To find the values of m and n such that the system has infinitely many solutions, we start by setting the determinant of the coefficient matrix to zero:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & m \end{vmatrix} = 0. \]
Expanding the determinant:
\[ D = 1 \cdot (5m - 10) - 1 \cdot (2m - 5) + 1 \cdot (4 - 5) = 3m - 6. \]
Setting \(D = 0\):
\[ 3m - 6 = 0 \implies m = 2. \]
Next, consider the augmented matrix determinant \(D_3\):
\[ D_3 = \begin{vmatrix} 1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & n \end{vmatrix} = 0. \]
Expanding \(D_3\) and setting it to zero gives:
\[ n = 7. \]
Substitute \(m = 2\) and \(n = 7\) into the given equation:
\[ m^2 + n^2 - mn = 2^2 + 7^2 - (2 \times 7) = 4 + 49 - 14 = 39. \]
Therefore, the correct answer is Option (4).
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: