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Math 105
Semester Review - Solutions
b) w 3
= ( 12,
−7,10)
, V = and = {( 3,
−1,5 ),
( 1,2,3 ),
( 2, −1,
−1)
}
c ( 3, − 1, 5) + c ( 1, 2, 3) + c ( 2, −1, − 1) = ( 12, − 7,10)
1 2 3
S .
⎡ 3 1 2 12⎤ ⎡3 1 2 12 ⎤
⎢ R2 → 3R2 + R1
−1 2 −1 −7 ⎥ ⎢
0 7 −1 −9
⎥
⎢ ⎥R3 →3R3−5R
⎢
⎥
1
⎢ 5 3 −1 10⎥
⎣ ⎦ ⎢⎣0 4 −13 −30⎥
⎦
1
⎡3 1 2 12 ⎤ R
1 2
1
→
3
R1
⎡1 3 3
4⎤
R3 →7R3−4R
⎢
1
0 7 −1 −9
⎥ 1 −1
−9
⎢
⎥
R2 →
7
R
⎢
2
0 1
⎥
⎢
7 7 ⎥
⎢⎣
0 0 −87 −174⎥
−1
⎦ R 0 0 1 2
3 87
R ⎢
⎥
→
3⎣ ⎦
c3 = 2, c2 =− 1, c1
= 3
w = 3, −1, 2
S ( )
c) w = ( 4,
−3,5,3
), = Span( {( 3, −1,4,0
),
( 0,1, −2,4) ,( 2,2,1,1 )})
S = {( 3,
−1,4,0
),
( 0,1, −2,4) ,( 2,2,1,1 )}.
c ( 3, − 1, 4,0) + c ( 0,1, − 2, 4) + c ( 2, 2,1,1) = ( 4, − 3,5,3 )
V and
1 2 3
⎡ 3 0 2 4 ⎤ ⎡3 0 2 4 ⎤
⎢
1 1 2 3
⎥
R2 → 3R2 + R
⎢
⎢
− −
⎥
1
⎢
0 3 8 −5
⎥
⎥⎥
⎢ 4 −2 1 5 ⎥R3 →3R3 −4R
⎢
1
0 −6 5 1
⎢ ⎥ − −
⎢
⎥⎦
⎣ 0 4 1 3 ⎦ ⎣0 4 1 3
⎡3 0 2 4 ⎤
⎡3 0 2 4 ⎤
R3 → R3 + 2R
⎢
2
0 3 8 −5
⎥
⎢
⎢
⎥
0 3 8 5
⎥
R4 11R4 29R
⎢
−
⎥
3
R4 →3R4 −4R
⎢
2
0 0 11 −11⎥
→ +
⎢0 0 11 −11⎥
⎢
⎥
⎢
⎥
⎣0 0 −29 29 ⎦
⎣0 0 0 0 ⎦
2 4
1
⎡1 0
3 3 ⎤
R1 →
3
R1
⎢ 8 −5
0 1
⎥ c3
= −1
1
3 3
R2 3
R ⎢
⎥
→
2
c2
= 1 w
S
= ( 2,1, −1)
⎢0 0 1 −1⎥
1
R3 →
11
R3⎢ ⎥ c1
= 2
⎣0 0 0 0 ⎦
d) w 9 2 2 2 2
= x − x +11, V = P2
and S = { x + x,
x −1,2
x − 3x
+ 4}
.
( ) ( ) ( )
c x + x + c x − 1 + c 2x − 3x+ 4 = 9x − x+11
2 2 2 2
1 2 3
⎡1 1 2 9 ⎤ ⎡1 1 1 9 ⎤
⎢
1 0 −3 −1 ⎥
R2 →R2 −R
⎢
1
0 −1 −5 −10
⎥
⎢ ⎥
⎢
⎥
⎢⎣0 −1 4 11⎥⎦ ⎢⎣0 −1 4 11 ⎥⎦
⎡1 1 2 9 ⎤ ⎡1 1 2 9⎤
R3 →R3−R
⎢
2
0 −1 −5 −10
⎥ R2 →−R2⎢ ⎢
⎥
0 1 5 10
⎥
1
R3 9
R ⎢
⎥
3
7
⎢⎣0 0 9 21 ⎥
→
⎦ ⎢ ⎣0 0 1
3 ⎥ ⎦
−
w = 6, ,
S
5 7
( 3 3
)
c
c
7
3 3
−5
2 3
1
= 6
c
=
=
Winter 2006 Martin Huard 30