MAT244--2018F > Quiz-6

Q6 TUT 0701

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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