Let A=[a_ij] be a 33 matrix of order 3, with a_ij=(√2)^i+j.

Question

Let $A=[a_{ij}]$ be a $3	imes3$ matrix of order 3, with $a_{ij}=(\sqrt2)^{i+j}$. If the sum of all the elements in the third row of $A^2$ is $\alpha+\beta\sqrt2$, $\alpha,\beta\in\mathbb{Z}$, then $\alpha+\beta$ is equal to:

Accepted Answer

$A=\begin{pmatrix}2 & 2\sqrt2 & 4\ 2\sqrt2 & 4 & 4\sqrt2\ 4 & 4\sqrt2 & 8\end{pmatrix}$ (using $a_{ij}=(\sqrt2)^{i+j}$).

Computing $A^2$'s third row: entry sums give (per the source) $4\big(14+14\sqrt2+28\big)$ as an intermediate combination, which simplifies to $\alpha+\beta\sqrt2$ with $\alpha=56+16\cdot\ldots$; the final combination yields
$$\alpha+\beta=224.$$

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