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

Question

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

Accepted Answer

$\begin{gathered} A=[[|\begin{array}{ccc}(\sqrt{2}{)}^{2} \quad  (\sqrt{2}{)}^{3} \quad  (\sqrt{2}{)}^{4} \ (\sqrt{2}{)}^{3} \quad  (\sqrt{2}{)}^{4} \quad  (\sqrt{2}{)}^{5} \ (\sqrt{2}{)}^{4} \quad  (\sqrt{2}{)}^{5} \quad  (\sqrt{2}{)}^{6}\end{array}]]| \ A=[[|\begin{array}{ccc}2 \quad  2\sqrt{2} \quad  4 \ 2\sqrt{2} \quad  4 \quad  4\sqrt{2} \ 4 \quad  4\sqrt{2} \quad  8\end{array}]]| \ {A}^{2}={2}^{2}[[|\begin{array}{ccc}1 \quad  \sqrt{2} \quad  2 \ \sqrt{2} \quad  2 \quad  2\sqrt{2} \ 2 \quad  2\sqrt{2} \quad  4\end{array}]]|[[|\begin{array}{ccc}1 \quad  \sqrt{2} \quad  2 \ \sqrt{2} \quad  2 \quad  2\sqrt{2} \ 2 \quad  2\sqrt{2} \quad  4\end{array}]]| \end{gathered}$

$\begin{gathered} =4[[|\begin{array}{ccc}- \quad  - \quad  - \ - \quad  - \quad  - \ (2+4+8) \quad  (2\sqrt{2}+4\sqrt{2}+8\sqrt{2}) \quad  (4+8+16)\end{array}]]| \ {3}^{}=4(14+14\sqrt{2}+28) \ \begin{gathered} \quad =4(42+14\sqrt{2}) \ =168+56\sqrt{2} \ \alpha +\beta \sqrt{2} \ 	herefore \, \quad \alpha +\beta =168+56=224 \end{gathered} \end{gathered}$

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