Linear Algebra - David Lay
1.1
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.
1.
x1 + 5 x2 = 7 −2 x1 − 7 x2 = −5
1 −2
5 −7
7 −5 x1 + 5 x2 = 7
Replace R2 by R2 + (2)R1 and obtain:
3x2 = 9 x1 + 5 x2 = 7 x2 = 3 x1
1 0 1 0 1 0
5 3 5 1 0 1
7 9 7 3 −8 3
Scale R2 by 1/3: Replace R1 by R1 + (–5)R2: The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
= −8 x2 = 3
2 x1 + 4 x2 = −4 5 x1 + 7 x2 = 11
2 5
4 7
−4 11 x1 + 2 x2 = −2
Scale R1 by 1/2 and obtain: Replace R2 by R2 + (–5)R1: Scale R2 by –1/3: Replace R1 by R1 + (–2)R2: The solution is (x1, x2) = (12, –7), or simply (12, –7).
5 x1 + 7 x2 = 11 x1 + 2 x2 = −2
1 5 1 0 1 0 1 0
2 7 2 −3 2 1 0 1
−2 11 −2 21 −2 −7 12 −7
−3x2 = 21 x1 + 2 x2 = −2 x2 = −7 x1
= 12 x2 = −7
1
2
CHAPTER 1
•
Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations: x1 + 5 x2 = 7 x1 − 2 x2 = −2
1 1
5 −2
7 −2 x1 + 5 x2 = 7
Replace R2 by R2 + (–1)R1 and obtain: Scale R2 by –1/7: Replace R1 by R1 + (–5)R2: The point of intersection is (x1, x2) = (4/7, 9/7).
−7 x2 = −9 x1 + 5 x2 = 7 x2 = 9/7 x1
1 0 1 0 1 0
5 −7 5 1 0 1
7 −9 7 9/7 4/7 9/7
= 4/7 x2 = 9/7
4. The point of intersection satisfies the system of two linear equations: x1 − 5 x2 = 1
3x1 − 7 x2 = 5
1 3
−5 −7
1 5 x1 − 5 x2 = 1
Replace R2 by R2 + (–3)R1 and obtain: Scale R2 by 1/8: Replace R1 by R1 + (5)R2: The point of intersection is (x1, x2) = (9/4, 1/4).
8 x2 = 2 x1 − 5 x2 = 1 x2 = 1/4 x1
1 0 1 0 1 0
−5 8
1 2
−5 1 1 1/4 0 1 9/4 1/4
= 9/4 x2 = 1/4
5. The system is already in
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.
1.
x1 + 5 x2 = 7 −2 x1 − 7 x2 = −5
1 −2
5 −7
7 −5 x1 + 5 x2 = 7
Replace R2 by R2 + (2)R1 and obtain:
3x2 = 9 x1 + 5 x2 = 7 x2 = 3 x1
1 0 1 0 1 0
5 3 5 1 0 1
7 9 7 3 −8 3
Scale R2 by 1/3: Replace R1 by R1 + (–5)R2: The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
= −8 x2 = 3
2 x1 + 4 x2 = −4 5 x1 + 7 x2 = 11
2 5
4 7
−4 11 x1 + 2 x2 = −2
Scale R1 by 1/2 and obtain: Replace R2 by R2 + (–5)R1: Scale R2 by –1/3: Replace R1 by R1 + (–2)R2: The solution is (x1, x2) = (12, –7), or simply (12, –7).
5 x1 + 7 x2 = 11 x1 + 2 x2 = −2
1 5 1 0 1 0 1 0
2 7 2 −3 2 1 0 1
−2 11 −2 21 −2 −7 12 −7
−3x2 = 21 x1 + 2 x2 = −2 x2 = −7 x1
= 12 x2 = −7
1
2
CHAPTER 1
•
Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations: x1 + 5 x2 = 7 x1 − 2 x2 = −2
1 1
5 −2
7 −2 x1 + 5 x2 = 7
Replace R2 by R2 + (–1)R1 and obtain: Scale R2 by –1/7: Replace R1 by R1 + (–5)R2: The point of intersection is (x1, x2) = (4/7, 9/7).
−7 x2 = −9 x1 + 5 x2 = 7 x2 = 9/7 x1
1 0 1 0 1 0
5 −7 5 1 0 1
7 −9 7 9/7 4/7 9/7
= 4/7 x2 = 9/7
4. The point of intersection satisfies the system of two linear equations: x1 − 5 x2 = 1
3x1 − 7 x2 = 5
1 3
−5 −7
1 5 x1 − 5 x2 = 1
Replace R2 by R2 + (–3)R1 and obtain: Scale R2 by 1/8: Replace R1 by R1 + (5)R2: The point of intersection is (x1, x2) = (9/4, 1/4).
8 x2 = 2 x1 − 5 x2 = 1 x2 = 1/4 x1
1 0 1 0 1 0
−5 8
1 2
−5 1 1 1/4 0 1 9/4 1/4
= 9/4 x2 = 1/4
5. The system is already in