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Let A = $\begin{pmatrix} 3 & -5 \\ 1 & -3 \end{pmatrix}$. Compute $A^{9}$. (Hint: Find a matrix P such that $P^{-1}AP$ is a diagonal matrix D and show that $A^{9}$= $PD^{9}P^{-1}$

Answer: $\begin{pmatrix} 768 & -1280 \\ 256 & -768 \end{pmatrix}$

I keep getting $\begin{pmatrix} -768 & 1280 \\ -256 & 768 \end{pmatrix}$ but could it be still right? I have D=$\begin{pmatrix} -2 & 0\\ 0& 2\end{pmatrix}$ and P =$\begin{pmatrix} 5&1 \\1 & 1 \end{pmatrix}$ with $P^{-1}$= $\begin{pmatrix} \frac{1}{4} & \frac{1}{-4} \\ \frac{1}{-4} & \frac{5}{4} \end{pmatrix}$

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    $\begingroup$ With that $P$ and $P^{-1}$, we have that $$P^{-1}AP = \left(\begin{array}{rr}2 & 0\\0 & -2\end{array}\right).$$Presumably, that is the source of your problems. Recheck your computation of $D$. $\endgroup$ – Arturo Magidin Jun 24 '12 at 18:30
  • $\begingroup$ You may want to consider registering, given that you've asked several questions. $\endgroup$ – Arturo Magidin Jun 24 '12 at 22:39
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Your answer is not correct. Please note that the eigenvectors should be corresponding to the eigenvalues. So, if you choose $$D=\left( \begin{array}{cc} -2 & 0 \\ 0& 2 \end{array} \right),$$ then your $P$ should be $$P=\left( \begin{array}{cc} 1 & 5 \\ 1& 1 \end{array} \right),$$

because $(1,1)$ is the eigenvector corresponding to $-2$.

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    $\begingroup$ This is correct. $\endgroup$ – Kerry Jun 24 '12 at 18:45
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With $P=\begin{pmatrix} 5&1 \\1 & 1 \end{pmatrix}$ we have $P^{-1}AP = \left(\begin{array}{rr}2 & 0\\0 & -2\end{array}\right)$ then you can use your formula $PA^{9}P^{-1}$ and calculate $PA^{9}P^{-1}$.

$ \left(\begin{array}{rr}2 & 0\\0 & -2\end{array}\right)^9=\left(\begin{array}{rr}\,\,2^9 & 0\,\,\,\,\,\,\,\\0 & \,(-2)^9\end{array}\right)$

Now you can calculate $A^9$

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