Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 19
Gauss’ method
⎛
⎝ 1 0 0 −2 1
⎞
1 1 −3 0 1
1 3 0 −4 0
⎠ −ρ1+ρ2
−→
−ρ 1+ρ 3
⎛
⎝ 1 0 0 −2 1
⎞
0 1 −3 2 0
0 3 0 −2 −1
⎠ −3ρ2+ρ3
−→
⎛
⎝ 1 0 0 −2 1
⎞
0 1 −3 2 0 ⎠
0 0 9 −8 −1
gives k = −(1/9) + (8/9)m, so s = −(1/3) + (2/3)m and t = 1 + 2m. The intersection is this.
⎛
{ ⎝ 1 ⎞ ⎛
1⎠ + ⎝ 0 ⎞
⎛
3⎠ (− 1 9 + 8 9 m) + ⎝ 2 ⎞
⎛
0⎠ m ∣ m ∈ R} = { ⎝ 1
⎞ ⎛
2/3⎠ + ⎝ 2
⎞
8/3⎠ m ∣ m ∈ R}
0 0
4
0 4
One.II.1.7
(a) The system
gives s = 6 and t = 8, so this is the solution set.
⎛
(b) This system
1 = 1
1 + t = 3 + s
2 + t = −2 + 2s
⎞
{ ⎝ 1 9 ⎠}
10
2 + t = 0
t = s + 4w
1 − t = 2s + w
gives t = −2, w = −1, and s = 2 so their intersection is this point.
⎛ ⎞
One.II.1.8
(a) The vector shown
⎝ 0 −2⎠
3
is not the result of doubling
instead it is
which has a parameter twice as large.
(b) The vector
⎛ ⎞ ⎛ ⎞
⎝ 2 0⎠ + ⎝ −0.5
1 ⎠ · 1
0 0
⎛ ⎞
2
⎛ ⎞
−0.5
⎛ ⎞
1
⎝0⎠ + ⎝ 1 ⎠ · 2 = ⎝2⎠
0 0
0
is not the result of adding
⎛
( ⎝ 2 ⎞ ⎛
0⎠ + ⎝ −0.5
⎞ ⎛
1 ⎠ · 1) + ( ⎝ 2 ⎞ ⎛
0⎠ + ⎝ −0.5
⎞
0 ⎠ · 1)
0 0
0 1