−− 4
 1 0 0 3 
R1 R1 4R3 −19
→− 0 1 0 4 .
R2 R2 + 3R3  1 
→ 0 0 1 4
The augmentedmatrix has been converted to reduced row–echelon form
and we read off the unique solution x = 3; y = 19 ; z = 1 .
− 4 4
1 1 1 2 10 1 1 1 2 10
− R2 R2 3R1 −
(b) 3 1 7 4 1 →− 0 4 10 2 29
− R3 R3 + 5R1 −−−
5 3 15 6 9 → 0 8 20 4 59
−−−−
 1 1 1  2 10 
−
R R + 2R 0 4 10 2 29 .
3 3 2 
→ 0 −0 0 −0 − 1
From the lastmatrix we see that the original system is inconsistent.
1
3 1 7 0 1 1 1 1
− 1 − 1
2 1 4 2 2 1 4 2
(c) − R1 R3