Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $P^{-1}$= $\begin{pmatrix} \frac{1}{4} & \frac{1}{-4} \\ \frac{1}{-4} & \frac{5}{4} \end{pmatrix}$

share|improve this question
3  
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 $D$. –  Arturo Magidin Jun 24 '12 at 18:30
    
You may want to consider registering, given that you've asked several questions. –  Arturo Magidin Jun 24 '12 at 22:39

2 Answers 2

up vote 3 down vote accepted

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 $-2$.

share|improve this answer
1  
This is correct. –  Kerry Jun 24 '12 at 18:45

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 $A^9$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.