Q6 TUT 0701

[Victor Ivrii]

Find the general solution of the given system of equations:
$$\mathbf{x}'=
\begin{pmatrix}
3 &2 &4\\
2 &0 &2\\
4 &2 &3
\end{pmatrix}\mathbf{x}.$$

[Guanyao Liang]

answer

[Qinger Zhang]

Here is my answer.

[cindy_wen]

here is my solution

[Tzu-Ching Yen]

det($M - rI$) gives

$ (3-r)(r^2 - 3r - 4) - 2(-2r -2) + 4(4 + 4r) = (r+1)((3-r)(r-4)-4 + 16) = -(r+1)(r^2-7r -8) = -(r+1)^2(r-8) $

Set this to be zero, $r = -1, 8$

Let $r = 8$, matrix is

$
    M=
   \left[ {\begin{array}{ccc}
    -5 & 2 & 4 \\
    2 & -8 & 2 \\
    4 & 2 & -5 \\
   \end{array} } \right]
$

By inspection solution is $x_1 = [2, 1, 2]$

Let $r = -1$

$
    M=
   \left[ {\begin{array}{ccc}
    4 & 2 & 4 \\
    2 & 1 & 2 \\
    4 & 2 & 4 \\
   \end{array} } \right]
 $

gives equation $r = -2s - 2t$ where solution is $[r, s, t]$. Hence two solutions are $x_2 = [1, -2, 0]$ and $x_3 = [0, -2, 1]$

General solution is therefore

$x = c_1e^{8t}x_1 + e^{-t}(c_2x_2 + c_3x_3)$

[Qi Cui]

$$det(A-\lambda I) = \left|
\begin {array}{ccc}
  {3- \lambda}&2&4\\
  2& {- \lambda}&2\\
  4&2&{3- \lambda}
\end {array}
\right| = 0$$
$${-\lambda}^3 + 6 {\lambda}^2+  15{\lambda}+8 = 0 $$
$$By\ long\ devision\ method, we\ get\ -({\lambda+1})^{2}(\lambda-8) = 0$$
$$\quad\therefore \lambda = -1,-1,8$$
$when\ \lambda = 8:$
$$(A-\lambda I)x = 0$$
$$\left[
\begin {array}{ccc}
  -5&2&4\\
  2&-8&2\\
  4&2&-5
\end {array}
\right]x = 0
$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  1&0&-1\\
  0&2&-1\\
  0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]= 0$$
$let x_3 = t:$
$$x_1= t$$
$$2x_2= t$$
$we\ have$:
$$\left[
\begin {array}{c}
  2\\
  1\\
  2
\end {array}
\right]$$
$When \lambda = -1:$
$$\left[
\begin {array}{ccc}
  4&2&4\\
  2&1&2\\
  4&2&4
\end {array}
\right] x=0$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  2&1&2\\
  0&0&0\\
  0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]=0$$
$let\ x_3=t,x_2=s:$
$$2x_1=-s-2t$$
$$x_1={{-1}\over {2}}s-t$$
$$x_2=s$$
$$x_3=t$$
$we\ have: $
$$\left[
\begin {array}{c}
  {{-1}\over {2}}s-t\\
  s\\
  t
\end {array}
\right] = t\left[
\begin {array}{c}
  -1\\
  0\\
  1
\end {array}
\right] +s \left[
\begin {array}{c}
  {{-1}\over {2}}\\
  1\\
  0
\end {array}
\right]  $$
$\quad\therefore we\ have\ eigenvector: \left[
\begin {array}{c}
  {{-1}\over {2}}\\
  1\\
  0
\end {array}
\right] , \left[
\begin {array}{c}
  -1\\
  0\\
  1
\end {array}
\right] $
$$\quad\therefore x(t) = c_1e^{8t}\left(
\begin {array}{c}
  2\\
  1\\
  2
\end {array}
\right)+c_2e^{-t} \left(
\begin {array}{c}
  {{-1}\over2}\\
  1\\
  0
\end {array}
\right)+c_3e^{-t}\left(
\begin {array}{c}
  -1\\
  0\\
  1
\end {array}
\right)$$

[Zoran]

i agree with the calculation from Qi Cui, but the final answer should be written as attached.

[Victor Ivrii]

Thomson
You wrote solution for $x_1$ (or $x_2$, or $x_3$); the same solution is for two remaining components albeit with different constants and we require the complete solution (in the vector form)

Qi Leave text out of MatJax

Cindy
I cannot identify you with Quercus name and as is the credit will go nowhere