
From 1st lens
1/v + 1/6 = 1/24
1/v = 1/24 - 1/6 = -1/8
v = -8 cm
From 2nd lens
1/v + 1/18 = 1/9
1/v = 1/9 - 1/18 = 1/18
v = 18 cm
So distance between object and its image:
6 + 10 + 18 = 34 cm

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: