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 \( \alpha + \beta\sqrt{2} \), where \(\alpha, \beta \in \mathbb{Z}\), then \(\alpha + \beta\) is equal to:
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When working with matrix exponentiation, identifying patterns in repeated elements simplifies calculations significantly.
Step 1: Matrix Element Identification. The matrix is defined as \( A = \begin{bmatrix} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{bmatrix} \), which simplifies to \( \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix} \).
Step 2: Matrix Squaring. The square of the matrix is \( A^2 = 2\sqrt{2} \begin{bmatrix} 2 & 2 & 4 \\ 2 & 2 & 4 \\ 2 & 2 & 4 \end{bmatrix} \).
Step 3: Third-Row Summation. The sum of the elements in the third row is calculated as \( 4(2 + 4 + 8) = 4(14\sqrt{2} + 28) \), resulting in \( 168 + 56\sqrt{2} \).
Step 4: Final Result. The sum \( \alpha + \beta \) equals \( 168 + 56 = 224 \).
