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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. | 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. | ||
| − | (rekblad 9/18) for #29 part 2 showing Q is indefinite, isn't it enough to just find two vectors that show Q > 0 and Q < 0 and also show that Q!=0 for x!=0 ? | + | (rekblad 9/18) for #29 part 2 showing Q is indefinite, isn't it enough to just find two vectors that show Q > 0 and Q < 0 and also show that Q!=0 for x!=0 ? (Actually, on second thought, I think Q indefinite => Q = 0 for some x!=0) |
Problem 18 on page 146 | Problem 18 on page 146 | ||
Revision as of 07:40, 18 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.
(rekblad 9/18) for #29 part 2 showing Q is indefinite, isn't it enough to just find two vectors that show Q > 0 and Q < 0 and also show that Q!=0 for x!=0 ? (Actually, on second thought, I think Q indefinite => Q = 0 for some x!=0)
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.
