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Solution

Given $A^n=A^{n-2}+A^2-I$ for $n\ge 3$

Let $S_n$ = sum of all elements of $A^n$

Taking sum on both sides:
$S_n=S_{n-2}+S_2-3$ (since sum of $I=3$)

Compute base values:
$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$
⇒ $S_1=3$

$A^2=\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix}$
⇒ $S_2=5$

So recurrence:
$S_n=S_{n-2}+2$

Even terms:
$S_2=5,\ S_4=7,\ S_6=9,\dots$
⇒ $S_{2k}=5+(k-1)\cdot2=2k+3$

For $n=50$, $k=25$:
$S_{50}=2(25)+3=53$