Linear Algebra Exercises-n-Answers.pdf

26 Linear Algebra, by Hefferon
One.III.1.8 Use ( Gauss-Jordan ) reduction. ( )
( )
(a) −(1/2)ρ 1+ρ 2 2 1 (1/2)ρ 1 1 1/2 −(1/2)ρ 2 +ρ 1 1 0
−→ −→
−→
0 5/2 (2/5)ρ 2
0 1
0 1
⎛
(b) −2ρ 1+ρ 2
−→ ⎝ 1 3 1 ⎞ ⎛
0 −6 2 ⎠ −(1/6)ρ 2
−→ ⎝ 1 3 1 ⎞ ⎛
0 1 −1/3⎠ (1/3)ρ 3+ρ 2
−→ ⎝ 1 3 0 ⎞ ⎛
0 1 0⎠ −3ρ 2+ρ 1
−→ ⎝ 1 0 0 ⎞
0 1 0⎠
ρ 1 +ρ 3
−(1/2)ρ
0 0 −2
3 −ρ
0 0 1
3 +ρ 1
0 0 1
0 0 1
(c)
⎛
−ρ 1 +ρ 2
−→ ⎝ 1 0 3 1 2
⎞ ⎛
0 4 −1 0 3 ⎠ −ρ 2+ρ 3
−→ ⎝ 1 0 3 1 2
⎞
0 4 −1 0 3 ⎠
−3ρ 1 +ρ 3
0 4 −1 −2 −4
0 0 0 −2 −7
⎛
(1/4)ρ 2
−→ ⎝ 1 0 3 1 2
⎞ ⎛
0 1 −1/4 0 3/2⎠ −ρ 3+ρ 1
−→ ⎝ 1 0 3 0 −3/2
⎞
0 1 −1/4 0 3/2 ⎠
−(1/2)ρ 3
0 0 0 1 7/2
0 0 0 1 7/2
(d)
⎛
ρ 1 ↔ρ 3
−→ ⎝ 1 5 1 5
⎞ ⎛
0 0 5 6⎠ ρ 2↔ρ 3
−→ ⎝ 1 5 1 5
⎞ ⎛
0 1 3 2⎠ (1/5)ρ 3
−→ ⎝ 1 5 1 19/5
⎞
0 1 3 2 ⎠
0 1 3 2 0 0 5 6 0 0 1 6/5
⎛
⎞ ⎛
⎞
1 3 0 −6/5
1 0 0 59/5
−3ρ 3 +ρ 2
−→ ⎝0 1 0 −8/5⎠ −5ρ 2+ρ 1
−→ ⎝0 1 0 −8/5⎠
−ρ 3 +ρ 1
0 0 1 6/5
0 0 1 6/5
One.III.1.9 For the “Gauss” halves, see the answers to Exercise 19.
(a) The “Jordan” half goes ( this way. )
( )
(1/2)ρ 1 1 1/2 −1/2 1/2 −(1/2)ρ 2 +ρ 1 1 0 −1/6 2/3
−→
−→
−(1/3)ρ 2
0 1 −2/3 −1/3
0 1 −2/3 −1/3
The solution set is this
⎛
{ ⎝ 2/3 ⎞ ⎛
−1/3⎠ + ⎝ 1/6 ⎞
2/3⎠ z ∣ z ∈ R}
0 1
(b) The second half is
⎛
ρ 3+ρ 2
−→ ⎝ 1 0 −1 0 1
⎞
0 1 2 0 3⎠
0 0 0 1 0
so the solution is this.
⎛ ⎞ ⎞
1 1
⎜3⎟
⎜−2⎟
⎛
{ ⎜ ⎟
⎝ ⎠ + ⎜
⎝
0
0
1
0
⎟
⎠ z ∣ z ∈ R}
(c) This Jordan half
⎛
⎞
1 0 1 1 0
ρ 2 +ρ 1
−→ ⎜0 1 0 1 0
⎟
⎝0 0 0 0 0⎠
0 0 0 0 0
gives
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 −1 −1
{ ⎜0
⎟
⎝0⎠ + ⎜ 0
⎟
⎝ 1 ⎠ z + ⎜−1
⎟
⎝ 0 ⎠ w ∣ z, w ∈ R}
0 0 1
(of course, the zero vector could be omitted from the description).
(d) The “Jordan” half ( )
−(1/7)ρ 2 1 2 3 1 −1 1
−→
0 1 8/7 2/7 −4/7 0
(
−2ρ 2+ρ 1 1 0 5/7 3/7 1/7 1
−→
0 1 8/7 2/7 −4/7 0
ends with this solution ⎛ set. ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 −5/7 −3/7 −1/7
0
{
⎜0
⎟
⎝0⎠ + −8/7
⎜ 1
⎟
⎝ 0 ⎠ c + −2/7
⎜ 0
⎟
⎝ 1 ⎠ d + 4/7
⎜ 0
⎟
⎝ 0 ⎠ e ∣ c, d, e ∈ R}
0 0
0
1
)