Q6 TUT 0501

[Victor Ivrii]

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

[Boyu Zheng]

\begin{equation*}
det
      \begin{pmatrix}
       -3\lambda    &0          &2 \\
       1         & -1\lambda    &0 \\
       -2         & -1            & -\lambda
      \end{pmatrix}
=-\lambda^3-4\lambda^2-7\lambda-6=-(\lambda+2)(\lambda+2\lambda+3)=0
\end{equation*}
$$
\lambda=-2,\lambda=\sqrt{2}\qquad i-1,\lambda=-\sqrt{2}\qquad i-1
$$

when $\lambda$=-2
\begin{equation*}
      \begin{pmatrix}
       -1          &0           &2 \\
       1         & 1             &0 \\
       -2         & -1            & 2
      \end{pmatrix}

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

when $\lambda=\sqrt{2}\qquad i-1$

   \begin{equation*}
         \begin{pmatrix}
          -2-\sqrt{2}\qquad i           &0           &2 \\
          1         & \sqrt{2}\qquad i            &0 \\
          -2         & -1            & -\sqrt{2}\qquad i+1
         \end{pmatrix}
         \begin{pmatrix}
          x_1    \\ x_2 \\ x_3
         \end{pmatrix}
         =0
   \end{equation*}
   $$
   x=\begin{pmatrix} \frac{2}{3}-\frac{i\sqrt{2}\qquad}{3}\\\frac{-1}{3}-\frac{i\sqrt{2}\qquad}{3} \\1   \end{pmatrix}t
   $$
\begin{equation*}
x(t)=c_1e^{-2t}
\begin{pmatrix}
2\\-2\\1
\end{pmatrix}
+c_2e^{-t}
\begin{pmatrix}
 \frac{2}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
 -\frac{1}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
 \cos\sqrt{2}\theta
\end{pmatrix}
+c_3e^{-t}i
\begin{pmatrix}
 \frac{2}{3}\sin \sqrt{2}\theta-\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
 -\frac{1}{3}\sin \sqrt{2}\theta+\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
 \sin\sqrt{2}\theta
\end{pmatrix}
\end{equation*}

[Ruiling Zhao]

Quiz 6

[Xiaoyuan Wang]

Answer

[Qi Cui]

$$det(A-\lambda I) = \left|
\begin {array}{ccc}
  {-3- \lambda}&0&2\\
  1& {-1- \lambda}&0\\
  -2&-1&{- \lambda}
\end {array}
\right| = 0$$
$${\lambda}^3 + 4 {\lambda}^2+  7{\lambda}+6 = 0 $$
$$By\ long\ devision\ method, we\ get\ (\lambda+2)({\lambda}^2+ 2{\lambda} +3) = 0$$
$$({\lambda}^2+ 2{\lambda} +3) : \lambda = {-2 \pm \sqrt{-8} \over 2} = -1\pm \sqrt{2}i$$
$$\quad\therefore \lambda = -2;   {-1\pm \sqrt{2}i}$$
$when\ \lambda = -2:$
$$(A-\lambda I)x = 0$$
$$\left[
\begin {array}{ccc}
  -1&0&2\\
  1&1&0\\
  -2&-1&2
\end {array}
\right]x = 0
$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  -2&-1&2\\
  -1&0&2\\
  -0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]= 0$$
$let x_3 = t:$
$$-2x_1-x_2+2t= 0$$
$$-x_1+2t= 0$$
$we\ have$:
$$span(\left[
\begin {array}{c}
  2\\
  -2\\
  1
\end {array}
\right])$$
$By\ similar\ procedure\ as\ above\ shown,we\ can\ get\ other\ two\ eigenvector\ for \lambda = -1\pm \sqrt{2}i : span(\left[
\begin {array}{c}
  2- \sqrt{2}i\\
  -1- \sqrt{2}i\\
  1
\end {array}
\right], \left[
\begin {array}{c}
  2+ \sqrt{2}i\\
  -1+ \sqrt{2}i\\
  1
\end {array}
\right] )$
$$\quad\therefore x(t) = c_1e^{-2t}\left(
\begin {array}{c}
  2\\
  -2\\
  1
\end {array}
\right)+c_2e^{(-1+ \sqrt{2}i)t }\left(
\begin {array}{c}
  2- \sqrt{2}i\\
  -1- \sqrt{2}i\\
  1
\end {array}
\right)+c_3e^{(-1- \sqrt{2}i)t}\left(
\begin {array}{c}
  2+ \sqrt{2}i\\
  -1+ \sqrt{2}i\\
  1
\end {array}
\right)$$

[Cheng Qian]

here is my solution in pdf using Latex

[Victor Ivrii]

No need to post solutions after a perfect one is posted