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. \[ \alpha + \beta = 168 + 56 = 224 \] 

Answer 2

Step 1: Write the given information.
We are given a \( 3 \times 3 \) matrix \( A = [a_{ij}] \), where:
\[ a_{ij} = (\sqrt{2})^{i + j} \] We have to find the sum of all elements in the third row of \( A^2 \), which is expressed as \( \alpha + \beta\sqrt{2} \), and then compute \( \alpha + \beta \).

Step 2: Write the matrix A explicitly.
Using \( a_{ij} = (\sqrt{2})^{i+j} \):
\[ A = \begin{bmatrix} (\sqrt{2})^{2} & (\sqrt{2})^{3} & (\sqrt{2})^{4} \\ (\sqrt{2})^{3} & (\sqrt{2})^{4} & (\sqrt{2})^{5} \\ (\sqrt{2})^{4} & (\sqrt{2})^{5} & (\sqrt{2})^{6} \end{bmatrix} = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix} \]

Step 3: Compute \( A^2 \).
Let us multiply \( A \times A \). We are interested only in the **third row** of \( A^2 \), which can be obtained as:
\[ \text{Third row of } A^2 = (\text{Third row of } A) \times A \] Third row of \( A \) = [4, 4√2, 8].

Step 4: Compute the product.
We calculate each element of the third row of \( A^2 \):

(1) Element (3,1):
\[ 4(2) + 4\sqrt{2}(2\sqrt{2}) + 8(4) = 8 + 16 + 32 = 56 \]
(2) Element (3,2):
\[ 4(2\sqrt{2}) + 4\sqrt{2}(4) + 8(4\sqrt{2}) = 8\sqrt{2} + 16\sqrt{2} + 32\sqrt{2} = 56\sqrt{2} \]
(3) Element (3,3):
\[ 4(4) + 4\sqrt{2}(4\sqrt{2}) + 8(8) = 16 + 32 + 64 = 112 \]

Thus, the third row of \( A^2 \) is:
\[ [56, 56\sqrt{2}, 112] \]

Step 5: Sum of all elements in the third row.
\[ \text{Sum} = 56 + 56\sqrt{2} + 112 = (168) + 56\sqrt{2} \] Hence, \( \alpha = 168 \) and \( \beta = 56 \).

Step 6: Compute \( \alpha + \beta \).
\[ \alpha + \beta = 168 + 56 = 224 \]

Final Answer:
\[ \boxed{224} \]