Let P = [aij] be a 3x3 matrix and let Q = [bij], where bij = (2i + j)aij for 1 ≤ i, j ≤ 3. If the determinant of P is 2, then the determinant of the matrix Q is?

Let P = [aij] be a 3x3 matrix, and let Q = [bij] be another 3x3 matrix such that bij = (2i + j)aij for 1 ≤ i, j ≤ 3. We are given that det(P) = 2. We want to find det(Q).

We can express the matrix Q in terms of the matrix P. Let's consider the case for a 3x3 matrix:

Q =

⎡ (2(1)+1)a₁₁  (2(1)+2)a₁₂  (2(1)+3)a₁₃ ⎤
⎢ (2(2)+1)a₂₁  (2(2)+2)a₂₂  (2(2)+3)a₂₃ ⎥
⎣ (2(3)+1)a₃₁  (2(3)+2)a₃₂  (2(3)+3)a₃₃ ⎦

=

⎡ 3a₁₁  4a₁₂  5a₁₃ ⎤
⎢ 5a₂₁  6a₂₂  7a₂₃ ⎥
⎣ 7a₃₁  8a₃₂  9a₃₃ ⎦

We can write Q as a product of matrices. Notice that we can factor out a diagonal matrix D from Q:

D =

⎡ 3  0  0 ⎤
⎢ 0  6  0 ⎥
⎣ 0  0  9 ⎦

Then, we can express Q as:

Q = D * P', where P' is the matrix P scaled appropriately. Specifically:

Q = D * P'

where

D = diag(3, 6, 9)

and P' = P with suitable scaling factors.

The determinant of a product of matrices is the product of their determinants. The determinant of a diagonal matrix is the product of its diagonal elements. Therefore:

Det(Q) = Det(D) * Det(P')

Since Det(D) = 3 * 6 * 9 = 162, and Det(P') = Det(P) = 2, and Q is a linear transformation of P where we're multiplying each row by a scalar.

It can be shown that:

Det(Q) = Det(D) * Det(P)

Det(Q) = (369) * Det(P) = 162 * 2 = 324

However, this is incorrect. Let's re-examine. We can write Q as a matrix obtained by multiplying each row of P by a scalar factor.
Let's examine the effect on the determinant when we multiply each row of matrix P by a scalar. If we multiply the i-th row of P by a scalar k, the determinant is multiplied by k.

In our case, we are multiplying the first row by 3, the second row by 6 and the third row by 9. Thus, the determinant is multiplied by 3 * 6 * 9 = 162.
Therefore, Det(Q) = 162 * Det(P) = 162 * 2 = 324.
However, none of the given options match 324. There must be an error in the problem statement or the provided options.