Author Topic: Q6 TOT 0301  (Read 1773 times)

Victor Ivrii

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Q6 TOT 0301
« on: November 17, 2018, 03:52:31 PM »
Find the general solution of the given system of equations:
$$\mathbf{x}'=
\begin{pmatrix}
1 &1 &2\\
1 &2 &1\\
2 &1 &1
\end{pmatrix}\mathbf{x}.$$

Guanyao Liang

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Re: Q6 TOT 0301
« Reply #1 on: November 17, 2018, 03:55:29 PM »
This is my answer!

Boyu Zheng

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Re: Q6 TOT 0301
« Reply #2 on: November 17, 2018, 04:10:39 PM »
\begin{equation*}
det
      \begin{pmatrix}
       1-\lambda    &1           &2 \\
       1         & 2-\lambda    &1 \\
       2         & 1            & 1-\lambda
      \end{pmatrix}
=-\lambda^3+4\lambda^2+\lambda-4=-(\lambda-1)(\lambda-4)(\lambda+1)=0
\end{equation*}
$$
\lambda=1,\lambda=4,\lambda=-1
$$

when $\lambda$=1
\begin{equation*}
      \begin{pmatrix}
       0           &1           &2 \\
       1         & 1             &1 \\
       2         & 1            & 0
      \end{pmatrix}
      \sim
         \begin{pmatrix}
          2           &1           &0 \\
          0         & 1             &2 \\
          1         & 1            & 1
         \end{pmatrix}
      \sim
            \begin{pmatrix}
             2           &0           &-2 \\
             -1         & 0             &1 \\
             1         & 1            & 1
            \end{pmatrix}
            \sim
               \begin{pmatrix}
                   2           &0           &-2 \\
                   0         & 0             &0 \\
                   1         & 1            & 1
               \end{pmatrix}
               \sim
                  \begin{pmatrix}
                   2           &0           &-2 \\
                   1         & 1             &1 \\
                   0         & 0            & 0
                  \end{pmatrix}
                  \sim
                     \begin{pmatrix}
       x_1            \\       x_2       \\       x_3      
      \end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=t,x_2=-2t
x=
   \begin{pmatrix}
       1           \\    -2 \\ 1
      \end{pmatrix}t
$$

 when $\lambda$=4
   \begin{equation*}
         \begin{pmatrix}
          -3           &1           &2 \\
          1         & -2             &1 \\
          2         & 1            & -3
         \end{pmatrix}
         \sim
         \begin{pmatrix}
          -3           &1           &2 \\
          0         & -5             &5 \\
          2         & 1            & -3
         \end{pmatrix}
         \begin{pmatrix}
          x_1    \\ x_2 \\ x_3
         \end{pmatrix}
         =0
   \end{equation*}
   $$
   x=\begin{pmatrix} 1   \\1 \\1   \end{pmatrix}t
   $$
when $\lambda$=-1
\begin{equation*}
         \begin{pmatrix}
          2           &1           &2 \\
          1         & 3             &1 \\
          2         & 1            & 2
         \end{pmatrix}
         \sim
         \begin{pmatrix}
          2           &1           &2 \\
          0         & 5             &0 \\
          0         & 0            & 0
         \end{pmatrix}
         \begin{pmatrix}
          x_1    \\ x_2 \\ x_3
         \end{pmatrix}
         =0
   \end{equation*}
$$
x=\begin{pmatrix}-1\\0\\1\end{pmatrix}
$$
$$
x(t)=c_1e^4t\begin{pmatrix}1\\1\\1\end{pmatrix}+c_2e^{-t}\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3e^t\begin{pmatrix}1\\-2\\1\end{pmatrix}
$$
« Last Edit: November 17, 2018, 04:12:53 PM by Boyu Zheng »

Chonghan Ma

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Re: Q6 TOT 0301
« Reply #3 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.

Mengfan Zhu

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Re: Q6 TOT 0301
« Reply #4 on: November 20, 2018, 01:25:22 AM »
Let's solve this question step by step!!!