### Author Topic: Q6 TUT 0201  (Read 2623 times)

#### Victor Ivrii

• Elder Member
• Posts: 2572
• Karma: 0
##### Q6 TUT 0201
« on: November 17, 2018, 03:52:00 PM »
Express the general solution of the given system of equations in terms of real-valued functions:
$$\mathbf{x}' = \begin{pmatrix} 1 &0 &0\\ 2 &1 &-2\\ 3 &2 &1 \end{pmatrix}\mathbf{x}.$$
« Last Edit: November 17, 2018, 04:02:23 PM by Victor Ivrii »

#### Guanyao Liang

• Jr. Member
• Posts: 13
• Karma: 12
##### Re: Q6 TUT 0201
« Reply #1 on: November 17, 2018, 03:54:30 PM »

#### Zhuojing Yu

• Newbie
• Posts: 3
• Karma: 2
##### Re: Q6 TUT 0201
« Reply #2 on: November 17, 2018, 04:11:25 PM »
Here is my solution and I think it is a quicker way. Sorry for the inconvenience of four photos.

#### Nikita Dua

• Jr. Member
• Posts: 14
• Karma: 0
##### Re: Q6 TUT 0201
« Reply #3 on: November 17, 2018, 04:14:07 PM »
My solution

#### Siran Wang

• Newbie
• Posts: 3
• Karma: 3
##### Re: Q6 TUT 0201
« Reply #4 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*}
« Last Edit: November 17, 2018, 05:24:36 PM by siran »

#### Cheng Qian

• Newbie
• Posts: 2
• Karma: 0
##### Re: Q6 TUT 0201
« Reply #5 on: November 17, 2018, 06:08:42 PM »
Here is my version in pdf using Latex.