Tips och lösning till U 22.35
SamverkanLinalgLIU
(Skillnad mellan versioner)
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+ | Matrisen <math> A=\left(\begin{array}{rr} 2&12\\1&3 \end{array}\right) </math> har egenvärdena | ||
+ | <math> \lambda_1=-1 </math> och <math> \lambda_2=6 </math>. | ||
+ | Tillhörande egenrum <math> E_{\lambda=-2}=[(-4,1)^t] </math>, <math> E_{\lambda=2}=[(3,1)^t] </math>, | ||
+ | så att <math> A=TDT^{-1} </math>. | ||
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+ | Systemet kan då skrivas | ||
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+ | |||
+ | <center><math> | ||
+ | \begin{array}{rcl} | ||
+ | \left(\begin{array}{r}\\{a_n}{b_n}\end{array}\right) &=&A\left(\begin{array}{r}{a_{n-1}}\\{b_{n-1}} \end{array}\right) | ||
+ | =A^n\left(\begin{array}{r}{a_0}\\{b_0}\end{array}\right) | ||
+ | =TD^nT^{-1}\left(\begin{array}{r}{a_0}\\{b_0}\end{array}\right)\\ | ||
+ | &=&\left(\begin{array}{rr}{-4}&3\\1&1 \end{array}\right) | ||
+ | \left(\begin{array}{rr}{(-1)^n}&0\\0&{6^n}\end{array}\right) | ||
+ | \frac{1}{7}\left(\begin{array}{rr}{-1}&3\\1&4 \end{array}\right) | ||
+ | \left(\begin{array}{r}2\\3 \end{array}\right). | ||
+ | \end{array} | ||
+ | </math></center> | ||
+ | |||
+ | |||
+ | Då är | ||
+ | |||
+ | |||
+ | <center><math> | ||
+ | \frac{1}{6^n} \left(\begin{array}{r}{a_n}{b_n}\end{array}\right) | ||
+ | =\frac{1}{7} \left(\begin{array}{cc}{-4}&3\\1&1 \end{array}\right) | ||
+ | \left(\begin{array}{rr}{(-1/6)^n}&0\\0&{1}\end{array}\right) \left(\begin{array}{rr}{-1}&3\\1&4 \end{array}\right) | ||
+ | \left(\begin{array}{r}2\\3 \end{array}\right)\rightarrow2 \left(\begin{array}{r}3\\1 \end{array}\right). | ||
+ | </math></center> | ||
+ | |||
+ | |||
+ | då <math> n\rightarrow\infty </math>. |
Versionen från 20 september 2010 kl. 10.00
Tips 1
Hej 1
Tips 2
Hej 2
Tips 3
Hej 3
Lösning
Matrisen \displaystyle A=\left(\begin{array}{rr} 2&12\\1&3 \end{array}\right) har egenvärdena \displaystyle \lambda_1=-1 och \displaystyle \lambda_2=6 . Tillhörande egenrum \displaystyle E_{\lambda=-2}=[(-4,1)^t] , \displaystyle E_{\lambda=2}=[(3,1)^t] , så att \displaystyle A=TDT^{-1} .
Systemet kan då skrivas
\begin{array}{rcl} \left(\begin{array}{r}\\{a_n}{b_n}\end{array}\right) &=&A\left(\begin{array}{r}{a_{n-1}}\\{b_{n-1}} \end{array}\right)
=A^n\left(\begin{array}{r}{a_0}\\{b_0}\end{array}\right) =TD^nT^{-1}\left(\begin{array}{r}{a_0}\\{b_0}\end{array}\right)\\ &=&\left(\begin{array}{rr}{-4}&3\\1&1 \end{array}\right) \left(\begin{array}{rr}{(-1)^n}&0\\0&{6^n}\end{array}\right) \frac{1}{7}\left(\begin{array}{rr}{-1}&3\\1&4 \end{array}\right) \left(\begin{array}{r}2\\3 \end{array}\right).
\end{array}
Då är
\frac{1}{6^n} \left(\begin{array}{r}{a_n}{b_n}\end{array}\right) =\frac{1}{7} \left(\begin{array}{cc}{-4}&3\\1&1 \end{array}\right)
\left(\begin{array}{rr}{(-1/6)^n}&0\\0&{1}\end{array}\right) \left(\begin{array}{rr}{-1}&3\\1&4 \end{array}\right) \left(\begin{array}{r}2\\3 \end{array}\right)\rightarrow2 \left(\begin{array}{r}3\\1 \end{array}\right).
då \displaystyle n\rightarrow\infty .