Linear Algebra Solutions 24

Linear Algebra Solutions 24 - a = 0 = d and hence c = 1 = b...

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

(ii) We next prove that if we impose the extra restriction that A 6 = 0 1 1 0 , then | a + d - 1 | < 1. This will then have the following consequence: A = P 1 0 0 a + d - 1 P - 1 A n = P 1 0 0 a + d - 1 n P - 1 = P 1 0 0 ( a + d - 1) n P - 1 P 1 0 0 0 P - 1 = b 1 c - 1 ‚• 1 0 0 0 - 1 b + c - 1 - 1 - c b = - 1 b + c b 0 c 0 ‚• - 1 - 1 - c b = - 1 b + c - b - b - c - c = 1 b + c b b c c , where we have used the fact that ( a + d - 1) n 0 as n → ∞ . We Frst prove the inequality | a + d - 1 | ≤ 1: a + d - 1 1 + d - 1 = d 1 a + d - 1 0 + 0 - 1 = - 1 . Next, if a + d - 1 = 1, we have a + d = 2; so a = 1 = d and hence c = 0 = b , contradicting our assumption that A 6 = I 2 . Also if a + d - 1 = - 1, then a + d = 0; so
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a = 0 = d and hence c = 1 = b and hence A = 1 1 . 22. The system is inconsistent: We work towards reducing the augmented matrix: 1 2 1 1 3 5 f f f f f f 4 5 12 R 2 R 2-R 1 R 3 R 3-3 R 1 1 2-1-1 f f f f f f 4 1 R 3 R 3-R 2 1 2-1 f f f f f f 4 1-1 . 27...
View Full Document

Ask a homework question - tutors are online