Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$, where $b_{ij} = 2^{i+j} a_{ij}$ for $1 \le i, j \le 3$. If the determinant of $P$ is $2$, then the determinant of the matrix $Q$ is

Let P = [aᵢⱼ] be a 3 × 3 matrix and let Q = [bᵢⱼ], where bᵢⱼ = 2^{i+j} aᵢⱼ for 1 ≤ i, j ≤ 3. If the determinant of P is 2, then the determinant of the matrix Q is $P = [a_{ij}]_{3 \times 3}, \quad \det(P) = 2$ $b_{ij} = 2^{i+j} a_{ij}, \quad 1 \le i, j \le 3$ $\text{To find: } \det(Q)$ $b_{ij} = 2^{i} \cdot a_{ij} \cdot 2^{j}$ $Q = D P D, \quad D = \operatorname{diag}(2,\ 2^2,\ 2^3)$ $\det(Q) = \det(D)\,\det(P)\,\det(D) = (\det D)^2 \det(P)$ $\det(D) = 2 \cdot 4 \cdot 8 = 2^{1} \cdot 2^{2} \cdot 2^{3} = 2^{6}$ $\det(Q) = (2^6)^2 \cdot 2 = 2^{12} \cdot 2 = 2^{13}$ $\text{row factors: } 2^1 \cdot 2^2 \cdot 2^3 = 2^6$ $\text{column factors: } 2^6 \Rightarrow \det Q = 2^6 \cdot 2^6 \cdot \det P = 2^{12}\cdot 2 = 2^{13}$ $\boxed{\det(Q) = 2^{13}} \quad \text{option (4)}$