Use the results of Exercise 2 and Newton's method to approximate the solutions of the nonlinear systems in Exercise 2 to within 10−6.
In exercise
a. 15x1 + x22 − 4x3 = 13,
x21 + 10x2 − x3 = 11,
x32 − 25x3 = −22.
b. 10x1 − 2x22 + x2 − 2x3 − 5 = 0,
8x22 + 4x23 − 9 = 0,
8x2 x3 + 4 = 0.
c. x31 + x21 x2 − x1 x3 + 6 = 0,
ex1 + ex2 − x3 = 0,
x22 − 2x1 x3 = 4.
d. x1 + cos(x1 x2 x3) − 1 = 0,
(1 − x1)1/4 + x2 + 0.05 x23− 0.15 x3 − 1 = 0,
−x21 − 0.1 x22 + 0.01 x2 + x3 − 1 = 0.