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Messages - Chonghan Ma

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Quiz-7 / Re: Q7 TUT 0101
« on: November 30, 2018, 04:29:45 PM »
(a)
Set x’ = 0 and y’=0
Then we have critical points (0,0), (-1,0)
(b)
J = \begin{bmatrix}2x+1 & 2y \\-y & 1-x \end{bmatrix}
Linear systems are shown with each critical point:
J(0,0) =  \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}
J(-1,0) =  \begin{bmatrix}-1 & 0 \\0 & 2 \end{bmatrix}
(c)
Eigenvalues are computed by det(A - tI)= 0
So that
At (0,0): t=1
Critical point is an unstable proper node or spiral point
At (-1,0): t=-1 and 2   
Critical point is an unstable saddle point

2
Quiz-7 / Re: Q7 TUT 5102
« on: November 30, 2018, 04:19:46 PM »
(a)
Set (2+x)(y-x)=0 and (4-x)(y+x)=0
Then we have critical points (0,0), (4,4), (-2,2)
(b)
J = \begin{bmatrix}-2-2x+y & 2+x \\4-2y-2x & 4-x \end{bmatrix}
Linear systems are shown with each critical point:
J(0,0) =  \begin{bmatrix}-2 & 2 \\4 & 4 \end{bmatrix}
J(-2,2) =  \begin{bmatrix}4 & 0 \\6 & 6 \end{bmatrix}
J(4,4) =  \begin{bmatrix}-6 & 6 \\-8 & 0 \end{bmatrix}
(c)
Eigenvalues are computed by det(A - tI)= 0
So that
At (0,0): t=1±√17}
Critical point is a saddle point and it is unstable
At (-2,2): t= 4 and 6
Critical point is an unstable node
At ((4,4): t=-3±√9 i   
Critical point is a stable spiral point

3
Quiz-6 / Re: Q6 TOT 0301
« on: November 17, 2018, 07:21:36 PM »
Sometimes we do not have to compute all the ref or rref. For example, when λ= 4, it is easy to observe that the sum of three columns of the matrix is 0. Sometimes it saves your time during the quiz if you can observe it directly.

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