Let the matrix A=pmatrix1&0&01&0&10&1&0pmatrix satisfy A^n=A^n-2+A^2-I for n3.

Question

Let the matrix $A=\begin{pmatrix}1&0&0\1&0&1\0&1&0\end{pmatrix}$ satisfy $A^n=A^{n-2}+A^2-I$ for $n\ge3$. Then the sum of all the elements of $A^{50}$ is:

Accepted Answer

$A^2=\begin{pmatrix}1&0&0\1&1&0\1&0&1\end{pmatrix}$. Using the recurrence $A^n=A^{n-2}+A^2-I$:

$A^3=A+A^2-I=\begin{pmatrix}1&0&0\2&0&1\1&1&0\end{pmatrix}$, $A^4=A^2+A^2-I=2A^2-I=\begin{pmatrix}1&0&0\2&1&0\2&0&1\end{pmatrix}$, and continuing the pattern, one finds
$$A^n=\begin{pmatrix}1&0&0\ \tfrac{n-1}2\cdot\text{(pattern)} &\ast&\ast\ \ast&\ast&\ast\end{pmatrix},\qquad A^{50}=\begin{pmatrix}1&0&0\25&1&0\25&0&1\end{pmatrix}.$$

Sum of all elements $=1+25+1+25+0+1=53$.

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