MAT244--2018F > Quiz-6

Q6 TUT 0701

<< < (2/2)

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

Navigation

[0] Message Index

[*] Previous page