For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. We set:
\[D = D_1 = D_2 = D_3 = 0\]
Calculating \( D_3 \):
\[D_3 = \begin{vmatrix}2 & 7 & 3 \\3 & 2 & 4 \\1 & \mu & -1\end{vmatrix}= 0\]
Expanding, we get:
\[2 \begin{vmatrix}2 & 4 \\\mu & -1\end{vmatrix}- 7 \begin{vmatrix}3 & 4 \\1 & -1\end{vmatrix}+ 3 \begin{vmatrix}3 & 2 \\1 & \mu\end{vmatrix}= 0\]
Solving for \( \mu \), we find:
\[\mu = -39\]
Now, calculating \( D \) with \( \lambda \) in place:
\[D = \begin{vmatrix}2 & 7 & \lambda \\3 & 2 & 5 \\1 & -39 & 32\end{vmatrix}= 0\]
Solving this determinant, we get:
\[\lambda = -1\]
Thus, \( \lambda - \mu = -1 - (-39) = 38 \).
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
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: