Solutions Manual For Linear Algebra, 5th Edition, Friedberg ; Solutions For Linear Algebra 5E ; All Chapters Covered

SOLUTIONS MANUAL

LINEAR ALGEBRA, 5TH EDITION

CHAPTER 1: VECTOR SPACES
1.1 INTRODUCTION


2. (b) x = (2, 4, 0) + t(−5, −10, 0) (d) x = (−2, −1, 5) + t(5, 10, 2)
3. (b) x = (3, −6, 7) + s(−5, 6, −11) + t(2, −3, −9)
(d) x = (1, 1, 1) + s(4, 4, 4) + t(−7, 3, 1)
4. (0, 0)


1.2 VECTOR SPACES
 
0 0 0 0
2. 0 0 0 0
0 0 0 0
   
1 −1 30 −20
4. (b) 3 −5 (d) −15 10
3 8 −5 −40
(f ) −x3 + 7x2 + 4 (h) 3x5 − 6x3 + 12x + 6
     
8 3 1 9 1 4 17 4 5
5. 3 0 0 + 3 0 0 =  6 0 0
3 0 0 1 1 0 4 1 0

16. Yes 18. No, (VS 1) fails. 19. No, (VS 8) fails.


1.3 SUBSPACES
 
   −2 7
0 3 10 2 −5  5
 0 −4 0
2. (b)  8 4 (d) 7 (f ) 
 1

1
−6 7 −8 3 6
4 −6
The trace is 12.
 
−4 0 6
(h)  0 1 −3
6 −3 5
The trace is 2.

8. (b) No (d) Yes (f ) No
9. W1 ∩ W3 = {(0, 0, 0)}, W1 ∩ W4 = W1 ,
W3 ∩ W4 = {(a1 , a2 , a3 ) ∈ R3 : a1 = −11a3 and a2 = −3a3 }



1

,1.4 LINEAR COMBINATIONS AND SYSTEMS OF LINEAR EQUATIONS

2. (b) (−2, −4, −3)
(d) {x3 (−8, 3, 1, 0) + (−16, 9, 0, 2): x3 ∈ R}
(f ) (3, 4, −2)
3. (a) (−2, 0, 3) = 4(1, 3, 0) − 3(2, 4, −1)
(b) (1, 2, −3) = 5(−3, 2, 1) + 8(2, −1, −1)
(d) No
(f ) (−2, 2, 2) = 4(1, 2, −1) + 2(−3, −3, 3)
4. (a) x3 − 3x + 5 = 3(x3 + 2x2 − x + 1) − 2(x3 + 3x2 − 1)
(b) No
(c) −2x3 − 11x2 + 3x + 2 = 4(x3 − 2x2 + 3x − 1) − 3(2x3 + x2 + 3x − 2)
(d) x3 + x2 + 2x + 13 = −2(2x3 − 3x2 + 4x + 1) + 5(x3 − x2 + 2x + 3)
(f ) No
5. (b) No (d) Yes (f ) No (h) No
11. The span of {x} is {0 } if x = 0 and is the line through the origin of R3 in the direction of x
if x 6= 0 .
17. if W is finite


1.5 LINEAR DEPENDENCE AND LINEAR INDEPENDENCE

2. (b) Linearly independent (d) Linearly dependent
(f ) Linearly independent (h) Linearly independent
(j) Linearly dependent
10. (1, 0, 0), (0, 1, 0), (1, 1, 0)


1.6 BASES AND DIMENSION

2. (b) Not a basis (d) Basis
3. (b) Basis (d) Basis
4. No, dim(P3 (R)) = 4. 5. No, dim(R3 ) = 3.
8. {u1 , u3 , u5 , u7 }
10. (b) 12 − 3x (d) 2x3 − x2 − 6x + 15
14. {(0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (1, 0, 1, 0, 0), (1, 0, 0, 1, 0)} and
{(−1, 0, 0, 0, 1), (0, 1, 1, 1, 0)}; dim(W1 ) = 4 and dim(W2 ) = 2.
16. dim(W) = 21 n(n + 1)


2

, 1.6 Bases and Dimension



18. Let σj be the sequence such that
(
1 i=j
σj (i) =
0 i 6= j.

Then {σj : j = 1, 2, . . .} is a basis for the vector space in Example 5 of Section 1.2.
22. W1 ⊆ W2
23. (a) v ∈ W1 (b) dim(W2 ) = dim(W1 ) + 1
25. mn
n
27. If n is even, then dim(W1 ) = dim(W2 ) =; and if n is odd,
2
n+1 n−1
then dim(W1 ) = and dim(W2 ) = .
2 2
32. (a) Take W1 = R3 and W2 = span({e1 }).
(b) Take W1 = span({e1 , e2 }) and W2 = span({e3 }).
(c) Take W1 = span({e1 , e2 }) and W2 = span({e2 , e3 }).
35. (b) dim(V) = dim(W) + dim(V/W)




3