A First Course in Linear Algebra, 2017a

559
4.11.14 Then a solution is
⎡
⎢
⎣
1
6√
6
1
3√
6
1
6√
6
0
⎤ ⎡ √
1
⎤ ⎡
6√
2 3
⎥
⎦ , ⎢ − 2 √
9√
2 3
√ ⎥
⎣ 5
18√
√ 2
√ 3 ⎦ , ⎢
⎣
2 3
1
9
5
√
111√
3
√ 37
1
333√
3 37
− 17
22
333
√
333√
√ 3
√ 37
3 37
⎤
⎥
⎦
4.11.15 The subspace is of the form ⎡
⎡
and a basis is ⎣
1
0
2
⎤
⎡
⎦, ⎣
0
1
3
⎤
⎣
x
y
2x + 3y
⎦. Therefore, an orthonormal basis is
⎡
⎣
⎤ ⎡
1
5√
5
√
0 ⎦, ⎣
5
2
5
− 3
1
⎤
⎦
√
35√
5
√ 14
14√
√ 5
√ 14
5 14
3
70
⎤
⎦
4.11.21
[ 2313
]
Solution is:
⎡
⎣
1 2
2 3
3 5
⎤
⎦
T ⎡
⎣
1 2
2 3
3 5
⎤
⎦ =
[ 14 23
23 38
⎡
1
⎤T ⎡
2
= ⎣ 2 3 ⎦ ⎣
3 5
[ ][ ] [
14 23 x 17
=
23 38 y 28
[ ][ ] [ 14 23 x 17
=
23 38 y 28
][ 14 23
23 38
1
2
4
]
]
,
⎤
⎦ =
][ x
y
[ 17
28
]
]
( ) ( )
4.12.7 The velocity is the sum of two vectors. 50⃗i + 300 √ ⃗
2
i +⃗j = 50 + 300 √ ⃗
2
i + 300 √
2
⃗j. The component in
the direction of North is then 300 √ = 150 √ 2 and the velocity relative to the ground is
2
(
50 + 300 )
√ ⃗i + 300 √ ⃗j
2 2
4.12.10 Velocity of plane for the first hour: [ 0 150 ] + [ 40 0 ] = [ 40 150 ] [
. After one hour it is at
√ ]
(40,150). Next the velocity of the plane is 150 1 3
+ [ 40 0 ] in miles per hour. After two hours
[ ]
2 2
it is then at (40,150)+150 + [ 40 0 ] = [ 155 75 √ 3 + 150 ] = [ 155.0 279.9 ]
1
2
√
3
2