TEST BANK FOR Linear Algebra with Applications 5th Edition By Otto Bretscher
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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.
1
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
2
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
[Solved] TEST BANK FOR Linear Algebra with Applications 5th Edition By Otto Bretscher
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- GradeMaster1
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