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

**Unformatted text preview: **1 2 3 2 d 4 3 4 5 are 1, d-4 and-4 d + 8. These are positive when, simultaneously, d > 4 and 2 > d , that is never. 15 ) We know that for any ~x 6 = ~ 0 that ~x T A~x > 0 and ~x T B~x > 0. So for any ~x 6 = ~ 0 we have ~x T ( A + B ) ~x = ~x T A~x + ~x T B~x > 0. 25 ) A = C T C = 3 0 1 2 3 1 0 2 = 9 3 3 5 . For 4 8 8 25 , the rst elimination step is to add-2 times row 1 to row 2. This is multi-plying by 1 0-2 1 on the left, so 4 8 8 25 = 1 0 2 1 4 8 0 9 = 1 0 2 1 4 0 0 9 1 2 0 1 so C = 2 0 0 3 1 2 0 1 = 2 4 0 3 ....

View
Full
Document