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

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CHAPTER 1

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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 “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 third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3. 6. One more step will put the system in triangular form. Replace R4 by its sum with –3 times R3, which 4 0 −1 1 −6 0 2 −7 0 4 . After that, the next step is to scale the fourth row by –1/5. produces 0 0 1 2 −3 0 0 −5 15 0 7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0 x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.

1.1

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Solutions

3

8. The standard row operations are:

1 0 0

−4 1 0

9 7 2

0 1 0 ~ 0 0 0

−4 1 0

9 7 1

0 1 0 ~ 0 0 0

−4 1 0

0 0 1

0 1 0 ~ 0 0 0

0 1 0

0 0 1

0 0 0

The solution set contains one solution: (0, 0, 0).

9. The system has already been reduced to triangular form. Begin by scaling the fourth row by 1/2 and then replacing R3 by R3 + (3)R4:

1 0 0 0 1 0 ~ 0 0

−1 1 0 0 −1 1 0 0

0 −3 1 0 0 0 1 0

0 0 −3 2 0 0 0 1

−4 1 −7 0 ~ −1 0 4 0 −4 1 8 0 ~ 5 0 2 0

−1 1 0 0 0 1 0 0 0 0 1 0

0 −3 1 0 0 0 0 1

0 0 −3 1 4 8 5 2

−4 1 7 0 ~ −1 0 2 0

−1 1 0 0

0 −3 1 0

0 0 0 1

−4 −7 5 2

Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2:

The solution set contains one solution: (4, 8, 5, 2).

10. The system has already been reduced to triangular form. Use the...