Question

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:

• A. 280
• B. 168
• C. 210
• D. 224

Hint:

When working with matrix exponentiation, identifying patterns in repeated elements simplifies calculations significantly.

Answer 1

The Correct Option is D

Step 1: Identifying the matrix elements. Matrix \( A = \begin{bmatrix} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{bmatrix} = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix} \)

Step 2: Squaring the matrix. \[ A^2 = 2\sqrt{2} \begin{bmatrix} 2 & 2 & 4 \\ 2 & 2 & 4 \\ 2 & 2 & 4 \end{bmatrix} \] 

Step 3: Finding the third-row elements. Sum of elements in the third row: \[ 4(2 + 4 + 8) = 4(14\sqrt{2} + 28) \] \[ = 168 + 56\sqrt{2} \] 

Step 4: Final Calculation. \[ \alpha + \beta = 168 + 56 = 224 \]