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add stupid calculations in la4 and update theo5

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christian
2020-05-27 10:25:05 +02:00
parent bdf972e31e
commit 9507d93385
4 changed files with 157 additions and 3 deletions
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sose2020/la/uebungen/la4.pdf
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sose2020/la/uebungen/la4.tex
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@@ -3,6 +3,8 @@
\title{Lineare Algebra 2: Übungsblatt 4}
\author{Dominik Daniel, Christian Merten}
\usepackage[]{gauss}
\begin{document}
\punkte[16]
@@ -57,4 +59,155 @@
\end{enumerate}
\end{aufgabe}
\begin{aufgabe}
\begin{enumerate}[(a)]
\item Kurze Rechung ergibt
\begin{align*}
P_A &= \begin{gmatrix}[p] t-10 & 11 & 11 & 32 \\
1 & t & 2 & - 4 \\
-1 & 1 & t-1 & 4 \\
-2 & 2 & 2 & t+6
\rowops
\swap{0}{1}
\end{gmatrix}
\sim
\begin{gmatrix}[p]
1 & t & 2 & - 4 \\
t-10 & 11 & 11 & 32 \\
-1 & 1 & t-1 & 4 \\
-2 & 2 & 2 & t+6
\rowops
\add[-(t-10)]{0}{1}
\add{0}{2}
\add[2]{0}{3}
\colops
\add[-t]{0}{1}
\add[-2]{0}{2}
\add[4]{0}{3}
\end{gmatrix} \\
&\sim
\begin{gmatrix}[p]
1 & 0 & 0 & 0 \\
0 & 11-t^2 + 10t & -8 + 4t \\
0 & t+1 & t+1 & 0 \\
0 & 2+2t & 6 & t-2
\rowops
\swap{1}{3}
\colops
\swap{1}{2}
\end{gmatrix}
\sim
\begin{gmatrix}[p]
1 & 0 & 0 & 0 \\
0 & 6 & 2+2t & t-2 \\
0 & t+1 & t+1 & 0 \\
0 & 31-2t & 11-t^2 + 10t & -8+4t
\rowops
\mult{1}{\cdot \frac{1}{6}}
\end{gmatrix} \\
&\sim
\begin{gmatrix}[p]
1 & 0 & 0 & 0 \\
0 & 1 & \frac{1}{3} + \frac{1}{3}t & \frac{1}{6}t - \frac{1}{3} \\
0 & t+1 & t+1 & 0 \\
0 & 31-2t & 11-t^2+10t & -8+4t
\rowops
\add[-(t+1)]{1}{2}
\add[-(31-2t)]{1}{3}
\colops
\add[-\left(\frac{1}{3} + \frac{1}{3}t \right)]{1}{2}
\add[-\left(\frac{1}{6}t - \frac{1}{3} \right)]{1}{3}
\end{gmatrix}
\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & -\frac{1}{3} t^2+\frac{2}{3} + \frac{1}{3}t & -\frac{1}{6}t^2 + \frac{1}{6}t + \frac{1}{3} \\
0 & \frac{2}{3} + \frac{1}{3}t - \frac{1}{3}t^2 & \frac{1}{3}t^2 - \frac{11}{6}t + \frac{7}{3}
\rowops
\add[-1]{1}{2}
\colops
\add[-\frac{1}{2}]{1}{2}
\end{gmatrix} \\
&\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & -\frac{1}{3}t^2 + \frac{1}{3}t + \frac{2}{3} & 0 \\
0 & 0 & \frac{1}{2}t^2 - 2t + 2
\rowops
\mult{1}{\cdot (-3)}
\mult{2}{\cdot 2}
\end{gmatrix}
\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & (t+1)(t-2) & 0 \\
0 & 0 & (t-2)^2
\rowops
\add{1}{2}
\colops
\add[-1]{1}{2}
\end{gmatrix} \\
&\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & (t+1)(t-2) & -(t+1)(t-2) \\
0 & (t+1)(t-2) & -3t + 6
\rowops
\swap{1}{2}
\colops
\swap{1}{2}
\end{gmatrix}
\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & -3t + 6 & (t-2)(t+1) \\
0 & -(t-2)(t+1) & (t-2)(t+1)
\rowops
\mult{1}{\cdot \left( -\frac{1}{3} \right)}
\colops
\mult{2}{\cdot 3}
\end{gmatrix} \\
&\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & t - 2 & -(t-2)(t+1) \\
0 & -(t-2)(t+1) & 3(t-2)(t+1)
\rowops
\add[t+1]{1}{2}
\colops
\add[t+1]{1}{2}
\end{gmatrix}
\sim
\begin{gmatrix}[p]
E_2 & 0 & 0 \\
0 & t-2 & 0 \\
0 & 0 & (t-2)^2(t+1)
\end{gmatrix}
.\end{align*}
Damit folgen als Invariantenteiler: $c_1 = c_2 = 1$, $c_3 = t- 2$ und $c_4 = (t-2)^2(t+1)$.
Die Determinantenteiler sind damit $d_1 = 1$, $d_2 = 1$, $d_3 = t-2$ und $d_4 = (t-2)^{3}(t+1)$.
\item Hier ist sofort ersichtlich:
\begin{align*}
\text{det}(P_B)
= \begin{gmatrix}[v]
t+5 & 3 & -5 \\
0 & t-1 & 1 \\
8 & 4 & t-7
\end{gmatrix}
= (t-1)^{3}
\neq
(t-2)(t-1)(t+1)
=
\begin{gmatrix}[v]
t+3 & -8 & -12 \\
-1 & t+2 & 3 \\
2 & -4 & t-7
\end{gmatrix}
= \text{det}(P_C)
.\end{align*}
Damit ist $d_3_{B} \neq d_3_{C}$, also sind nach Invariantenteilersatz
$B$ und $C$ nicht ähnlich.
\end{enumerate}
\end{aufgabe}
\end{document}