Use the modified algorithm from Exercise 7 to solve the following linear systems.
a. 4x1 − x2+ x3 = −1,
−x1 + 3x2 = 4,
x1 +2x3 = 5.
b. 4x1 + 2x2+2x3 = 0,
2x1 + 6x2+2x3 = 1,
2x1 + 2x2+5x3 = 0.
c. 4x1 + 2x3 + x4 = −2,
3x2 − x3 + x4 = 0,
2x1 − x2 + 6x3 + 3x4 = 7,
x1 + x2 + 3x3 + 8x4 = −2.
d. 4x1+ x2 + x3+ x4 = 2,
x1+3x2 − x4 = 2,
x1 + 2x3+ x4 = 1,
x1− x2 + x3+4x4 = 1.