A 2-dimensional system of linear difference equations that describes Jujukan Fibonacci is

( F k + 2 F k + 1 ) = ( 1 1 1 0 ) ( F k + 1 F k ) {\displaystyle {F_{k+2} \choose F_{k+1}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{F_{k+1} \choose F_{k}}}

or

F → k + 1 = A F → k . {\displaystyle {\vec {F}}_{k+1}=A{\vec {F}}_{k}.\,}

eigenvalues of matriks A are φ {\displaystyle \varphi \,\!} and ( 1 − φ ) {\displaystyle (1-\varphi )\,\!} , and elements of eigenvectors of A, ( φ 1 ) {\displaystyle {\varphi \choose 1}} and ( 1 − φ ) {\displaystyle {1 \choose -\varphi }} , are dalam nisbah-nisbah φ {\displaystyle \varphi \,\!} and ( 1 − φ ) . {\displaystyle (1-\varphi \,\!).}

matriks ini mempunyai determinant of −1, and thus it is a 2×2 unimodular matriks. This property can be understood in terms of pecahan berterusan representation for nisbah keemasan:

φ = 1 + 1 1 + 1 1 + 1 ⋱ . {\displaystyle \varphi =1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{\;\;\ddots \,}}}}}}\;.}

Nombor Fibonacci occur as nisbah pertembungan pecahan berterusan yang berterusan φ {\displaystyle \varphi \,\!} , dan matriks yang dibentuk daripada from successive convergents of any pecahan berterusan mempunyai determinant of +1 or −1.

Perwakilan matriks memberikan ungkapan tertutup nombor Fibonacci yang berikut:

( 1 1 1 0 ) n = ( F n + 1 F n F n F n − 1 ) . {\displaystyle {\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{n}={\begin{pmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{pmatrix}}.}

Taking determinant of kedua-dua belah persamaan ini menyerlahkan identiti Cassini

( − 1 ) n = F n + 1 F n − 1 − F n 2 . {\displaystyle (-1)^{n}=F_{n+1}F_{n-1}-F_{n}^{2}.\,}

Additionally, since A n A m = A m + n {\displaystyle A^{n}A^{m}=A^{m+n}} untuk sebarang square matriks A {\displaystyle A} , following identities can be derived:

F n 2 + F n − 1 2 = F 2 n − 1 , {\displaystyle {F_{n}}^{2}+{F_{n-1}}^{2}=F_{2n-1},\,} F n + 1 F m + F n F m − 1 = F m + n . {\displaystyle F_{n+1}F_{m}+F_{n}F_{m-1}=F_{m+n}.\,}

Untuk first one of these, there is a related identity:

( 2 F n − 1 + F n ) F n = ( F n − 1 + F n + 1 ) F n = F 2 n . {\displaystyle (2F_{n-1}+F_{n})F_{n}=(F_{n-1}+F_{n+1})F_{n}=F_{2n}.\,}

Untukanother way to derive F 2 n + k {\displaystyle F_{2n+k}} formulas see "EWD note" by Dijkstra[9].