# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 03:57:36 PM

Title: Q6 TUT 0701
Post by: Victor Ivrii on November 17, 2018, 03:57:36 PM
Find the general solution of the given system of equations:
$$\mathbf{x}'= \begin{pmatrix} 3 &2 &4\\ 2 &0 &2\\ 4 &2 &3 \end{pmatrix}\mathbf{x}.$$
Title: Re: Q6 TUT 0701
Post by: Guanyao Liang on November 17, 2018, 04:05:13 PM
Title: Re: Q6 TUT 0701
Post by: Qinger Zhang on November 17, 2018, 04:07:11 PM
Title: Re: Q6 TUT 0701
Post by: cindy_wen on November 17, 2018, 04:24:19 PM
here is my solution
Title: Re: Q6 TUT 0701
Post by: Tzu-Ching Yen 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
$x = c_1e^{8t}x_1 + e^{-t}(c_2x_2 + c_3x_3)$
Title: Re: Q6 TUT 0701
Post by: Qi Cui on November 17, 2018, 07:10:29 PM
$$det(A-\lambda I) = \left| \begin {array}{ccc} {3- \lambda}&2&4\\ 2& {- \lambda}&2\\ 4&2&{3- \lambda} \end {array} \right| = 0$$
$${-\lambda}^3 + 6 {\lambda}^2+ 15{\lambda}+8 = 0$$
$$By\ long\ devision\ method, we\ get\ -({\lambda+1})^{2}(\lambda-8) = 0$$
$$\quad\therefore \lambda = -1,-1,8$$
$when\ \lambda = 8:$
$$(A-\lambda I)x = 0$$
$$\left[ \begin {array}{ccc} -5&2&4\\ 2&-8&2\\ 4&2&-5 \end {array} \right]x = 0$$
$$By\ row\ operation, we\ get: \left[ \begin {array}{ccc} 1&0&-1\\ 0&2&-1\\ 0&0&0 \end {array} \right]\left[ \begin {array}{c} x_1\\ x_2\\ x_3 \end {array} \right]= 0$$
$let x_3 = t:$
$$x_1= t$$
$$2x_2= t$$
$we\ have$:
$$\left[ \begin {array}{c} 2\\ 1\\ 2 \end {array} \right]$$
$When \lambda = -1:$
$$\left[ \begin {array}{ccc} 4&2&4\\ 2&1&2\\ 4&2&4 \end {array} \right] x=0$$
$$By\ row\ operation, we\ get: \left[ \begin {array}{ccc} 2&1&2\\ 0&0&0\\ 0&0&0 \end {array} \right]\left[ \begin {array}{c} x_1\\ x_2\\ x_3 \end {array} \right]=0$$
$let\ x_3=t,x_2=s:$
$$2x_1=-s-2t$$
$$x_1={{-1}\over {2}}s-t$$
$$x_2=s$$
$$x_3=t$$
$we\ have:$
$$\left[ \begin {array}{c} {{-1}\over {2}}s-t\\ s\\ t \end {array} \right] = t\left[ \begin {array}{c} -1\\ 0\\ 1 \end {array} \right] +s \left[ \begin {array}{c} {{-1}\over {2}}\\ 1\\ 0 \end {array} \right]$$
$\quad\therefore we\ have\ eigenvector: \left[ \begin {array}{c} {{-1}\over {2}}\\ 1\\ 0 \end {array} \right] , \left[ \begin {array}{c} -1\\ 0\\ 1 \end {array} \right]$
$$\quad\therefore x(t) = c_1e^{8t}\left( \begin {array}{c} 2\\ 1\\ 2 \end {array} \right)+c_2e^{-t} \left( \begin {array}{c} {{-1}\over2}\\ 1\\ 0 \end {array} \right)+c_3e^{-t}\left( \begin {array}{c} -1\\ 0\\ 1 \end {array} \right)$$
Title: Re: Q6 TUT 0701
Post by: Zoran 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.
Title: Re: Q6 TUT 0701
Post by: Victor Ivrii 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