TEST BANK FOR Linear Algebra, 4th Edition By THOMAS POLASKI JUDITH MCDONALD

CHAPTER 1 Linear Equations in Linear Algebra 1
CHAPTER 2 Matrix Algebra 87
CHAPTER 3 Determinants 167
CHAPTER 4 Vector Spaces 197
CHAPTER 5 Eigenvalues and Eigenvectors 273
CHAPTER 6 Orthogonality and Least Squares 357
CHAPTER 7 Symmetric Matrices and Quadratic Forms 405
CHAPTER 8 The Geometry of Vector Spaces 453

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. 1 2
1 2
5 7
2 7 5
x x
x x
+ =
− − = −
1 5 7
2 7 5
􀂪 􀂺
􀂫− − − 􀂻 􀂬 􀂼
Replace R2 by R2 + (2)R1 and obtain: 1 2
2
5 7
3 9
x x
x
+ =
=
1 5 7
0 3 9
􀂪 􀂺
􀂫 􀂻
􀂬 􀂼
Scale R2 by 1/3: 1 2
2
5 7
3
x x
x
+ =
=
1 5 7
0 1 3
􀂪 􀂺
􀂫 􀂻
􀂬 􀂼
Replace R1 by R1 + (–5)R2: 1
2
8
3
x
x
= −
=
1 0 8
0 1 3
􀂪 − 􀂺
􀂫 􀂻
􀂬 􀂼
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2. 1 2
1 2
3 6 3
5 7 10
x x
x x
+ =−
+ =
3 6 3
5 7 10
􀂪 − 􀂺
􀂫 􀂻
􀂬 􀂼
Scale R1 by 1/3 and obtain: 1 2
1 2
2 1
5 7 10
x x
x x
+ =−
+ =
1 2 1
5 7 10
􀂪 − 􀂺
􀂫 􀂻
􀂬 􀂼
Replace R2 by R2 + (–5)R1: 1 2
2
2 1
3 15
x x
x
+ =−
− =
1 2 1
0 3 15
􀂪 − 􀂺
􀂫 − 􀂻 􀂬 􀂼
Scale R2 by –1/3: 1 2
2
2 1
5
x x
x
+ =−
= −
1 2 1
0 1 5
􀂪 − 􀂺
􀂫 − 􀂻 􀂬 􀂼
Replace R1 by R1 + (–2)R2: 1
2
9
5
x
x
=
= −
1 0 9
0 1 5
􀂪 􀂺
􀂫 − 􀂻 􀂬 􀂼
The solution is (x1, x2) = (9, –5), or simply (9, –5).
2 CHAPTER 1 • Linear Equations in Linear Algebra

3. The point of intersection satisfies the system of two linear equations:
1 2
1 2
2 4
1
x x
x x
+ =
− =
1 2 4
1 1 1
􀂪 􀂺
􀂫 − 􀂻 􀂬 􀂼
Replace R2 by R2 + (–1)R1 and obtain: 1 2
2
2 4
3 3
x x
x
+ =
− = −
1 2 4
0 3 3
􀂪 􀂺
􀂫 − − 􀂻 􀂬 􀂼
Scale R2 by –1/3: 1 2
2
2 4
1
x x
x
+ =
=
1 2 4
0 1 1
􀂪 􀂺
􀂫 􀂻
􀂬 􀂼
Replace R1 by R1 + (–2)R2: 1
2
2
1
x
x
=
=
1 0 2
0 1 1
􀂪 􀂺
􀂫 􀂻
􀂬 􀂼
The point of intersection is (x1, x2) = (2, 1).
4. The point of intersection satisfies the system of two linear equations:
1 2
1 2
2 13
3 2 1
x x
x x
+ =−
− =
1 2 13
3 2 1
􀂪 − 􀂺
􀂫 − 􀂻 􀂬 􀂼
Replace R2 by R2 + (–3)R1 and obtain: 1 2
2
2 13
8 40
x x
x
+ = −
− =
1 2 13
0 8 40
􀂪 − 􀂺
􀂫 − 􀂻 􀂬 􀂼
Scale R2 by –1/8: 1 2
2
2 13
5
x x
x
+ = −
= −
1 2 13
0 1 5
􀂪 − 􀂺
􀂫 − 􀂻 􀂬 􀂼
Replace R1 by R1 + (–2)R2: 1
2
3
5
x
x
= −
= −
1 0 3
0 1 5
􀂪 − 􀂺
􀂫 − 􀂻 􀂬 􀂼
The point of intersection is (x1, x2) = (–3, –5).
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 –4 times R3, and then replace R1 by its sum with 3 times R3.
6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which
produces
1 6 4 0 1
0 2 7 0 4
0 0 1 2 3
0 0 0 7 14
􀂪 − − 􀂺
􀂫 − 􀂻 􀂫 􀂻
􀂫 − 􀂻
􀂫 􀂻 􀂬 − 􀂼
. After that, the next step is to scale the fourth row by –1/7.
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.