Let A=[a_i j] be a matrix of order 3 \times 3 , with a_i j=(\sqrt2)^i+j . If the sum of all the elements in the third row of A^2 is \alpha+\beta \sqrt2, \alpha, \beta \in Z , then \alpha+\beta is equal to :

Matrices and Determinants PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q39 with Solution

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Correct answer: C

Given $a_{ij}=(\sqrt{2})^{i+j}, \quad i,j\in\{1,2,3\}.$

(\sqrt2)^2 & (\sqrt2)^3 & (\sqrt2)^4\\ (\sqrt2)^3 & (\sqrt2)^4 & (\sqrt2)^5\\ (\sqrt2)^4 & (\sqrt2)^5 & (\sqrt2)^6 \end{bmatrix} =\begin{bmatrix} 2 & 2\sqrt2 & 4\\ 2\sqrt2 & 4 & 4\sqrt2\\ 4 & 4\sqrt2 & 8 \end{bmatrix}.$$ We need the **sum of all elements in the third row of** $A^2$. --- 2. **Use the structure of the matrix** Observe that $$a_{ij}=(\sqrt2)^i(\sqrt2)^j.$$ So if we define $$u=\begin{bmatrix}(\sqrt2)^1\\(\sqrt2)^2\\(\sqrt2)^3\end{bmatrix} =\begin{bmatrix}\sqrt2\\2\\2\sqrt2\end{bmatrix},$$ then $$A=uu^T.$$ Hence, $$A^2=(uu^T)(uu^T)=u(u^Tu)u^T=(u^Tu)A.$$ Now, $$u^Tu=(\sqrt2)^2+2^2+(2\sqrt2)^2=2+4+8=14.$$ Thus, $$A^2=14A.$$ --- 3. **Find the third row of $A$** The third row of $A$ is $$[4,\;4\sqrt2,\;8].$$ So the third row of $A^2=14A$ is $$14[4,\;4\sqrt2,\;8]=[56,\;56\sqrt2,\;112].$$ --- 4. **Sum the elements of the third row** Required sum $$=56+56\sqrt2+112=168+56\sqrt2.$$ So $$\alpha=168,\quad \beta=56.$$ Therefore, $$\alpha+\beta=168+56=224.$$ --- 5. **Match with the options** $$224$$ corresponds to **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Derived answer: **C** So they agree.

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Correct answer: C

Given $a_{ij}=(\sqrt{2})^{i+j}, \quad i,j\in\{1,2,3\}.$

(\sqrt2)^2 & (\sqrt2)^3 & (\sqrt2)^4\\ (\sqrt2)^3 & (\sqrt2)^4 & (\sqrt2)^5\\ (\sqrt2)^4 & (\sqrt2)^5 & (\sqrt2)^6 \end{bmatrix} =\begin{bmatrix} 2 & 2\sqrt2 & 4\\ 2\sqrt2 & 4 & 4\sqrt2\\ 4 & 4\sqrt2 & 8 \end{bmatrix}.$$ We need the **sum of all elements in the third row of** $A^2$. --- 2. **Use the structure of the matrix** Observe that $$a_{ij}=(\sqrt2)^i(\sqrt2)^j.$$ So if we define $$u=\begin{bmatrix}(\sqrt2)^1\\(\sqrt2)^2\\(\sqrt2)^3\end{bmatrix} =\begin{bmatrix}\sqrt2\\2\\2\sqrt2\end{bmatrix},$$ then $$A=uu^T.$$ Hence, $$A^2=(uu^T)(uu^T)=u(u^Tu)u^T=(u^Tu)A.$$ Now, $$u^Tu=(\sqrt2)^2+2^2+(2\sqrt2)^2=2+4+8=14.$$ Thus, $$A^2=14A.$$ --- 3. **Find the third row of $A$** The third row of $A$ is $$[4,\;4\sqrt2,\;8].$$ So the third row of $A^2=14A$ is $$14[4,\;4\sqrt2,\;8]=[56,\;56\sqrt2,\;112].$$ --- 4. **Sum the elements of the third row** Required sum $$=56+56\sqrt2+112=168+56\sqrt2.$$ So $$\alpha=168,\quad \beta=56.$$ Therefore, $$\alpha+\beta=168+56=224.$$ --- 5. **Match with the options** $$224$$ corresponds to **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Derived answer: **C** So they agree.