| Line 14: | Line 14: | ||
I am not sure on the 48/100Y1 portion of equation 1. | I am not sure on the 48/100Y1 portion of equation 1. | ||
| + | |||
| + | Need help with P356 #29 and 30. | ||
| + | This is my understanding of #29-> Positive definiteness of Prob23: [x1 x2]^T * [4 Sqrt(3), Sqrt(3) 2] * [x1 x2] >0 for all X(vector) not equal to 0(vector). So, 4 > 0, and the det([4 Sqrt(3), Sqrt(3) 2]) >0. Therefore, it is positive definite. And for Prob19: [x1 x2]^T * [1 12, 12 -6] * [x1 x2], 1 > 0 and det ([1 12, 12 -6]) < 0. Therefore, it it indefinite. | ||
[[2010 MA 527 Bell|Back to the MA 527 start page]] | [[2010 MA 527 Bell|Back to the MA 527 start page]] | ||
Revision as of 23:21, 17 September 2010
Homework 4 work area
Collaborate on HWK 4 here.
Section 8.4 #29 Does anybody have any thoughts about the solution to the second part of number 29 and the proof for #30? For the positive definite case and negative definite case, finding the determinate seems sufficient. I'm not sure how to show the indefinite case.
Problem 18 on page 146
Do I have the time rate of change equations correct:
Y1' = 48/100Y1 + 16/400Y2 - 64/100Y1 Y2' = 64/100Y1 - 64/100Y2
I am not sure on the 48/100Y1 portion of equation 1.
Need help with P356 #29 and 30. This is my understanding of #29-> Positive definiteness of Prob23: [x1 x2]^T * [4 Sqrt(3), Sqrt(3) 2] * [x1 x2] >0 for all X(vector) not equal to 0(vector). So, 4 > 0, and the det([4 Sqrt(3), Sqrt(3) 2]) >0. Therefore, it is positive definite. And for Prob19: [x1 x2]^T * [1 12, 12 -6] * [x1 x2], 1 > 0 and det ([1 12, 12 -6]) < 0. Therefore, it it indefinite.
