MAT244--2018F > Quiz-6

Q6 TUT 0201

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Victor Ivrii:
 Express the general solution of the given system of equations in terms of real-valued functions:
$$\mathbf{x}' = \begin{pmatrix}
1 &0 &0\\
2 &1 &-2\\
3 &2 &1
\end{pmatrix}\mathbf{x}.$$

Guanyao Liang:
This is my answer

Zhuojing Yu:
Here is my solution and I think it is a quicker way. Sorry for the inconvenience of four photos.

Nikita Dua:
My solution

Siran Wang:
 \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*}

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