### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Chonghan Ma

Pages: [1]
1
##### 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
##### MAT244--Lectures & Home Assignments / Re: Repeated eigenvalues in 3x3 matrix homogeneous equation
« on: November 29, 2018, 10:51:46 PM »
Set det\begin{bmatrix}1-\lambda & -3 & 3 \\3 & -5-\lambda & 3 \\6 & -6 & 4-\lambda\end{bmatrix}=0
then,
We have  λ3-12λ-16 = 0
Therefore, the eigenvalues are λ = 4 , -2
λ = 4:
Null\begin{bmatrix}-3 & -3 & 3 \\3 & -9 & 3 \\6 & -6 & 0\end{bmatrix} = span\begin{bmatrix}1\\1\\2\end{bmatrix}
λ = -2:
Null\begin{bmatrix}3 & -3 & 3 \\3 & -3 & 3 \\6 & -6 & 6\end{bmatrix}=span{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\0\end{bmatrix}}

Then the solution should be

y(t) = c1e4t\begin{bmatrix}1\\1\\2\end{bmatrix} + c2e-2t\begin{bmatrix}1\\0\\1\end{bmatrix} + c3e-2t\begin{bmatrix}-1\\1\\0\end{bmatrix}

4
##### 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.

Pages: [1]