Author Topic: Q6 TUT 0701  (Read 1740 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2553
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q6 TUT 0701
« on: November 17, 2018, 03:57:36 PM »
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

  • Jr. Member
  • **
  • Posts: 13
  • Karma: 12
    • View Profile
Re: Q6 TUT 0701
« Reply #1 on: November 17, 2018, 04:05:13 PM »
answer
« Last Edit: November 17, 2018, 07:19:28 PM by Guanyao Liang »

Qinger Zhang

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 0
    • View Profile
Re: Q6 TUT 0701
« Reply #2 on: November 17, 2018, 04:07:11 PM »
Here is my answer.
« Last Edit: November 17, 2018, 04:11:59 PM by Qinger Zhang »

cindy_wen

  • Newbie
  • *
  • Posts: 1
  • Karma: 2
    • View Profile
Re: Q6 TUT 0701
« Reply #3 on: November 17, 2018, 04:24:19 PM »
here is my solution

Tzu-Ching Yen

  • Full Member
  • ***
  • Posts: 31
  • Karma: 22
    • View Profile
Re: Q6 TUT 0701
« Reply #4 on: November 17, 2018, 04:59:10 PM »
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
$x = c_1e^{8t}x_1 + e^{-t}(c_2x_2 + c_3x_3)$
« Last Edit: November 18, 2018, 07:07:16 PM by Thomson Yen »

Qi Cui

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 11
    • View Profile
Re: Q6 TUT 0701
« Reply #5 on: November 17, 2018, 07:10:29 PM »
$$det(A-\lambda I) = \left|
\begin {array}{ccc}
  {3- \lambda}&2&4\\
  2& {- \lambda}&2\\
  4&2&{3- \lambda}
\end {array}
\right| = 0$$
$${-\lambda}^3 + 6 {\lambda}^2+  15{\lambda}+8 = 0 $$
$$By\ long\ devision\ method, we\ get\ -({\lambda+1})^{2}(\lambda-8) = 0$$
$$\quad\therefore \lambda = -1,-1,8$$
$when\ \lambda = 8:$
$$(A-\lambda I)x = 0$$
$$\left[
\begin {array}{ccc}
  -5&2&4\\
  2&-8&2\\
  4&2&-5
\end {array}
\right]x = 0
$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  1&0&-1\\
  0&2&-1\\
  0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]= 0$$
$let x_3 = t:$
$$x_1= t$$
$$2x_2= t$$
$we\ have$:
$$\left[
\begin {array}{c}
  2\\
  1\\
  2
\end {array}
\right]$$
$When \lambda = -1:$
$$\left[
\begin {array}{ccc}
  4&2&4\\
  2&1&2\\
  4&2&4
\end {array}
\right] x=0$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
  2&1&2\\
  0&0&0\\
  0&0&0
\end {array}
\right]\left[
\begin {array}{c}
  x_1\\
  x_2\\
  x_3
\end {array}
\right]=0$$
$let\ x_3=t,x_2=s:$
$$2x_1=-s-2t$$
$$x_1={{-1}\over {2}}s-t$$
$$x_2=s$$
$$x_3=t$$
$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

  • Newbie
  • *
  • Posts: 2
  • Karma: 0
    • View Profile
Re: Q6 TUT 0701
« Reply #6 on: November 18, 2018, 05:47:55 PM »
i agree with the calculation from Qi Cui, but the final answer should be written as attached.

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2553
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Q6 TUT 0701
« Reply #7 on: November 25, 2018, 09:32:47 AM »
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
« Last Edit: November 25, 2018, 09:36:32 AM by Victor Ivrii »