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### Messages - Siran Wang

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1
##### Term Test 2 / Re: TT2-P3
« on: November 22, 2018, 02:24:55 PM »
a)
\begin{equation*}
A-\lambda I=\begin{pmatrix}
2-\lambda & 1\\
-3 & -2-\lambda
\end{pmatrix}
\end{equation*}

\begin{equation*}
\det(A-\lambda I)=(2-\lambda)(-2-\lambda)+3=0
\end{equation*}
\begin{equation*}
\end{equation*}
\begin{equation*}
\lambda_1=1~~~~\lambda_2=-1
\end{equation*}

When $\lambda_1=1$
\begin{equation*}
x_1+x_2=0
\end{equation*}
The first eigenvector is $\begin{pmatrix} 1\\ -1 \end{pmatrix}$

When $\lambda_1=-1$
\begin{equation*}
3x_1+x_2=0
\end{equation*}
The second eigenvector is $\begin{pmatrix} 1\\ -3 \end{pmatrix}$
So, the general solution is $y=C_1e^t\begin{pmatrix} 1\\ -1 \end{pmatrix}+C_2e^{-t}\begin{pmatrix} 1\\ -3 \end{pmatrix}$

graph in attachment

c)
\begin{equation*}
\phi=\begin{pmatrix}
e^t&e{-t}\\
-e^t&-3e^{-t}
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
e^t&e{-t}\\
-e^t&-3e^{-t}
\end{pmatrix}\begin{pmatrix}
U_1\\
U_2
\end{pmatrix}=\begin{pmatrix}
\frac{4}{e^t+e^{-t}}\\
\frac{-12}{e^t+e^{-t}}
\end{pmatrix}
\end{equation*}
\begin{equation*}
e^tU_1+e^{-t}U_2=\frac{4}{e^t+e^{-t}}
\end{equation*}
\begin{equation*}
-e^tU_1-3e^{-t}U_2= \frac{-12}{e^t+e^{-t}}
\end{equation*}
\begin{equation*}
U_2= \frac{4}{1+e^{-2t}}-e^{2t}U_1
\end{equation*}
\begin{equation*}
-e^tU_1-3e^{-t}(\frac{4}{1+e^{-2t}}-e^{2t}U_1)=\frac{-12}{e^t+e^{-t}}
\end{equation*}
\begin{equation*}
U_1=0~~~U_2=\frac{4}{1+e^{-2t}}
\end{equation*}
\begin{equation*}
V_1=\int U_1dt=\int0dt=0
\end{equation*}
\begin{equation*}
V_2=\int U_2dt=\int\frac{4}{1+e^{-2t}}dt=2ln(1+e^{2t})
\end{equation*}
\begin{equation*}\begin{pmatrix}
x_1\\
x_2
\end{pmatrix}=\begin{pmatrix}
C_1e^t+C_2e^{-t}+2e{-t}ln(1+e^{2t}\\
-C_1e^t-3C_2e^{-t}-6e{-t}ln(1+e^{2t}
\end{pmatrix}
\end{equation*}
since $\begin{equation*}x(0)=\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{equation*}$
\begin{equation*}
C_1+C_2+2ln2=0~~~-C_1-3C_2-6ln2=0
\end{equation*}\begin{equation*}
C_1=0~~~C_2=-2ln2
\end{equation*}

2
##### Quiz-6 / Re: Q6 TUT 0201
« on: November 17, 2018, 04:53:37 PM »
\begin{equation*}
A-\lambda I=\begin{pmatrix}
1-\lambda & 0 & 0\\
2 & 1-\lambda & -2\\
3 & 2 & 1-\lambda
\end{pmatrix}
\end{equation*}

\begin{equation*}
\det(A-\lambda I)=(1-\lambda)[(1-\lambda)^2+4]=0
\end{equation*}
\begin{equation*}
\end{equation*}
\begin{equation*}
\lambda_1=1~~~~\lambda_2=1+2i~~~~\lambda_2=1-2i
\end{equation*}

When $\lambda_1=1$
\begin{equation*}
\begin{pmatrix}
0 & 0 & 0\\
2 & 0 & -2\\
3 & 2 & 0
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=0
\end{equation*}
let x1 = t
\begin{equation*}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=t
\begin{pmatrix}
1\\
-3/2\\
1
\end{pmatrix}
\end{equation*}

When $\lambda_2=1+2i$
\begin{equation*}
\begin{pmatrix}
-2i & 0 & 0\\
2 & -2i & -2\\
3 & 2 & -2i
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=0
\end{equation*}
let x2 = t
\begin{equation*}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=t
\begin{pmatrix}
0\\
i\\
1
\end{pmatrix}
\end{equation*}

When $\lambda_3=1-2i$
\begin{equation*}
\begin{pmatrix}
2i & 0 & 0\\
2 & 2i & -2\\
3 & 2 & 2i
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=0
\end{equation*}
let x2 = t
\begin{equation*}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=t
\begin{pmatrix}
0\\
1\\
i
\end{pmatrix}
\end{equation*}

\begin{equation*}
e^{(1+2i)t}\begin{pmatrix}
0\\
i\\
1
\end{pmatrix}=\begin{pmatrix}
0\\
i\\
1
\end{pmatrix}e^te^{2it}=\begin{pmatrix}
0\\
icos2t-sin2t\\
cos2t+isin2t
\end{pmatrix}e^t=e^t\begin{pmatrix}
0\\
-sin2t\\
cos2t
\end{pmatrix}+ie^t\begin{pmatrix}
0\\
cos2t\\
sin2t
\end{pmatrix}
\end{equation*}

\begin{equation*}
e^{(1-2i)t}\begin{pmatrix}
0\\
1\\
i
\end{pmatrix}=\begin{pmatrix}
0\\
1\\
i
\end{pmatrix}e^te^{-2it}=\begin{pmatrix}
0\\
cos(-2t)+isin(-2t)\\
-icos(-2t)-sin(-2t)
\end{pmatrix}e^t=e^t\begin{pmatrix}
0\\
cos(-2t)\\
-sin(-2t)
\end{pmatrix}+ie^t\begin{pmatrix}
0\\
sin(-2t)\\
cos(-2t)
\end{pmatrix}
\end{equation*}

so, the real-valued function is
\begin{equation*}
x(t)=C_1e^t\begin{pmatrix}
1\\
-3/2\\
1
\end{pmatrix}+C_2e^t\begin{pmatrix}
0\\
-sin2t\\
cos2t
\end{pmatrix}+C_3e^t\begin{pmatrix}
0\\
cos(-2t)\\
sin(2t)
\end{pmatrix}
\end{equation*}

3
##### Quiz-6 / Re: Q6 TUT 5101
« on: November 17, 2018, 04:23:23 PM »
(a)
\begin{equation*}
A-\lambda I=\begin{pmatrix}
2-\lambda & -5\\
\alpha & -2-\lambda
\end{pmatrix}
\end{equation*}
(b)
\begin{equation*}
\det(A-\lambda I)=(2-\lambda)(-2-\lambda)+5\alpha=\lambda^2+5\alpha-4=0
\end{equation*}
\begin{equation*}
\lambda=\frac{0\pm \sqrt{0^2-4(5\alpha-4)}}{2}=\frac{0\pm \sqrt{-4(5\alpha-4)}}{2}=\frac{0\pm 2\sqrt{-(5\alpha-4)}}{2}
\end{equation*}
\begin{equation*}
\lambda_1=\sqrt{4-5\alpha}~~~~\lambda_2=-\sqrt{4-5\alpha}
\end{equation*}

\begin{equation*}\begin{split}
b^2-4ac&\geq0\\
4-5\alpha&\geq0\\
-5\alpha&\geq-4\\
\alpha&\leq\frac{4}{5}
\end{split}
\end{equation*}
when $\alpha<\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are two different real numbers.
when $\alpha>\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are solutions with complex numbers.
when $\alpha=\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are repeated roots, which are 0.
(c) in attachments

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