How to compute the nth number of a general Fibonacci sequence with matrix multiplication? If we want to compute the nth Fibonacci number we just power the matrix
$M =
\left[\begin{array}{cc}
1 & 1 \\
1 & 0
\end{array}\right]$
 $n$ times and we get $M =\left[
\begin{array}{cc}
F_{n+1} & F_n \\
F_n & F_{n-1}
\end{array}\right]$
But, How should be the elements arranged when I want to find the $n$th term of the fibonnaci series whose starting elements are $a$ and $b$?
 A: If $a_{n} = a_{n-1} + a_{n-2}$, then
$$
{\begin{bmatrix}
  1 & 1 \\
  1 & 0 \\
\end{bmatrix}}^n
\begin{bmatrix}
  a_1 \\
  a_0 \\
\end{bmatrix} = 
\begin{bmatrix}
a_{n+1} \\
a_n
\end{bmatrix}
$$
for any $a_0, a_1$.
Any linear recurrence relation can be solved using matrix exponentiation, e.g.,  $a_{n}=a_{n-1}-3a_{n-3}+5a_{n-5}-7$.
See this blog post: Recurrence relation and matrix exponentiation.
A: If you have a sequence, say $G_{-1} = a, G_0 = b$, and $G_{n+1} = G_{n} + G_{n-1}$, note that
$$\left[\begin{array}{cc}
1 & 1 \\
1 & 0
\end{array}\right]\cdot\left[\begin{array}{c} G_n \\ G_{n-1} \end{array}\right] = \left[\begin{array}{c} G_n + G_{n-1} \\ G_n \end{array}\right] = \left[\begin{array}{c} G_{n+1} \\ G_n \end{array}\right].$$ 
Then it isn't too hard to show that
$$\left[\begin{array}{cc}
1 & 1 \\
1 & 0
\end{array}\right]^n\cdot\left[\begin{array}{c} b \\ a \end{array}\right] = \left[\begin{array}{c} G_{n} \\ G_{n-1} \end{array}\right].$$ 
A: Just multiply your power matrix by the vector $[a,b]$ and you will get the $n+1$th and $n$th terms of your sequence as the resultant vector.
