\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix} \] \[ A^3 = \begin{bmatrix} 3 & -6 \\ 6 & -3 \end{bmatrix} \] \[ A^4 = \begin{bmatrix} 0 & -9 \\ -9 & -9 \end{bmatrix} \] \[ A^5 = \begin{bmatrix} -9 & -9 \\ 9 & -18 \end{bmatrix} \] \[ A^8 = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \] \[ A^{13} = A^8 \times A^5 = \begin{bmatrix} 81 & 81 \\ -81 & 0 \end{bmatrix} \times \begin{bmatrix} -9 & -9 \\ 9 & -18 \end{bmatrix} \] \[ A^{13} = \left[ (-81)(-9) + (81 \times 9) \right] \quad \left[ (-81)(9) \right] \] \[ \text{Sum of diagonal} = (81 \times 27) = 34^3 \times 3^7 \] \[ \Rightarrow n = 7 \]
The given matrix is:
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}. \]
Compute successive powers of \(A\):
\[ A^2 = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix}. \]
\[ A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -6 \\ 6 & -3 \end{bmatrix}. \]
\[ A^4 = A^3 \cdot A = \begin{bmatrix} 3 & -6 \\ 6 & -3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix}. \]
\[ A^5 = A^4 \cdot A = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -9 & -9 \\ 9 & -18 \end{bmatrix}. \]
\[ A^6 = A^5 \cdot A = \begin{bmatrix} -9 & -9 \\ 9 & -18 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -27 & 0 \\ 0 & -27 \end{bmatrix}. \]
\[ A^7 = A^6 \cdot A = \begin{bmatrix} -27 & 0 \\ 0 & -27 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 36 & -27 \\ -27 & 36 \end{bmatrix}. \]
Observe that the diagonal elements of \(A^7\) are \(36\) and \(36\). Their sum is:
\[ \text{Sum of diagonal elements} = 36 + 36 = 72 = 3^2 \cdot 3^5 = 3^7. \]
Thus:\[ n = 7. \]
Let $ A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\4 & 6 + 2p & 8 + 3p + 2q \\6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix} $ If $ \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n, \, m, n \in \mathbb{N}, $ then $ m + n $ is equal to:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
