Determinant of Scaled Matrix Entries with Power Factors

To find the determinant of matrix $Q$, we write the matrices $P$ and $Q$ explicitly:

$$P = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

The entries of matrix $Q = [b_{ij}]$ are given by $b_{ij} = 2^{i+j} a_{ij} = 2^i \cdot 2^j \cdot a_{ij}$ for $1 \le i, j \le 3$. Thus, the matrix $Q$ is:

$$Q = \begin{bmatrix} 2^{1+1} a_{11} & 2^{1+2} a_{12} & 2^{1+3} a_{13} \\ 2^{2+1} a_{21} & 2^{2+2} a_{22} & 2^{2+3} a_{23} \\ 2^{3+1} a_{31} & 2^{3+2} a_{32} & 2^{3+3} a_{33} \end{bmatrix}$$

Using the properties of determinants:

1. We can factor out $2^1$ from the first row ($R_1$), $2^2$ from the second row ($R_2$), and $2^3$ from the third row ($R_3$):

$$\det(Q) = 2^1 \cdot 2^2 \cdot 2^3 \begin{vmatrix} 2^1 a_{11} & 2^2 a_{12} & 2^3 a_{13} \\ 2^1 a_{21} & 2^2 a_{22} & 2^3 a_{23} \\ 2^1 a_{31} & 2^2 a_{32} & 2^3 a_{33} \end{vmatrix}$$

1. Next, from the resulting determinant, we can factor out $2^1$ from the first column ($C_1$), $2^2$ from the second column ($C_2$), and $2^3$ from the third column ($C_3$):

$$\det(Q) = (2^1 \cdot 2^2 \cdot 2^3)(2^1 \cdot 2^2 \cdot 2^3) \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$

Combining the powers of $2$:

$\det(Q) = 2^{1+2+3} \cdot 2^{1+2+3} \cdot \det(P) = 2^6 \cdot 2^6 \cdot \det(P) = 2^{12} \det(P)$

Given that $\det(P) = 2$, we obtain:

$\det(Q) = 2^{12} \cdot 2 = 2^{13}$

Hence, the correct option is (D).