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Math 105
Semester Review - Solutions
2. ku = ( kx1, ky1,
kz1)
∈W
f) ( )
No. If = ( 4,0,0)
∈
then ku ( 0,0,0)
{ , , :2 3 8 0, , , }
2 kx − 3 ky + kz = k 2x − 3y + z = k0=
0
since ( ) ( ) ( )
W= xyz x− y+ z− = xyz∈ V
u W and k = 0 ,
= ∉W
since
Geometrically, W is the plane 2x− 3y+ z = 8.
{ }
g) W = ( a,3a− 4, a+ 1 ):
a∈
No. If u = ( 1, −1, 2)
∈W
then ku ( 0,0,0)
and
= ∉W
3
V =
k = 0 ,
since if
1 1 1 1 1 1
3
=
20 ⋅ −30 ⋅ + 0−8≠
0
( 0,0,0 ) ( a,3a 4, a 1)
= − + , then
4
we have a = 0, a= and a =− 1 which is impossible.
3
22. For each of the subsets W in ,
i) Find a basis for W.
ii) Find the dimension of W
iii) Give a geometrical interpretation of W.
W = 2 a+ b, a, a−2 b : a,
b∈
{ }
a) ( )
Since ( 2 a+ b, a, a− 2b) = a( 2,1,1) + b( 1,0, − 2)
Then if B = ( 2,1,1 ), ( 1, 0. − 2 we have W span ( B)
{ )}
3
= .
Since B is linearly independent (the two vectors are not multiples of each other)
then B is a basis for W.
dim W = 2
( )
i j k
n = × − = = − −
( 2,1,1) ( 1,0, 2) 2 1 1 ( 2,5, 1)
1 0 −2
Ergo, W is the plane π :2x− 5y+ z = 0.
{ }
b) W= ( a− 2b+ 3 c,3a+ b+ 2 ca , + 4b−3 c)
: abc , , ∈
Since ( a b c a b c a b c) a( ) b( ) c(
then if B = {( 1,3,1 ),( −2,1, 4 ),( 3, 2, − 3)
} we have W = span ( B)
c ( 1,3,1) + c ( − 2,1, 4) + c ( 3, 2, − 3) = ( 0, 0, 0)
If
− 2 + 3 ,3 + + 2 , + 4 − 3 = 1,3,1 + − 2,1, 4 + 3, 2, − 3)
1 2 3
⎡1 −2 3 0⎤ ⎡1 −2 3 0⎤
⎢ 2 2
3
1
3 1 2 0
⎥R →R − R ⎢
0 7 −7 0
⎥
⎢ ⎥R3 →R3−R
⎢
⎥
1
⎢1 4 −3 0⎥
⎣ ⎦ ⎢⎣0 6 −6 0⎥⎦
⎡1 −2 3 0⎤
⎡1 −2 3 0⎤
R3 →7R3−6R
⎢
1
0 7 −7 0
⎥ 1
⎢
⎥
R2 7
R
⎢
2
0 1 1 0
⎥
→
⎢
−
⎥
⎢⎣
0 0 0 0⎥⎦
⎢⎣
0 0 0 0⎥⎦
1, then 3, 2, − 3 = 1,3,1 − − 2,1, 4
t = ( ) ( ) ( )
c
c
c
3
2
1
= t
= t
= −t
Winter 2006 Martin Huard 20