Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 03:54:44 PM
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Express the general solution of the given system of equations in terms of real-valued functions:
$$\mathbf{x}' = \begin{pmatrix}
-3 &0 &2\\
1 &-1 &0\\
-2 &-1 &0
\end{pmatrix}\mathbf{x}.$$
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\begin{equation*}
det
\begin{pmatrix}
-3\lambda &0 &2 \\
1 & -1\lambda &0 \\
-2 & -1 & -\lambda
\end{pmatrix}
=-\lambda^3-4\lambda^2-7\lambda-6=-(\lambda+2)(\lambda+2\lambda+3)=0
\end{equation*}
$$
\lambda=-2,\lambda=\sqrt{2}\qquad i-1,\lambda=-\sqrt{2}\qquad i-1
$$
when $\lambda$=-2
\begin{equation*}
\begin{pmatrix}
-1 &0 &2 \\
1 & 1 &0 \\
-2 & -1 & 2
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}=0
\end{equation*}
$$
\text{let } x_3=t,x_1=2t,x_2=-2t,
x=
\begin{pmatrix}
2 \\ -2 \\ 1
\end{pmatrix}t
$$
when $\lambda=\sqrt{2}\qquad i-1$
\begin{equation*}
\begin{pmatrix}
-2-\sqrt{2}\qquad i &0 &2 \\
1 & \sqrt{2}\qquad i &0 \\
-2 & -1 & -\sqrt{2}\qquad i+1
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$
x=\begin{pmatrix} \frac{2}{3}-\frac{i\sqrt{2}\qquad}{3}\\\frac{-1}{3}-\frac{i\sqrt{2}\qquad}{3} \\1 \end{pmatrix}t
$$
\begin{equation*}
x(t)=c_1e^{-2t}
\begin{pmatrix}
2\\-2\\1
\end{pmatrix}
+c_2e^{-t}
\begin{pmatrix}
\frac{2}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
-\frac{1}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
\cos\sqrt{2}\theta
\end{pmatrix}
+c_3e^{-t}i
\begin{pmatrix}
\frac{2}{3}\sin \sqrt{2}\theta-\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
-\frac{1}{3}\sin \sqrt{2}\theta+\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
\sin\sqrt{2}\theta
\end{pmatrix}
\end{equation*}
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Quiz 6
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Answer
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$$det(A-\lambda I) = \left|
\begin {array}{ccc}
{-3- \lambda}&0&2\\
1& {-1- \lambda}&0\\
-2&-1&{- \lambda}
\end {array}
\right| = 0$$
$${\lambda}^3 + 4 {\lambda}^2+ 7{\lambda}+6 = 0 $$
$$By\ long\ devision\ method, we\ get\ (\lambda+2)({\lambda}^2+ 2{\lambda} +3) = 0$$
$$({\lambda}^2+ 2{\lambda} +3) : \lambda = {-2 \pm \sqrt{-8} \over 2} = -1\pm \sqrt{2}i$$
$$\quad\therefore \lambda = -2; {-1\pm \sqrt{2}i}$$
$when\ \lambda = -2:$
$$(A-\lambda I)x = 0$$
$$\left[
\begin {array}{ccc}
-1&0&2\\
1&1&0\\
-2&-1&2
\end {array}
\right]x = 0
$$
$$By\ row\ operation, we\ get: \left[
\begin {array}{ccc}
-2&-1&2\\
-1&0&2\\
-0&0&0
\end {array}
\right]\left[
\begin {array}{c}
x_1\\
x_2\\
x_3
\end {array}
\right]= 0$$
$let x_3 = t:$
$$-2x_1-x_2+2t= 0$$
$$-x_1+2t= 0$$
$we\ have$:
$$span(\left[
\begin {array}{c}
2\\
-2\\
1
\end {array}
\right])$$
$By\ similar\ procedure\ as\ above\ shown,we\ can\ get\ other\ two\ eigenvector\ for \lambda = -1\pm \sqrt{2}i : span(\left[
\begin {array}{c}
2- \sqrt{2}i\\
-1- \sqrt{2}i\\
1
\end {array}
\right], \left[
\begin {array}{c}
2+ \sqrt{2}i\\
-1+ \sqrt{2}i\\
1
\end {array}
\right] )$
$$\quad\therefore x(t) = c_1e^{-2t}\left(
\begin {array}{c}
2\\
-2\\
1
\end {array}
\right)+c_2e^{(-1+ \sqrt{2}i)t }\left(
\begin {array}{c}
2- \sqrt{2}i\\
-1- \sqrt{2}i\\
1
\end {array}
\right)+c_3e^{(-1- \sqrt{2}i)t}\left(
\begin {array}{c}
2+ \sqrt{2}i\\
-1+ \sqrt{2}i\\
1
\end {array}
\right)$$
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here is my solution in pdf using Latex
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No need to post solutions after a perfect one is posted