Let the matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ satisfy $$A^n = A^{n-2} + A^2 - I$$ for $$n \geq 3$$. Then the sum of all the elements of $$A^{50}$$ is :

Let $$\Sigma(M)$$ denote the sum of all the elements of a matrix $$M$$.
We have $$A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ and the relation

$$A^{n} = A^{\,n-2} + A^{2} - I \quad\text{for}\; n \ge 3$$ $$-(1)$$

Applying $$\Sigma$$ on both sides of $$(1)$$ gives a relation for the sums:

$$\Sigma\!\left(A^{n}\right) = \Sigma\!\left(A^{\,n-2}\right) + \Sigma\!\left(A^{2}\right) - \Sigma(I)$$ $$-(2)$$

Because $$\Sigma$$ is linear (sum of elements of a sum is the sum of the individual sums), $$(2)$$ is valid for every $$n \ge 3$$.

First compute the required base values.

Case 1: $$A^{0}=I$$
$$I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ so $$\Sigma(I)=1+1+1=3$$.

Case 2: $$A^{1}=A$$
Row sums: $$1,\;2,\;1 \;\Rightarrow\; \Sigma(A)=1+2+1=4$$.

Case 3: $$A^{2}$$

Compute $$A^{2}=A\! \cdot\! A$$:

$$A^{2}= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$
Row sums: $$1,\;2,\;2 \;\Rightarrow\; \Sigma(A^{2})=1+2+2=5$$.

Substituting $$\Sigma(A^{2})=5$$ and $$\Sigma(I)=3$$ into $$(2)$$ gives the recurrence for $$S_n=\Sigma(A^{n})$$:

$$S_n = S_{\,n-2} + 5 - 3 = S_{\,n-2} + 2 \quad\text{for}\; n \ge 3$$ $$-(3)$$

Now build the sequence.

Even powers
$$S_0 = 3$$ (already obtained)
Using $$(3)$$ repeatedly:
$$S_2 = S_0 + 2 = 3 + 2 = 5$$
$$S_4 = S_2 + 2 = 5 + 2 = 7$$
Continuing, every step of two in the index adds 2 to the sum, so

$$S_{2k} = 3 + 2k \quad\text{for}\; k \ge 0$$ $$-(4)$$

Odd powers
$$S_1 = 4$$ (already obtained)
Similarly,

$$S_3 = S_1 + 2 = 4 + 2 = 6$$
$$S_5 = S_3 + 2 = 6 + 2 = 8$$
Hence

$$S_{2k+1} = 4 + 2k \quad\text{for}\; k \ge 0$$ $$-(5)$$

We need $$S_{50}$$. Since $$50 = 2 \times 25$$ is even, use $$(4)$$ with $$k = 25$$:

$$S_{50} = 3 + 2 \times 25 = 3 + 50 = 53$$.

Therefore, the sum of all the elements of $$A^{50}$$ is $$53$$.

Option A is correct.