Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 33
Maple thought for perhaps twenty seconds and gave this reply.
[ −d p + q c
−
−b c + a d , −b p + a q ]
−b c + a d
Topic: Input-Output Analysis
1 These answers were given by Octave.
(a) s = 33 379, a = 43 304
(b) s = 37 284, a = 43 589
(c) s = 37 411, a = 43 589
2 Octave gives these answers.
(a) s = 24 244, a = 30 309
(b) s = 24 267, a = 30 675
Topic: Accuracy of Computations
1 Sceintific notation is convienent to express the two-place restriction. We have .25 × 10 2 + .67 × 10 0 =
.25 × 10 2 . The 2/3 has no apparent effect.
2 The reduction
−3ρ 1 +ρ 2 x + 2y = 3
−→
−8 = −7.992
gives a solution of (x, y) = (1.002, 0.999).
3 (a) The fully accurate solution is that x = 10 and y = 0.
(b) The four-digit conclusion is quite
(
different.
)
−(.3454/.0003)ρ 1+ρ 2 .0003 1.556 1.569
−→ =⇒ x = 10460, y = −1.009
0 1789 −1805
4 (a) For the first one, first, (2/3) − (1/3) is .666 666 67 − .333 333 33 = .333 333 34 and so (2/3) +
((2/3) − (1/3)) = .666 666 67 + .333 333 34 = 1.000 000 0. For the other one, first ((2/3) + (2/3)) =
.666 666 67 + .666 666 67 = 1.333 333 3 and so ((2/3) + (2/3)) − (1/3) = 1.333 333 3 − .333 333 33 =
.999 999 97.
(b) The first equation is .333 333 33 · x + 1.000 000 0 · y = 0 while the second is .666 666 67 · x +
2.000 000 0 · y = 0.
5 (a) This calculation
⎛
−(2/3)ρ 1+ρ 2
−→ ⎝ 3 2 1 6
⎞
0 −(4/3) + 2ε −(2/3) + 2ε −2 + 4ε⎠
−(1/3)ρ 1 +ρ 3
0 −(2/3) + 2ε −(1/3) − ε −1 + ε
⎛
−(1/2)ρ 2+ρ 3
−→ ⎝ 3 2 1 6
⎞
0 −(4/3) + 2ε −(2/3) + 2ε −2 + 4ε⎠
0 ε −2ε −ε
gives a third equation of y−2z = −1. Substituting into the second equation gives ((−10/3)+6ε)·z =
(−10/3) + 6ε so z = 1 and thus y = 1. With those, the first equation says that x = 1.
(b) The solution with two digits kept
⎛
⎝ .30 × ⎞
101 .20 × 10 1 .10 × 10 1 .60 × 10 1
.10 × 10 1 .20 × 10 −3 .20 × 10 −3 .20 × 10 1 ⎠
.30 × 10 1 .20 × 10 −3 −.10 × 10 −3 .10 × 10 1
−(2/3)ρ 1 +ρ 2
−→
−(1/3)ρ 1 +ρ 3
−(.67/1.3)ρ 2 +ρ 3
−→
comes out to be z = 2.1, y = 2.6, and x = −.43.
⎛
⎞
.30 × 10 1 .20 × 10 1 .10 × 10 1 .60 × 10 1
⎝ 0 −.13 × 10 1 −.67 × 10 0 −.20 × 10 1 ⎠
0 −.67 × 10 0 −.33 × 10 0 −.10 × 10 1
⎛
⎝ .30 × ⎞
101 .20 × 10 1 .10 × 10 1 .60 × 10 1
0 −.13 × 10 1 −.67 × 10 0 −.20 × 10 1 ⎠
0 0 .15 × 10 −2 .31 × 10 −2