(i) Use Gaussian elimination and three-digit rounding arithmetic to approximate the solutions to the following linear systems. (ii) Then use one iteration of iterative refinement to improve the approximation, and compare the approximations to the actual solutions.
a. 0.03x1 + 58.9x2 = 59.2,
5.31x1 − 6.10x2 = 47.0
Actual solution (10, 1)t .
b. 3.3330x1 + 15920x2 + 10.333x3 = 7953,
2.2220x1 + 16.710x2 + 9.6120x3 = 0.965,
−1.5611x1 + 5.1792x2 − 1.6855x3 = 2.714
Actual solution (1, 0.5,−1)t .
c. 1.19x1 + 2.11x2 − 100x3 + x4 = 1.12,
14.2x1 − 0.122x2 + 12.2x3 − x4 = 3.44,
100x2 − 99.9x3 + x4 = 2.15,
15.3x1 + 0.110x2 − 13.1x3 − x4 = 4.16
Actual solution (0.17682530, 0.01269269,−0.02065405,−1.18260870)t .
d. πx1 − ex2 + √2x3 − √3x4 = √11,
π2x1 + ex2 − e2x3 + 3/7x4 = 0,
√5x1 − √6x2 + x3 − √2x4 = π,
π3x1 + e2x2 − √7x3 + 1/9 x4 = √2.
Actual solution (0.78839378,−3.12541367, 0.16759660, 4.55700252)t .