Homepage › Solution manuals › Gilbert Strang › Introduction to Linear Algebra › Exercise 2.2.21

Exercise 2.2.21

Answers

(a) Pivots $2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}$ in the equations $2 x + y = 0, \frac{3}{2} y + z = 0, \frac{4}{3} z + t = 0, \frac{5}{4} t = 5$ after elimination. Back substitution gives $t = 4, z = - 3, y = 2, x = - 1$ .

(b) If the off-diagonal entries change from $+ 1$ to $- 1$ , the pivots are the same. The solution is $(1, 2, 3, 4)$ instead of $(- 1, 2, - 3, 4)$ .

mathchakra

2021-11-20 15:07

Exercise 2.2.21

Answers

Comments

Add answer