JEE Mathematics 2025 Question with Solution

Given: $A = [a_{ij}]$ is a matrix of order $3 \times 3$ with $a_{ij} = (\sqrt{2})^{i+j}$.

Find: The value of $\alpha + \beta$ if the sum of all elements in the third row of $A^2$ is $\alpha + \beta\sqrt{2}$.

Write the matrix explicitly:

Only the third row of $A^2$ is needed. So multiply the third row of $A$ by the matrix $A$.

Third row of $A$ is $[4, 4\sqrt{2}, 8]$.

Now compute each entry of the third row of $A^2$:

Hence, the third row of $A^2$ is

So the sum of its elements is

Therefore, $\alpha = 168$ and $\beta = 56$.

Thus,

Therefore, the correct option is D.