Understanding power method for finding dominant eigenvalues 
The power method aims to find the eigenvalue with the largest magnitude. Does magnitude still have the same meaning in this context? If so, can't we tell from the outset which eigenvalue is the largest? 
And for $\lambda_1^{(1)}$, they got $\frac{61}{13}$,  why isn't it $\frac{13}{61}$?
Because we have $
\begin{bmatrix}-2 & -3 \\
       6 & 7
\end{bmatrix}
\begin{bmatrix}5\\
       -13
\end{bmatrix}=\lambda
\begin{bmatrix}-29\\
       61 
\end{bmatrix}
$
So I set up my equations as $\lambda61=13$
 A: Magnitude means the same as it usually does; we just say that because if the eigenvalues are, say, $-4$ and $1$, power iteration will find $-4$, not $1$, even though $1$ is the "largest" in the usual ordering. I'm not sure I understand your question about being able to tell, because we often don't know anything about the eigenvalues in advance when using these methods.
As for it being $\frac{61}{13}$, suppose $x_1$ were exactly an eigenvector with eigenvalue $\lambda$, then $(Ax_1)_2$ would be $\lambda (x_1)_2$ which would be $13 \lambda$. So you divide the new vector's component by the old vector's component to estimate the eigenvalue, not the other way around. (Note that this is not the only reasonable way to approximate the eigenvalue, another way would be to consider $\frac{\| x_1 \|}{\| x_0 \|}$, which would approximate $|\lambda|$.)
A: Suppose you start with an eigenvector $x_0 = v$, and $[x_0]_k \neq 0$.
Then $x_1 = Ax_0 = \lambda x_0$ and so $\lambda = {[x_1]_k \over [x_0]_k}$.
The hope is that you start with $x_0$ having some component in the direction of the dominant eigenvector. After enough iterations, the components corresponding to the other eigenvalues fade away.
