Let A = $$[a_{ij}]$$ be a matrix of order $$3 \times 3$$, with $$a_{ij}$$ = $$(\sqrt{2})^{i+j}$$. If the sum of all the elements in the third row of $$A^{2}$$ is $$\alpha + \beta\sqrt{2}, \quad \alpha,\beta \in \mathbb{Z}$$, then $$\alpha + \beta$$ is equal to:

The matrix $$A = [a_{ij}]$$ is defined by $$a_{ij} = (\sqrt{2})^{i+j}$$.

$$ A = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix} $$

We seek the sum of all elements in the third row of $$A^2$$.

Notice that the sum of the entries in the third row of $$A^2$$ can be written as the dot product of the third row of $$A$$ with the vector of column sums of $$A$$ (i.e., $$A^2 \cdot \mathbf{1}$$ gives row sums of $$A^2$$, which equals $$A \cdot (A\mathbf{1})$$).

The sums of the columns of $$A$$ are

Column 1: $$2 + 2\sqrt{2} + 4 = 6 + 2\sqrt{2}$$

Column 2: $$2\sqrt{2} + 4 + 4\sqrt{2} = 4 + 6\sqrt{2}$$

Column 3: $$4 + 4\sqrt{2} + 8 = 12 + 4\sqrt{2}$$

Hence the sum of the third row of $$A^2$$ is

$$=4(6+2\sqrt{2}) + 4\sqrt{2}(4+6\sqrt{2}) + 8(12+4\sqrt{2})$$

$$=24 + 8\sqrt{2} + 16\sqrt{2} + 48 + 96 + 32\sqrt{2}$$

$$=(24 + 48 + 96) + (8 + 16 + 32)\sqrt{2}$$

$$=168 + 56\sqrt{2}$$

So $$\alpha = 168$$ and $$\beta = 56$$.

$$\alpha + \beta = 168 + 56 = 224$$

The correct answer is Option 2: 224.