Let $$P = \begin{bmatrix} \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}$$, $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ and $$Q = PAP^T$$. If $$P^TQ^{2007}P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ then $$2a + b - 3c - 4d$$ is equal to

Observe that $$P$$ is an orthogonal matrix ($$P P^T = I$$), which means $$P^T = P^{-1}$$.

The expression $$P^T Q^{2007} P$$ simplifies using the property $$Q = PAP^T$$:

$$P^T (P A P^T)^{2007} P = P^T (P A^{2007} P^T) P = (P^T P) A^{2007} (P^T P) = A^{2007}$$

For $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$, the $$n$$-th power is $$A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$$.

Thus, $$A^{2007} = \begin{bmatrix} 1 & 2007 \\ 0 & 1 \end{bmatrix}$$, giving:

• $$a = 1$$
• $$b = 2007$$
• $$c = 0$$
• $$d = 1$$

Final Calculation:

$$2a + b - 3c - 4d = 2(1) + 2007 - 3(0) - 4(1) = 2 + 2007 - 4 = \mathbf{2005}$$

Correct Option: (B)