Linear Algebra Exercises-n-Answers.pdf

18 Linear Algebra, by Hefferon
⎛ ⎞ ⎛ ⎞
( ( )
2 −1
One.II.1.1 (a) (b) (c) ⎠ (d) ⎠
1)
2
One.II.1.2
⎝ 4 0
−3
(a) No, their canonical positions are different.
( ) (
1 0
−1 3)
⎝ 0 0
0
(b) Yes, their canonical positions are the same. ⎛
⎝ 1 ⎞
−1⎠
3
One.II.1.3
That line is this set.
Note that this system
⎛ ⎞ ⎛
−2
{ ⎜ 1
⎟
⎝ 1 ⎠ + ⎜
⎝
0
7
9
−2
4
⎞
−2 + 7t = 1
1 + 9t = 0
1 − 2t = 2
0 + 4t = 1
has no solution. Thus the given point is not in the line.
One.II.1.4
(a) Note that
⎛ ⎞ ⎛
2
⎜2
⎟
⎝2⎠ − ⎜
⎝
0
and so the plane is this set.
(b) No; this system
has no solution.
⎛
{ ⎜
⎝
1
1
5
−1
1
1
5
−1
⎞
⎛
⎟
⎠ = ⎜
⎝
⎞
⎛
⎟
⎠ + ⎜
⎝
1
1
−3
1
1
1
−3
1
⎞
⎟
⎠
⎞ ⎛
⎟
⎠ t + ⎜
⎝
⎟
⎠ t ∣ t ∈ R}
⎛ ⎞ ⎛
3
⎜1
⎟
⎝0⎠ − ⎜
⎝
4
2
0
−5
5
⎞
1 + 1t + 2s = 0
1 + 1t = 0
5 − 3t − 5s = 0
−1 + 1t + 5s = 0
One.II.1.5 The vector ⎛ ⎞
2
⎝0⎠
3
is not in the line. Because
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ 2 0⎠ − ⎝ −1
0 ⎠ = ⎝ 3 0⎠
3 −4 7
1
1
5
−1
⎞
⎛
⎟
⎠ = ⎜
⎝
⎟
⎠ s ∣ t, s ∈ R}
that plane can be described in this way.
⎛
{ ⎝ −1 ⎞ ⎛
0 ⎠ + m ⎝ 1 ⎞ ⎛
1⎠ + n ⎝ 3 ⎞
0⎠ ∣ m, n ∈ R}
4 2 7
One.II.1.6
The points of coincidence are solutions of this system.
t = 1 + 2m
t + s = 1 + 3k
t + 3s = 4m
2
0
−5
5
⎞
⎟
⎠