 #1
 805
 8
Homework Statement
The problem:
Attached as TheProblem.jpg.
Solution:
x^{T} A x is 0 when (x_{1}, x_{2}, x_{3}) = (0,1,0) because of the zero on the diagonal. Actually, x^{T} A x goes negative for x = (1, 10, 0) because the second pivot is negative.
Homework Equations
x^{T} A x
The Attempt at a Solution
I computed the product and got 9x_{2}^{2} + 4 x_{1}x_{1} + 4x_{1}^{2}.
I'm thinking that I might need to show that that obtained polynomial must be below zero since we want to show that matrix A is not a positive definite matrix but, I don't know where to go from here.
Any input would be appreciated!
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