= Two-coefficient reduction of matrix powers
{title2=$A^n=a_nA+b_nI$}
Consider the real <matrix>
$$
A=\begin{pmatrix}2&1\\1&1\end{pmatrix},\qquad
\chi_A(t)=(t-2)(t-1)-1=t^2-3t+1.
$$
The <Cayley-Hamilton theorem> gives $A^2=3A-I$. Consequently
$$
A^3=3A^2-A=8A-3I,\qquad
A^4=21A-8I,\qquad A^5=55A-21I.
$$
Thus
$$
\boxed{A^5=\begin{pmatrix}89&55\\55&34\end{pmatrix}.}
$$
All four entries come from just two scalar coefficients. If $A^n=a_nA+b_nI$, multiplying once by $A$ and replacing $A^2$ gives the <linear recurrence relation>
$$
a_{n+1}=3a_n+b_n,\qquad b_{n+1}=-a_n,
\qquad(a_0,b_0)=(0,1).
$$
Equivalently $a_{n+2}=3a_{n+1}-a_n$, with $a_0=0,a_1=1$. The appearance of <Fibonacci numbers> is explained by $A=Q^2$, where $Q=\begin{pmatrix}1&1\\1&0\end{pmatrix}$ advances the <Fibonacci recurrence>. No <eigenvectors> need to be computed.
\Image[/cayley-hamilton-matrix-power-reduction.png]
{title=Cayley–Hamilton reduces powers of a chosen two-by-two matrix to two integer coefficients; the fifth power is 55A − 21I}
The same identity computes an inverse. Rearrange it as $A(3I-A)=I$, giving
$$
\boxed{A^{-1}=3I-A=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}.}
$$
This works because the constant term of this <characteristic polynomial> is nonzero and the coefficients lie in a <field>. Generally, when $c_0\ne0$ over a <field>, rearranging the <Cayley-Hamilton theorem> expresses $A^{-1}$ as a <polynomial evaluated at a matrix>. Over a general <commutative ring>, a nonzero constant need not be a <unit>; invertibility requires that extra condition.
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