Assignment Solutions_a6s

3 1 4 0 4 1 1 0 4 7 13 0 4 1 13 0 4 1 0 thus x4 0 x3

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 | 1 | 7 3 0 1 4|7 −1 | 1 1 −1 0 −1 | 1 1 | 3 1 −4 −1 | −3 → 0 0 −3 | 1 2 −1 −3 | 1 7|4 0 3 1 7| 4 −1 | 1 1 −1 0 −1 | 1 −1 | −3 1 −4 −1 | − 3 → 0 0 −1 | 7 0 14 −2 | 14 10 | 13 0 0 13 10 | 13 −1 | 1 1 −1 0 −1 | 1 0 −1 | − 3 1 −4 −1 | − 3 → 0 −12 | 1 0 1 −12 | 1 10 | 13 0 0 0 166 | 0 −1 | 1 −1 | −3 −12 | 1 1| 0 4 −1 0 1 −1 −3 1 4 0 4 −1 1 0 −4 7 13 0 −4 1 13 0 −4 1 0 Thus x4 = 0, x3 − 12x4 = 1, so x3 = 1 + 12x4 = 1, x2 − 4x3 − x4 = −3, so x2 = −3 + 4x3 + x4 = −...
View Full Document

This note was uploaded on 02/13/2014 for the course MATH 2050 taught by Professor Dryuan during the Fall '12 term at Memorial University.

Ask a homework question - tutors are online