Understanding the effect of Earth's rotation on gravity.
Due to the rotation of the Earth, the effective acceleration due to gravity is given by: \[ g_{\text{eff}} = g - \omega^2 R \cos^2 \theta \] Where \( \theta \) is the angle made with the equator.
At the poles, where \( \theta = 90^\circ \), the change in gravity is zero because \( \cos(90^\circ) = 0 \). This shows no effect on the poles. For the equator, where \( \theta = 0^\circ \), the change in gravity is maximum: \[ g_{\text{eff}} = g - \omega^2 R \] Thus, the change in gravity is maximum at the equator and zero at the poles.
This contradicts Statement II, but Statement I is correct.

Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: