4.6E: Exercises

Last updated
Save as PDF

Page ID: 18479

This page is a draft and is under active development.

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

In Exercises $(4.6E.1)$ to $(4.6E.16)$, find the general solution.

Exercise $\PageIndex{1}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & 2 \\ {-5} & 5 \end{array} \right] {\bf y} }$

Exercise $\PageIndex{2}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-11} & 4 \\ {-26} & 9 \end{array} \right] {\bf y} }$

Answer: $\left|\begin{array}{cc}-11-\lambda&4\\-26&9-\lambda
\end{array}\right|=(\lambda+1)^2+4$ .
The augmented matrix of
$\left(A-\left(-1+2i\right)I\right){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}-10-2i&4&\vdots&0\\
-26&10-2i&\vdots&0 \end{array}\right]$ ,
which is row equivalent to
$\left[\begin{array}{cccr} 1&{-5+i\over13}&\vdots&0\\
0&0&\vdots&0
\end{array}\right]$ .
Therefore,$x_1=(5-i)x_2/13$ . Taking $x_2=13$ yields the eigenvector
${\bf x}=\left[\begin{array}{c}5-i\\13\end{array}\right]$ .
Taking real and imaginary parts of
$e^{-t}(\cos2t+i\sin2t)
\left[\begin{array}{c}5-i\\13\end{array}\right]$ yields
$$ {\bf y}=\displaystyle{c_1e^{-t}\left[\begin{array}{c}{5\cos2t+\sin2t}\\{13\cos2t}\end{array}\right]
+c_2e^{-t}\left[\begin{array}{c}{5\sin2t-\cos2t}\\{13\sin2t}\end{array}\right]}.
$$

Exercise $\PageIndex{3}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ {-4} & 5 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{4}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & {-6} \\ 3 & {-1} \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc} 5-\lambda&-6\\ 3&-1-\lambda
\end{array}\right|=(\lambda-2)^2+9$ .
Hence, $\lambda=2+3i$ is an eigenvalue of $A$ . The associated
eigenvectors satisfy $\left(A-\left(2+3i\right)I\right){\bf x}={\bf
0}$ . The augmented matrix of this system is
$\left[\begin{array}{cccr} 3-3i&-6&\vdots&0\\
3&-3-3i&\vdots&0 \end{array}\right]$ ,
which is row equivalent to
$\left[\begin{array}{cccr} 1&-1-i&\vdots&0\\
0&0&\vdots&0
\end{array}\right]$ .
Therefore,$x_1=(1+i)x_2$ . Taking $x_2=1$ yields $x_1=1+i$ , so
${\bf x}=\left[\begin{array}{c}1+i\\1\end{array}\right]$
is an eigenvector. Taking real and imaginary parts of
$e^{2t}(\cos3t+i\sin3t)
\left[\begin{array}{c}1+i\\1\end{array}\right]$ yields
${\bf y}=\displaystyle{c_1e^{2t}\left[\begin{array}{c}{\cos3t-\sin3t}\\{\cos3t}\end{array}\right]
+c_2e^{2t}\left[\begin{array}{c}{\sin3t+\cos3t}\\{\sin3t}\end{array}\right]}$ .

Exercise $\PageIndex{5}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & {-3} & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{6}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & 3 & 1 \\ 1 & {-5} & {-3} \\ {-3} & 7 & 3 \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{ccc}-3-\lambda&3&1\\1&-5-\lambda&-3\\-3
&7&3-\lambda\end{array}\right|=-(\lambda+1)\left((\lambda+2)^2+4\right)$ .
The augmented matrix of $ (A+I){\bf x=0}$ is
$\left[\begin{array}{rrrcr}-2&3&1&\vdots&0\\1&-4&-3&\vdots&0\\
-3&7&4&\vdots&0\end{array}\right]$ ,
which is row equivalent to
$\left[\begin{array}{rrrcr} 1&0&1&\vdots&0\\ 0&1&1&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$ .
Therefore,$x_1=x_2=-x_3$ . Taking $x_3=1$ yields ${\bf y}_1= \left[ \begin{array}{c}{-1}\\{-1}\\1\end{array}\right]e^{-t}$ .

The augmented matrix of
$\left(A-(-2+2i)I\right){\bf x=0}$ is
$\left[\begin{array}{ccccc}-1-2i&3&1&\vdots&0\\1&
-3-2i&-3&\vdots&0\\-3&7&5-2i&\vdots&0
\end{array}\right]$ ,
which is row equivalent to
$\left[\begin{array}{ccccc} 1&0&-{1+i\over2}&\vdots&0\\
0&1&{1-i\over2}&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$ .
Therefore,$x_1=\displaystyle{(1+i)\over2}x_3$ and $x_2=-\displaystyle{(1-i)\over2}x_3$ .
Taking $x_3=2$ yields the eigenvector
${\bf x}_2=\left[\begin{array}{c}1+i\\-1+i\\2\end{array}\right]$ .
The real and imaginary parts of
$e^{-2t}(\cos2t+i\sin2t)\left[\begin{array}{c}1+i\\-1+i\\2\end{array}\right]$
are
${\bf y}_2=e^{-2t}\left[\begin{array}{c}{\cos2t-\sin2t}\\{-\cos2t-\sin2t}\\{2\cos2t}\end{array}\right]$
and
${\bf y}_3=e^{-2t}\left[\begin{array}{c}{\sin2t+\cos2t}\\{-\sin2t+\cos2t}\\{2\sin2t}\end{array}\right]$ .
Therefore,
$
\displaystyle{{\bf y}=c_1\left[\begin{array}{c}{-1}\\{-1}\\1\end{array}\right]e^{-t}+c_2e^{-2t}\left[\begin{array}{c}{\cos2t-\sin 2t}\\{-\cos2t-\sin 2t}\\{2\cos2t}\end{array}\right]+c_3e^{-2t}\left[\begin{array}{c} {\sin2t+\cos2t}\\{-\sin2t+\cos2t}\\{2\sin2t}\end{array}\right]}.
$

Exercise $\PageIndex{7}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 2 & 1 & {-1} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{8}$

$ \displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & 1 & {-3} \\ 4 & {-1} & 2 \\ 4 & {-2} & 3 \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{ccc}-3-\lambda&1&-3\\4&-1-\lambda&2\\4
&-2&3-\lambda\end{array}\right|=-(\lambda-1)\left((\lambda+1)^2+4\right)$.
The augmented matrix of $(A-I){\bf x=0}$ is
$\left[\begin{array}{rrrcr}-4&1&-3&\vdots&0\\4&-2&2&\vdots&0\\
4&-2&2&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{rrrcr} 1&0&1&\vdots&0\\ 0&1&1&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=x_2=-x_3$. Taking $x_3=1$ yields
${\bf y}_1=\left[\begin{array}{c}{-1}\\{-1}\\1\end{array} \right]e^t$. The augmented matrix of
$\left(A-(-1+2i)I\right){\bf x=0}$ is
$\left[\begin{array}{crccr}-2-2i&1&-3&\vdots&0\\4&-2i&2&\vdots&0\\
4&-2&4-2i&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{ccccc} 1&0&{1-i\over2}&\vdots&0\\
0&1&-1&\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=-\displaystyle{1-i\over2}x_3$ and $x_2=x_3$.
Taking $x_3=2$ yields the eigenvector
${\bf x}_2=\left[\begin{array}{c}-1+i\\2\\2\end{array}\right]$.
The real and imaginary parts of
$e^{-t}(\cos2t+i\sin2t)\left[\begin{array}{c}-1+i\\2\\2\end{array}\right]$
are ${\bf y}_2=e^{-t}\left[\begin{array}{c}{-\sin2t-\cos2t}\\{2\cos2t}\\{2\cos2t} \end{array} \right]$
and ${\bf y}_3=e^{-t}\left[\begin{array}{c} {\cos2t-\sin2t}\\{2\sin2t}\\{2\sin2t}\end{array} \right]$.
Therefore,
$$
\displaystyle{{\bf y}=c_1\left[\begin{array}{c}{-1}1\\1 \end{array}\right]e^{-t}+c_2e^{-t} \left[\begin{array}{c}{-\sin 2t-\cos2t}\\{2\cos2t}\\{2\cos2t}\end{array}\right]+c_3e^{-t}\left[\begin{array}{c} {\cos2t-\sin2t}\\{2\sin2t}\\{2\sin2t} \end{array}\right]}.
$$

Exercise $\PageIndex{9}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & {-4} \\ {10} & 1 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{10}$

$\displaystyle{{\bf y}' = {1\over3} \left[ \begin{array} \\ 7 & {-5} \\ 2 & 5 \end{array} \right] {\bf y}}$

Answer: $\displaystyle{1\over3}\left|\begin{array}{cc}7-3\lambda&-5\\2&5-3\lambda
\end{array}\right|=\displaystyle{\left(\lambda-2\right)^2+1}$.
The augmented matrix of
$\displaystyle{\left(A-(2+i)I\right)}{\bf x}={\bf 0}$ is
$\displaystyle{1\over3}\left[\begin{array}{cccr}1-3i&-5&\vdots&0\\2&-1-3i&\vdots&0
\end{array}\right]$, which is row equivalent to
$\left[\begin{array}{cccr} 1&-{1+3i\over2}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$. Therefore,$x_1=\displaystyle{1+3i\over2}x_2$. Taking
$x_2=2$ yields the eigenvector
${\bf x}=\left[\begin{array}{c}1+3i\\2\end{array}\right]$.
Taking real and imaginary parts of
$e^{2t}(\cos t+i\sin t)
\left[\begin{array}{c}1+3i\\2\end{array}\right]$
yields
$$
\displaystyle{{\bf y}=c_1e^{2t}\left[\begin{array}{c}\cos t-3\sin t
\\2\cos t\end{array}\right]+
c_2e^{2t}\left[\begin{array}{c}
\sin t+3\cos t\\ 2\sin t\end{array}\right]}.
$$

Exercise $\PageIndex{11}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & 2 \\ {-5} & 1 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{12}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {34} & {52} \\ {-20} & {-30} \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}34-\lambda&52\\-20&-30-\lambda
\end{array}\right|=(\lambda-2)^2+16$. The augmented matrix of
$\left(A-\left(2+4i\right)I\right){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}32-4i&52&\vdots&0\\-20&-32-4i&\vdots&0
\end{array}\right]$, which is row equivalent to
$\left[\begin{array}{cccr} 1&{8+i\over5}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$. Therefore,$x_1=-\displaystyle{(8+i)\over5}x_2$. Taking
$x_2=5$ yields the eigenvector ${\bf
x}=\left[\begin{array}{c}-8-i\\5\end{array}\right]$.
Taking real and imaginary parts of $e^{2t}(\cos4t+i\sin
4t)\left[\begin{array}{c}-8-i\\5\end{array}\right]$ yields $\displaystyle{{\bf
y}=c_1e^{2t}\left[\begin{array}{c}\sin4t-8\cos4t\\
5\cos4t\end{array}\right]+c_2e^{2t}\left[\begin{array}{c}
-\cos4t-8\sin4t\\ 5\sin4t\end{array}\right]}$.

Exercise $\PageIndex{13}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & 1 & 2 \\ 1 & 0 & {-1} \\ {-1} & {-2} & {-1} \end{array} \right] {\bf y}}$

Exercise $\PageIndex{14}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & {-4} & {-2} \\ {-5} & 7 & {-8} \\ {-10} & {13} & {-8} \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{ccc}3-\lambda&-4&-2\\-5&7-\lambda&-8\\-10
&13&-8-\lambda\end{array}\right|=-(\lambda+2)\left((\lambda-2)^2+9\right)$.
The augmented matrix of $(A+2I){\bf x=0}$ is
$\left[\begin{array}{rrrcr}5&-4&-2&\vdots&0\\-5&9&-8&\vdots&0\\
-10&13&-6&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&-2&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=x_2=2x_3$. Taking $x_3=1$ yields
${\bf y}_1=\left[\begin{array}{c}2\\2\\1\end{array}\right]e^{-2t}$.
The augmented matrix of
$\left(A-(2+3i)I\right){\bf x=0}$ is
$\left[\begin{array}{ccccr}1-3i&-4&-2&\vdots&0\\-5&5-3i&-8&\vdots&0\\
-10&13&-10-3i&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{ccccc} 1&0&1-i&\vdots&0\\
0&1&-i&\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=-(1-i)x_3$ and $x_2=ix_3$.
Taking $x_3=1$ yields the eigenvector
${\bf x}_2=\left[\begin{array}{c}-1+i\\i\\1\end{array}\right]$.
The real and imaginary parts of $e^{2t}(\cos3t+i\sin3t)
\left[\begin{array}{c}-1+i\\i\\1\end{array}\right]$ are
${\bf y}_2=e^{2t}\left[\begin{array}{c}-\cos3t-\sin3t\\-\sin3t\\
\cos3t\end{array}\right]$ and ${\bf y}_3=
c_3e^{2t}\left[\begin{array}{c}-\sin3t+\cos3t\\\cos3t\\
\sin3t\end{array}\right]$. Therefore,
$$
\displaystyle{{\bf
y}=c_1\left[\begin{array}{c} 2\\ 2\\1 \end{array}\right]e^{-2t}+
c_2e^{2t}\left[\begin{array}{c}-\cos3t-\sin3t\\-\sin3t\\
\cos3t\end{array}\right]+c_3e^{2t}\left[\begin{array}{c}
-\sin3t+\cos3t\\\cos3t\\\sin3t\end{array}\right]}.
$$

Exercise $\PageIndex{15}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 6 & 0 & {-3} \\ {-3} & 3 & 3 \\ 1 & {-2} & 6 \end{array} \right] {\bf y}'}$

Exercise $\PageIndex{16}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & 2 & {-2} \\ 0 & 2 & {-1} \\ 1 & 0 & 0 \end{array} \right] {\bf y}'}$

Answer: $\left|\begin{array}{ccc}1-\lambda&2&-2\\0&2-\lambda&-1\\1
&0&-\lambda\end{array}\right|=-(\lambda-2)\left((\lambda-1)^2+1\right)$.
The augmented matrix of $(A-I){\bf x=0}$ is
$\left[\begin{array}{rrrcr}0&2&-2&\vdots&0\\0&
1&-1&\vdots&0\\ 1&0&-1&\vdots&0
\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{rrrcr} 1&0&-1&\vdots&0\\ 0&1&-1&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=x_2=1$. Taking $x_3=1$ yields
${\bf y}_1=\left[\begin{array}{c}1\\1\\1\end{array}\right]e^t$.
The augmented matrix of
$\left(A-(1+i)I\right){\bf x=0}$ is
$\left[\begin{array}{ccccc}-i&2&-2&\vdots&0\\0&
1-i&-1&\vdots&0\\ 1&0&-1-i&\vdots&0
\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{ccccc} 1&0&-1-i&\vdots&0\\ 0&1&-{1+i\over2}&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=(1+i)x_3$ and $x_2=\displaystyle{(1+i)\over2}x_3$.
Taking $x_3=2$ yields the eigenvector
${\bf x}_2=\left[\begin{array}{c}2+2i\\1+i\\2\end{array}\right]$.
The real and imaginary parts of
$e^{4t}(\cos t+i\sin t)\left[\begin{array}{c}2+2i
\\1+i\\2\end{array}\right]$ are
${\bf y}_2=e^t\left[\begin{array}{c} 2\cos t-2\sin t\\\cos t-\sin
t\\ 2\cos t\end{array}\right]$ and
${\bf y}_3=
c_3e^t\left[\begin{array}{ccc} 2\sin t+2\cos t\\\cos t+\sin t\\
2\sin t\end{array}\right]$. Therefore,
$$
\displaystyle{{\bf y}=c_1\left[\begin{array}{c} 1\\1\\1
\end{array}\right]e^t+
c_2e^t\left[\begin{array}{c} 2\cos t-2\sin t
\\\cos t-\sin t\\ 2\cos t\end{array}\right]+
c_3e^t\left[\begin{array}{ccc} 2\sin t+2\cos t\\\cos t+\sin t\\
2\sin t\end{array}\right]}.
$$

In Exercises $(4.6E.17)$ to $(4.6E.24)$, solve the initial value problem.

Exercise $\PageIndex{17}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-6} \\ 3 & {-2} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] } $

Exercise $\PageIndex{18}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 7 & {15} \\ {-3} & 1 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 5 \\ 1 \end{array} \right] } $

Answer: $\left|\begin{array}{cc}7-\lambda&15\\-3&1-\lambda
\end{array}\right|=(\lambda-4)^2+36$. The augmented matrix of
$\left(A-\left(4+6i\right)I\right){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}3-6i&15&\vdots&0\\-3&-3-6i&\vdots&0
\end{array}\right]$, which is row equivalent to
$\left[\begin{array}{cccr} 1&1+2i&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$. Therefore,$x_1=-(1+2i)x_2$. Taking $x_2=1$ yields
the eigenvector ${\bf
x}=\left[\begin{array}{c}-1-2i\\1\end{array}\right]$.
Taking real and imaginary parts of
$e^{4t}(\cos6t+i\sin6t)\left[\begin{array}{c}-1-2i\\1\end{array}\right]$
yields ${\bf y}=\displaystyle{c_1e^{4t}\left[\begin{array}{c}{2\sin6t-\cos6t} \\ {\cos6t} \end{array} \right]
+c_2e^{4t}\left[\begin{array}{c}{-2\cos6t-\sin6t} \\ {\sin6t} \end{array} \right]}$. Now ${\bf
y}(0)=\left[\begin{array}{c}5\\1\end{array} \right]\Rightarrow
\left[\begin{array}{cc}{-1}\\{-2}\\10\end{array}\right]\left[\begin{array}{c}{c_1}\\{c_2}\end{array} \right]=\left[\begin{array}{c}5\\1 \end{array} \right]$, so $c_1=1$, $c_2=-3$,
and $\displaystyle{{\bf
y}=e^{4t}\left[\begin{array}{c}5\cos6t+5\sin6t\\\cos6t-3\sin6t
\end{array}\right]}$.

Exercise $\PageIndex{19}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 7 & {-15} \\ 3 & {-5} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ {17} \\ 7 \end{array} \right]} $

Exercise $\PageIndex{20}$

$\displaystyle{{\bf y}' = {1\over 6} \left[ \begin{array} \\ 4 & {-2} \\ 5 & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 1 \\ {-1} \end{array} \right] }$

Answer

$\displaystyle{1\over6}\left|\begin{array}{cc}4-6\lambda&-2\\5&2-6\lambda
\end{array}\right|=\displaystyle{\left(\lambda-{1\over2}\right)^2+{1\over4}}$.
The augmented matrix of
$\displaystyle{\left(A-{1+i\over2}I\right)}{\bf x}={\bf 0}$ is
$\displaystyle{1\over6}\left[\begin{array}{cccr}1-3i&-2&\vdots&0\\5&-1-3i&\vdots&0
\end{array}\right]$, which is row equivalent to
$\left[\begin{array}{cccr} 1&-{1+3i\over5}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$. Therefore,$x_1=\displaystyle{1+3i\over5}x_2$. Taking
$x_2=5$ yields the eigenvector
${\bf x}=\left[\begin{array}{c}1+3i\\5\end{array}\right]$.
Taking real and imaginary parts of
$e^{t/2}(\cos t/2+i\sin
t/2)\left[\begin{array}{c}1+3i\\5\end{array}\right]$
yields ${\bf y}=\displaystyle{c_1e^{t/2}\left[\begin{array}{c}{\cos t/2-3\sin t/2}\\{5\cos t/2} \end{array}\right]
+c_2e^{t/2}\left[\begin{array}{c}{\sin t/2+3\cos t/2}\\{5\sin t/2} \end{array}\right] }$.

Now ${\bf
y}(0)=\left[\begin{array}{c}1\\{-1}\end{array}\right]\Rightarrow
\left[\begin{array}{cc}1&3\\5&0\end{array}\right]\left[\begin{array}{c}{c_1}\\{c_2}\end{array}\right]=\left[\begin{array}{c}1\\{-1}\end{array}\right]$, so
$c_1=-\displaystyle{1\over5}$,
$c_2=\displaystyle{2\over5}$, and
${\bf y}=\displaystyle{e^{t/2}\left[\begin{array}{c}\cos(t/2)+
\sin(t/2)\\-\cos(t/2)+2\sin(t/2)
\end{array}\right]}$.

Exercise $\PageIndex{21}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & 2 & {-1} \\ {-3} & 2 & 2 \\ 1 & 3 & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 4 \\ 0 \\ 6 \end{array} \right]}$

Exercise $\PageIndex{22}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & 4 & 0 \\ 8 & {10} & {-20} \\ 2 & 3 & {-2} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 8 \\ 6 \\ 5 \end{array} \right] }$

Answer: $\left|\begin{array}{ccc}4-\lambda&4&0\\8&10-\lambda&-20\\2
&3&-2-\lambda\end{array}\right|=-(\lambda-8)\left((\lambda-2)^2+4\right)$.
The augmented matrix of $(A-8I){\bf x=0}$ is
$\left[\begin{array}{rrrcr}0&4&0&\vdots&0\\8&6&-20&\vdots&0\\
2&3&-6&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&-2&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=x_2=2x_3$. Taking $x_3=2$ yields
${\bf y}_1=\left[\begin{array}{c}2\\2\\1\end{array}\right]e^{8t}$.
The augmented matrix of
$\left(A-(2+2i)I\right){\bf x=0}$ is
$\left[\begin{array}{ccccc}2-2i&4&0&\vdots&0\\8&
8-2i&-20&\vdots&0\\2&3&-4-2i&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{ccccc} 1&0&-2+2i&\vdots&0\\
0&1&-2i&\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=(2-2i)x_3$ and $x_2=2ix_3$.
Taking $x_3=1$ yields the eigenvector
${\bf x}_2=\left[\begin{array}{c}2-2i\\2i\\1\end{array}\right]$.
The real and imaginary parts of
$e^{2t}(\cos2t+i\sin
2t)\left[\begin{array}{c}2-2i\\2i\\1\end{array}\right]$
are ${\bf y}_2=e^{2t}\left[\begin{array}{c}2\cos2t+2\sin2t
\\-2\sin2t\\2\cos2t\end{array}\right]$ and
${\bf y}_3=e^{2t}\left[\begin{array}{c}2\sin2t+2\cos2t
\\2\cos2t\\\sin2t\end{array}\right]$, so the general solution
is ${\bf y}=c_1\left[\begin{array}{c}2\\2\\1 \end{array}\right]e^{8t}+c_2
e^{2t}\left[\begin{array}{c}2\cos2t+2\sin2t
\\-2\sin2t\\2\cos2t\end{array}\right]+
c_3e^{2t}\left[\begin{array}{c}2\sin2t+2\cos2t
\\2\cos2t\\\sin2t\end{array}\right]$.
Now ${\bf y}(0)=\left[\begin{array}{c}8\\6\\5 \end{array} \right]\Rightarrow \left[\begin{array}{ccc}2\\2\\{-2}&2\\0\\2&1\\\\10 \end{array} \right]
\left[\begin{array}{c}{c_1}\\{c_2}\\{c_3}\end{array} \right]=\left[\begin{array}{c}8\\6\\5 \end{array} \right]$, so $c_1=2$, $c_2=3$,
$c_3=1$, and $\displaystyle{{\bf y}=\left[\begin{array}{c}4\\4\\2\end{array} \right]e^{8t}
+e^{2t}\left[\begin{array}{c}4\cos2t+8\sin2t\\
-6\sin2t+2\cos2t\\3\cos2t+\sin2t
\end{array}\right]}$.

Exercise $\PageIndex{23}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & {15} & {-15} \\ {-6} & {18} & {-22} \\ {-3} & {11} & {-15} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ {15} \\ {17} \\ {10} \end{array} \right]}$

Exercise $\PageIndex{24}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-4} & 4 \\ {-10} & 3 & {15} \\ 2 & {-3} & 1 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ {16} \\ {14} \\ 6 \end{array} \right] }$

Answer: $\left|\begin{array}{ccc}4-\lambda&-4&4\\-10&3-\lambda&15\\2
&-3&1-\lambda\end{array}\right|=-(\lambda-8)(\lambda^2+16)$.
The augmented matrix of $(A-8I){\bf x=0}$ is
$\left[\begin{array}{rrrcr}-4&-4&4&\vdots&0\\-10&-5&15&\vdots&0\\
2&-3&-7&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&1&
\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=2x_3$ and $x_2=-x_3$. Taking $x_3=1$ yields
${\bf y}_1=\left[\begin{array}{c}2\\{-1}\\1\end{array}\right]e^{8t}$.
The augmented matrix of
$\left(A-4iI\right){\bf x=0}$ is
$\left[\begin{array}{ccccr}4-4i&-4&4&\vdots&0\\-10&3-4i&15&\vdots&0\\
2&-3&1-4i&\vdots&0\end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{ccccc} 1&0&-1+i&\vdots&0\\
0&1&-1+2i&\vdots&0\\ 0&0&0&\vdots&0\end{array}\right]$.
Therefore,$x_1=(1-i)x_3$ and $x_2=(1-2i)x_3$.
Taking $x_3=1$ yields the eigenvector
${\bf x}_2=\left[\begin{array}{c}1-i\\1-2i\\1\end{array}\right]$.
The real and imaginary parts of
$(\cos4t+i\sin
4t)\left[\begin{array}{c}1-i\\1-2i\\1\end{array}\right]$
are ${\bf y}_2=\left[\begin{array}{c}\cos4t+\sin4t
\\\cos4t+2\sin4t\\\cos4t\end{array}\right]$ and
${\bf y}_3=\left[\begin{array}{c}\sin4t-\cos4t
\\\sin4t-2\cos4t\\\sin4t\end{array}\right]$, so the general solution
is ${\bf y}=c_1 \left[\begin{array}{c}2\\{-1}\\1\end{array}\right]e^{8t}+
c_2\left[\begin{array}{c}\cos4t+\sin4t
\\\cos4t+2\sin4t\\\cos4t\end{array}\right]
+c_3\left[\begin{array}{c}\sin4t-\cos4t
\\\sin4t-2\cos4t\\\sin4t\end{array}\right]$.
Now ${\bf
y}(0)=\left[\begin{array}{c}{16}\\{14}\\6 \end{array}\right]\Rightarrow\left[\begin{array}{ccc}2&1&{-1}\\{-1}&1&{-2}\\1&1&0\end{array}\right]
\left[\begin{array}{c}{c_1}\\{c_2}\\{c_3}\end{array}\right]=\left[\begin{array}{c}{16}\\{14}\\6\end{array}\right]$, so $c_1=3$, $c_2=3$,
$c_3=-7$, and $\displaystyle{{\bf y}=\left[\begin{array}{c}6\\{-3}\\3\end{array}\right]e^{8t}
+\left[\begin{array}{c}10\cos4t-4\sin4t\\17\cos4t-\sin4t\\3\cos4t-7\sin4t
\end{array}\right]}$.

Exercise $\PageIndex{25}$

Suppose an $n\times n$ matrix $A$ with real entries has a complex eigenvalue $\lambda=\alpha+i\beta$ ($\beta\ne0$) with associated eigenvector ${\bf x}={\bf u}+i{\bf v}$, where ${\bf u}$ And ${\bf v}$ have real components. Show that ${\bf u}$ And ${\bf v}$ Are both nonzero.

Exercise $\PageIndex{26}$

Verify that

\begin{eqnarray*}
{\bf y}_1 = e^{\alpha t} ({\bf u} \cos \beta t - {\bf v} \sin \beta t) \quad \mbox{and} \quad {\bf y}_2 = e^{\alpha t} ({\bf u} \sin \beta t + {\bf v} \cos \beta t),
\end{eqnarray*}

are the real and imaginary parts of

\begin{eqnarray*}
e^{\alpha t} (\cos \beta t + i \sin \beta t) ({\bf u} + i {\bf v}).
\end{eqnarray*}

Exercise $\PageIndex{27}$

Show that if the vectors ${\bf u}$ And ${\bf v}$ Are not both ${\bf 0}$ And $\beta\ne0$ then the vector functions

\begin{eqnarray*}
{\bf y}_1 = e^{\alpha t}({\bf u} \cos \beta t - {\bf v} \sin \beta t) \quad \mbox{and} \quad {\bf y}_2 = e^{\alpha t} ({\bf u} \sin \beta t + {\bf v} \cos \beta t)
\end{eqnarray*}

are linearly independent on every interval.

Hint: There are two cases to consider: $\{{\bf u},{\bf v}\}$ linearly independent, and $\{{\bf u},{\bf v}\}$ linearly dependent. In either case, exploit the linear independence of $\{\cos\beta t,\sin\beta t\}$ on every interval.

Exercise $\PageIndex{28}$

Suppose ${\bf u}=\displaystyle{ \left[ \begin{array} \\ {u_1} \\ {u_2} \end{array} \right] }$ And ${\bf v}=\displaystyle{ \left[ \begin{array} \\ {v_1} \\ {v_2} \end{array} \right]}$ Are not orthogonal; that is, $({\bf u},{\bf v})\ne0$.

(a) Show that the quadratic equation

\begin{eqnarray*}
({\bf u}, {\bf v}) k^2 + (\| {\bf v} \|^2 - \| {\bf u} \|^2) k - ({\bf u}, {\bf v}) = 0
\end{eqnarray*}

has a positive root $k_1$And a negative root $k_2=-1/k_1$.

(b) Let ${\bf u}_1^{(1)}={\bf u}-k_1{\bf v}$, ${\bf v}_1^{(1)}={\bf v}+k_1{\bf u}$, ${\bf u}_1^{(2)}={\bf u}-k_2{\bf v}$, and ${\bf v}_1^{(2)}={\bf v}+k_2{\bf u}$, so that $({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0$, from the discussion given above. Show that

\begin{eqnarray*}
{\bf u}_1^{(2)} = {{\bf v}_1^{(1)} \over k_1} \quad \mbox{and} \quad {\bf v}_1^{(2)} = -{{\bf u}_1^{(1)} \over k_1}.
\end{eqnarray*}

(c) Let ${\bf U}_1$, ${\bf V}_1$, ${\bf U}_2$, and ${\bf V}_2$ be unit vectors in the directions of ${\bf u}_1^{(1)}$, ${\bf v}_1^{(1)}$, ${\bf u}_1^{(2)}$, and ${\bf v}_1^{(2)}$, respectively. Conclude from part (a) that ${\bf U}_2={\bf V}_1$ And ${\bf V}_2=-{\bf U}_1$, and that therefore the counterclockwise angles from ${\bf U}_1$ to ${\bf V}_1$ And from ${\bf U}_2$ to ${\bf V}_2$ Are both $\pi/2$ or both $-\pi/2$.

Answer

(a) From the quadratic formula the roots are
\begin{eqnarray*}
k_1&=&{\|{\bf u}\|^2-\|{\bf v}^2\|+\sqrt{(\|{\bf u}\|^2-\|{\bf
v}^2\|)^2+4({\bf u},{\bf v})^2}\over2({\bf u},{\bf v})}\\
k_2&=&{\|{\bf u}\|^2-\|{\bf v}^2\|-\sqrt{(\|{\bf u}\|^2-\|{\bf
v}^2\|)^2+4({\bf u},{\bf v})^2}\over2({\bf u},{\bf v})}.
\end{eqnarray*}
Clearly $k_1>0$ and $k_2<0$. Moreover,
$$
k_1k_2={(\|{\bf u}\|^2-\|{\bf v}^2\|)^2-\left[(\|{\bf u}\|^2-\|{\bf
v}^2\|)^2+4({\bf u},{\bf v})^2\right]\over4({\bf u},{\bf v})^2}=-1.
$$

(b) Since $k_2=-1/k_1$,
\begin{eqnarray*}
{\bf u}_1^{(2)}&=&{\bf u}-k_2{\bf v}={\bf u}+{1\over k_1}{\bf
v}={1\over k_1}({\bf v}+k_1{\bf u})={1\over k_1}{\bf v}_1^{(1)}\\
{\bf v}_1^{(2)}&=&{\bf v}+k_2{\bf u}={\bf v}-{1\over k_1}{\bf
u}=-{1\over k_1}({\bf u}-k_1{\bf v})=-{1\over k_1}{\bf u}_1^{(1)}.
\end{eqnarray*}

In Exercises $(4.6E.29)$ to $(4.6E.32)$, find vectors ${\bf U}$ And ${\bf V}$ parallel to the axes of symmetry of the trajectories, and plot some typical trajectories.

Exercise $\PageIndex{29}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & {-5} \\ 5 & {-3} \end{array} \right]{\bf y}}$

Exercise $\PageIndex{30}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-15} & {10} \\ {-25} & {15} \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}-15-\lambda&10\\-25&15-\lambda
\end{array}\right|=\lambda^2+25$.
The augmented matrix of $(A-5iI){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}-15-5i&10&\vdots&0\\
-25&15-5i&\vdots&0 \end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{cccr} 1&{-3+i\over5}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$.
Therefore,$x_1=\displaystyle{(3-i)\over5}x_2$. Taking $x_2=5$
yields the eigenvector
${\bf x}=\left[\begin{array}{c}3-i\\5\end{array}\right]$,
so ${\bf u}=\displaystyle{\left[\begin{array}{c}3\\5\end{array}\right]}$ and ${\bf v}=\displaystyle{\left[\begin{array}{c}\\{-1}\\0\end{array}\right]}$.
The quadratic equation is $-3k^2-33k+3=0$, with positive root
$k\approx.0902$.
Routine calculations yield
${\bf U}\approx\displaystyle{\left[\begin{array}{c}{.5257}\\{.8507}\end{array}\right]}$,
${\bf V}\approx\displaystyle{\left[\begin{array}{c}{-.8507}\\{.5257}\end{array}\right]}$.

Exercise $\PageIndex{31}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-4} & 8 \\ {-4} & 4 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{32}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & {-15} \\ 3 & 3 \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}-3-\lambda&-15\\3&3-\lambda
\end{array}\right|=\lambda^2+36$.
The augmented matrix of $(A-6iI){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}-3-6i&-15&\vdots&0\\
3&3-6i&\vdots&0 \end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{cccr} 1&1-2i&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$.
Therefore,$x_1=-(1-2i)x_2$. Taking $x_2=1$
yields the eigenvector
${\bf x}=\left[\begin{array}{c}-1+2i\\1\end{array}\right]$,
so ${\bf u}=\displaystyle{\left[\begin{array}{c}{-1}\\1\end{array}\right]}$ and ${\bf v}=\displaystyle{\left[\begin{array}{c}2\\0\end{array}\right]}$.
The quadratic equation is $-2k^2+2k+2=0$, with positive root
$k\approx1.6180$.
Routine calculations yield
${\bf U}\approx\displaystyle{\left[\begin{array}{c}{-.9732}\\{.2298}\end{array}\right]}$,
${\bf V}\approx\displaystyle{\left[\begin{array}{c}{.2298}\\{.9732}\end{array}\right]}$.

In Exercises $(4.6E.33)$ to $(4.6E.40)$, find vectors ${\bf U}$ And ${\bf V}$ parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory.

Exercise $\PageIndex{33}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-5} & 6 \\ {-12} & 7 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{34}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & {-12} \\ 6 & {-7} \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}5-\lambda&-12\\6&-7-\lambda
\end{array}\right|=(\lambda+1)^2+36$.
The augmented matrix of $(A-(-1+6i)I){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}6-6i&-12&\vdots&0\\
6&-6-6i&\vdots&0 \end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{cccr} 1&-(1+i)&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$.
Therefore,$x_1=(1+i)x_2$. Taking $x_2=1$
yields the eigenvector
${\bf x}=\left[\begin{array}{c}1+i\\1\end{array}\right]$,
so ${\bf u}=\displaystyle{\left[\begin{array}{c}1\\1\end{array}\right]}$ and ${\bf v}=\displaystyle{\left[\begin{array}{c}1\\0\end{array}\right]}$.
The quadratic equation is $k^2-k-1=0$, with positive root
$k\approx1.6180$.
Routine calculations yield
${\bf U}\approx\displaystyle{\left[\begin{array}{c}{-.5257}\\{.8507}\end{array}\right]}$,
${\bf V}\approx\displaystyle{\left[\begin{array}{c}{.8507}\\{.5257}\end{array}\right]}$

Exercise $\PageIndex{35}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-5} \\ 9 & {-2} \end{array} \right] {\bf y}}$

Exercise $\PageIndex{36}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-4} & 9 \\ {-5} & 2 \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}-4-\lambda&9\\-5&2-\lambda
\end{array}\right|=(\lambda+1)^2+36$.
The augmented matrix of $(A-(-1+6i)I){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}-3-6i&9&\vdots&0\\
-5&3-6i&\vdots&0 \end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{cccr} 1&-{3-6i\over5}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$.
Therefore,$x_1=\displaystyle{3-6i\over5}x_2$. Taking $x_2=5$
yields the eigenvector
${\bf x}=\left[\begin{array}{c}3-6i\\5\end{array}\right]$,
so ${\bf u}=\displaystyle{\left[\begin{array}{c}3\\5\end{array}\right]}$ and ${\bf v}=\displaystyle{\left[\begin{array}{c}{-6}\\0\end{array}\right]}$.
The quadratic equation is $-18k^2+2k+18=0$, with positive root
$k\approx1.0571$.
Routine calculations yield
${\bf U}\approx\displaystyle{\left[\begin{array}{c}{.8817}\\{.4719}\end{array}\right]}$,
${\bf V}\approx\displaystyle{\left[\begin{array}{c}{-.4719}\\{.8817}\end{array}\right]}$.

Exercise $\PageIndex{37}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & {10} \\ {-10} & {-1} \end{array} \right] {\bf y}}$

Exercise $\PageIndex{38}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & {-5} \\ {20} & {-1} \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}-1-\lambda&-5\\20&-1-\lambda
\end{array}\right|=(\lambda+1)^2+100$.
The augmented matrix of $(A-(-1+10i)I){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}-10i&-5&\vdots&0\\
20&-10i&\vdots&0 \end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{cccr} 1&-{i\over2}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$.
Therefore,$x_1=\displaystyle{i\over2}x_2$. Taking $x_2=2$
yields the eigenvector
${\bf x}=\left[\begin{array}{c}i\\2\end{array}\right]$,
so ${\bf u}=\displaystyle{\left[\begin{array}{c}0\\2}\end{array}\right]}$ and ${\bf v}=\displaystyle{\left[\begin{array}{c}1\\0\end{array}\right]}$.
Since $({\bf u},{\bf v})=0$ we just normalize ${\bf u}$ and ${\bf v}$
to obtain
${\bf U}=\displaystyle{\left[\begin{array}{c}0\\1\end{array}\right]}$, ${\bf V}=\displaystyle{\left[\begin{array}{c}1\\0\end{array}\right]}$.

Exercise $\PageIndex{39}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & {10} \\ {-10} & 9 \end{array} \right] {\bf y}}$

Exercise $\PageIndex{40}$

$\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & 6 \\ {-12} & 5 \end{array} \right] {\bf y}}$

Answer: $\left|\begin{array}{cc}-7-\lambda&6\\-12&5-\lambda
\end{array}\right|=(\lambda+1)^2+36$.
The augmented matrix of $(A-(-1+6i)I){\bf x}={\bf 0}$ is
$\left[\begin{array}{cccr}-6-6i&6&\vdots&0\\
-12&6-6i&\vdots&0 \end{array}\right]$,
which is row equivalent to
$\left[\begin{array}{cccr} 1&-{1-i\over2}&\vdots&0\\ 0&0&\vdots&0
\end{array}\right]$.
Therefore,$x_1=\displaystyle{1-i\over2}x_2$. Taking $x_2=2$
yields the eigenvector
${\bf x}=\left[\begin{array}{c}1-i\\2\end{array}\right]$,
so ${\bf u}=\displaystyle{\left[\begin{array}{c}1\\2\end{array}\right]}$ and ${\bf v}=\displaystyle{\left[\begin{array}{c}{-1}\\0\end{array}\right]}$.
The quadratic equation is $-k^2-4k+1=0$, with positive root
$k\approx.2361$.
Routine calculations yield
${\bf U}\approx\displaystyle{\left[\begin{array}{c}{.5257}\\{.8507}\end{array}\right]}$,
${\bf V}\approx\displaystyle{\left[\begin{array}{c}{-.8507}\\{.5257}\end{array}\right]}$.

Search

Text Color

Text Size

Margin Size

Font Type

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)

Exercise \(\PageIndex{4}\)

Exercise \(\PageIndex{5}\)

Exercise \(\PageIndex{6}\)

Exercise \(\PageIndex{7}\)

Exercise \(\PageIndex{8}\)

Exercise \(\PageIndex{9}\)

Exercise \(\PageIndex{10}\)

Exercise \(\PageIndex{11}\)

Exercise \(\PageIndex{12}\)

Exercise \(\PageIndex{13}\)

Exercise \(\PageIndex{14}\)

Exercise \(\PageIndex{15}\)

Exercise \(\PageIndex{16}\)

Exercise \(\PageIndex{17}\)

Exercise \(\PageIndex{18}\)

Exercise \(\PageIndex{19}\)

Exercise \(\PageIndex{20}\)

Exercise \(\PageIndex{21}\)

Exercise \(\PageIndex{22}\)

Exercise \(\PageIndex{23}\)

Exercise \(\PageIndex{24}\)

Exercise \(\PageIndex{25}\)

Exercise \(\PageIndex{26}\)

Exercise \(\PageIndex{27}\)

Exercise \(\PageIndex{28}\)

Exercise \(\PageIndex{29}\)

Exercise \(\PageIndex{30}\)

Exercise \(\PageIndex{31}\)

Exercise \(\PageIndex{32}\)

Exercise \(\PageIndex{33}\)

Exercise \(\PageIndex{34}\)

Exercise \(\PageIndex{35}\)

Exercise \(\PageIndex{36}\)

Exercise \(\PageIndex{37}\)

Exercise \(\PageIndex{38}\)

Exercise \(\PageIndex{39}\)

Exercise \(\PageIndex{40}\)