Let $A=\left[a_{ij}\right]$ 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 $\alpha+\beta\sqrt{2}$, $\alpha, \beta \in \mathbb{Z}$, then $\alpha+\beta$ is equal to:

Step 1: Define the matrix A

First, we define the elements of the matrix $A$ using the given rule $a_{ij} = (\sqrt{2})^{i+j}$. This means each element is determined by the sum of its row and column indices.

Step 2: Calculate the elements of A

Now, we calculate each element $a_{ij}$ by substituting the values of $i$ and $j$ into the formula $(\sqrt{2})^{i+j}$. For example, $a_{11} = (\sqrt{2})^{1+1} = (\sqrt{2})^2 = 2$, and $a_{12} = (\sqrt{2})^{1+2} = (\sqrt{2})^3 = 2\sqrt{2}$. We do this for all 9 elements.

Step 3: Calculate the third row of A squared

To find the elements of the third row of $A^2$, we use the matrix multiplication rule. The element $(A^2)_{3j}$ is the dot product of the third row of $A$ and the $j$ -th column of $A$.

Step 4: Compute the elements of the third row of A squared

We calculate each element of the third row of $A^2$: $(A^2)_{31}$, $(A^2)_{32}$, and $(A^2)_{33}$. This involves multiplying the elements of the third row of $A$ by the corresponding elements of the first, second, and third columns of $A$, respectively, and summing the products.

Step 5: Sum the elements of the third row of A squared

Now, we sum the three elements we just calculated for the third row of $A^2$. This gives us the total sum in the form $\alpha + \beta\sqrt{2}$.

Step 6: Determine $\alpha$ and $\beta$ and calculate their sum

By comparing the sum $168 + 56\sqrt{2}$ with $\alpha + \beta\sqrt{2}$, we identify $\alpha = 168$ and $\beta = 56$. Finally, we calculate the sum $\alpha + \beta$.

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