TEST BANK FOR Linear Algebra with Applications 5th Edition By Otto Bretscher

Section 1.1
1.1.1

x + 2y = 1
2x + 3y = 1

−2 × 1st equation →

x + 2y = 1
−y = −1

÷(−1) →

x + 2y = 1
y = 1

−2 × 2nd equation
→

x = −1
y = 1

, so that (x, y) = (−1, 1).
1.1.2

4x + 3y = 2
7x + 5y = 3

÷4
→

x + 3
4y = 1
2
7x + 5y = 3

−7 × 1st equation →

x + 3
4y = 1
2
−1
4y = −1
2

×(−4) →

x + 3
4y = 1
2
y = 2

−3
4 × 2nd equation →

x = −1
y = 2

,
so that (x, y) = (−1, 2).
1.1.3

2x + 4y = 3
3x + 6y = 2

÷2
→

x + 2y = 3
2
3x + 6y = 2

−3 × 1st equation →

x + 2y = 3
2
0 = −5
2

.
So there is no solution.
1.1.4

2x + 4y = 2
3x + 6y = 3

÷2
→

x + 2y = 1
3x + 6y = 3

−3 × 1st equation →

x + 2y = 1
0 = 0

This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation
x + 2y = 1 gives us x = 1 − 2y = 1 − 2t. Therefore the general solution is (x, y) = (1 − 2t, t), where t is an
arbitrary real number.
1.1.5

2x + 3y = 0
4x + 5y = 0

÷2
→

x + 3
2y = 0
4x + 5y = 0

−4 × 1st equation →

x + 3
2y = 0
−y = 0

÷(−1) →

x + 3
2y = 0
y = 0

−3
2 × 2nd equation
→

x = 0
y = 0

,
so that (x, y) = (0, 0).
1.1.6


x + 2y + 3z = 8
x + 3y + 3z = 10
x + 2y + 4z = 9

 −I
−I →


x + 2y + 3z = 8
y = 2
z = 1

 −2(II)
→


x + 3z = 4
y = 2
z = 1

 −3(III)
→


x = 1
y = 2
z = 1

, so that (x, y, z) = (1, 2, 1).
1.1.7


x + 2y + 3z = 1
x + 3y + 4z = 3
x + 4y + 5z = 4

 −I
−I →


x + 2y + 3z = 1
y + z = 2
2y + 2z = 3

 −2(II)
−2(II) →


x + z = −3
y + z = 2
0 = −1


This system has no solution.
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Copyright c 2013 Pearson Education, Inc.
Chapter 1
1.1.8


x + 2y + 3z = 0
4x + 5y + 6z = 0
7x + 8y + 10z = 0

 −4(I)
−7(I) →


x + 2y + 3z = 0
−3y − 6z = 0
−6y − 11z = 0

 ÷(−3) →


x + 2y + 3z = 0
y + 2z = 0
−6y − 11z = 0

 −2(II)
+6(II) →


x − z = 0
y + 2z = 0
z = 0


+III
−2(III) →


x = 0
y = 0
z = 0

,
so that (x, y, z) = (0, 0, 0).
1.1.9


x + 2y + 3z = 1
3x + 2y + z = 1
7x + 2y − 3z = 1

 −3(I)
−7(I) →


x + 2y + 3z = 1
−4y − 8z = −2
−12y − 24z = −6

 ÷(−4) →


x + 2y + 3z = 1
y + 2z = 1
2
−12y − 24z = −6

 −2(II)
+12(II) →


x − z = 0
y + 2z = 1
2
0 = 0


This system has infinitely many solutions: if we choose z = t, an arbitrary real number, then we get x = z = t
and y = 1
2 − 2z = 1
2 − 2t. Therefore, the general solution is (x, y, z) =
􀀀
t, 1
2 − 2t, t

, where t is an arbitrary real
number.
1.1.10


x + 2y + 3z = 1
2x + 4y + 7z = 2
3x + 7y + 11z = 8

 −2(I)
−3(I) →


x + 2y + 3z = 1
z = 0
y + 2z = 5

 Swap :
II ↔ III →


x + 2y + 3z = 1
y + 2z = 5
z = 0

 −2(II)
→


x − z = −9
y + 2z = 5
z = 0


+III
−2(III) →


x = −9
y = 5
z = 0

,
so that (x, y, z) = (−9, 5, 0).
1.1.11

x − 2y = 2
3x + 5y = 17

−3(I) →

x − 2y = 2
11y = 11

÷11 →

x − 2y = 2
y = 1

+2(II)
→

x = 4
y = 1

,
so that (x, y) = (4, 1). See Figure 1.1.
Figure 1.1: for Problem 1.1.11.
1.1.12

x − 2y = 3
2x − 4y = 6

−2(I) →

x − 2y = 3
0 = 0

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Copyright c 2013 Pearson Education, Inc.
Section 1.1
This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation
x − 2y = 3 gives us x = 3 + 2y = 3 + 2t. Therefore the general solution is (x, y) = (3 + 2t, t), where t is an
arbitrary real number. (See Figure 1.2.)
Figure 1.2: for Problem 1.1.12.
1.1.13

x − 2y = 3
2x − 4y = 8

−2(I) →

x − 2y = 3
0 = 2

, which has no solutions. (See Figure 1.3.)
Figure 1.3: for Problem 1.1.13.
1.1.14 The system reduces to


x + 5z = 0
y − z = 0
0 = 1

, so that there is no solution; no point in space belongs to all three
planes.
Compare with Figure 2b.
1.1.15 The system reduces to


x = 0
y = 0
z = 0

 so the unique solution is (x, y, z) = (0, 0, 0). The three planes intersect at
the origin.
1.1.16 The system reduces to


x + 5z = 0
y − z = 0
0 = 0

 , so the solutions are of the form (x, y, z) = (−5t, t, t), where t is an
arbitrary number. The three planes intersect in a line; compare with Figure 2a.
1.1.17

x + 2y = a
3x + 5y = b

−3(I) →

x + 2y = a
−y = −3a + b

÷(−1) →

x + 2y = a
y = 3a − b

−2(II)
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Chapter 1

x = −5a + 2b
y = 3a − b

, so that (x, y) = (−5a + 2b, 3a − b).
1.1.18


x + 2y + 3z = a
x + 3y + 8z = b
x + 2y + 2z = c

 −I
−I →


x + 2y + 3z = a
y + 5z = −a + b
−z = −a + c

 −2(II)
→


x − 7z = 3a − 2b
y + 5z = −a + b
−z = −a + c


÷(−1) →


x − 7z = 3a − 2b
y + 5z = −a + b
z = a − c


+7(III)
−5(III) →


x = 10a − 2b − 7c
y = −6a + b + 5c
z = a − c

,
so that (x, y, z) = (10a − 2b − 7c, −6a + b + 5c, a − c).
1.1.19 The system reduces to


x + z = 1
y − 2z = −3
0 = k − 7

.
a. The system has solutions if k − 7 = 0, or k = 7.
b. If k = 7 then the system has infinitely many solutions.
c. If k = 7 then we can choose z = t freely and obtain the solutions
(x, y, z) = (1 − t,−3 + 2t, t).
1.1.20 The system reduces to


x − 3z = 1
y + 2z = 1
(k2 − 4)z = k − 2


This system has a unique solution if k2 − 4 6= 0, that is, if k 6= ±2.
If k = 2, then the last equation is 0 = 0, and there will be infinitely many solutions.
If k = −2, then the last equation is 0 = −4, and there will be no solutions.
1.1.21 Let x, y, and z represent the three numbers we seek. We set up a system of equations and solve systematically
(although there are short cuts):

x +y = 24
x +z = 28
y+ z = 30

−(I) →

x +y = 24
−y +z = 4
y+ z = 30

÷(−1) →

x +y = 24
y −z = −4
y+ z = 30

−(II)
−(II)
→

x +z = 28
y −z = −4
2z = 34

÷2 →

x +z = 28
y −z = −4
z = 17

−(III)
+(III) →

x = 11
y = 13
z = 17

We see that x = 11, y = 13, and z = 17.
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