Linear Algebra and its Applications [Solutions Manual] 3rd ed - D. Lay WW.pdf

文档介绍：SOLUTIONS
Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand for
row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
xx+=57 157
1. 12
−−=−−−−
27xx12 5 275
xx+=57 157
Replace R2 by R2 + (2)R1 and obtain: 12
= 
39x2 039
xx+=57 157
Scale R2 by 1/3: 12
= 
x2 3 013
x =−8 10− 8
Replace R1 by R1 + (–5)R2: 1
= 
x2 3 01 3
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
24xx+=− 4 24− 4
2. 12
+= 
57xx12 11 5711
xx+=−2212− 2
Scale R1 by 1/2 and obtain: 12
+= 
57xx12 11 5711
xx+=−22 122−
Replace R2 by R2 + (–5)R1: 12
−= −
321x2 0321
xx+=−2212− 2
Scale R2 by –1/3: 12
=−−
x2 7 01 7
x = 12 10 12
Replace R1 by R1 + (–2)R2: 1
=−−
x2 7 01 7
The solution is (x1, x2) = (12, –7), or simply (12, –7).
1
2 CHAPTER 1 • Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations:
xx+=57 157
12
−=−−−
xx1222122
xx+=57157
Replace R2 by R2 + (–1)R1 and obtain: 12
−=−−−
79x2 079
xx+=57 15 7
Scale R2 by –1/7: 12
= 
x2 9/7 019/7
x = 4/7 104/7
Replace R1 by R1 + (–5)R2: 1
= 
x2 9/7 019/7
The point of intersection is (x1, x2) = (4/7, 9/7).
4. The point of intersection satisfies the system of two linear equations:
xx−=51 151−
12
−= −
37xx12 5 375
xx−=51 151−
Replace R2 by R2 + (–3)R1 and obtain: 12
= 
82x2 082
xx−=51 15− 1
Scale R2 by 1/8: 12
= 
x2 1/4 011/4
x = 9/4 109/4
Replace R1 by R1 + (5)R2: 1
= 
x2 1/4 011/4
The point of intersection is (x1, x2) = (9/4, 1/4).
5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do not
contain the variable x4. The next two steps should be to use the variable x3 in the th