If the system of linear equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = μ has infinitely many solutions, then
Step-by-Step Solution
Key Concept: For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matrix must satisfy the condition for consistency (rank(A) = rank(A|B) < number of variables).
The system is: x + y + z = 6, x + 2y + 3z = 10, x + 2y + \lambda z = \mu. The coefficient matrix is A = [[1, 1, 1], [1, 2, 3], [1, 2, \lambda]]. For infinitely many solutions, det(A) = 0. det(A) = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) = 2\lambda - 6 - \lambda + 3 = \lambda - 3. Setting \lambda - 3 = 0 gives \lambda = 3. Substituting \lambda = 3, the third equation becomes x + 2y + 3z = \mu. Comparing this with the second equation x + 2y + 3z = 10, for the system to have infinitely many solutions, we must have \mu = 10.
Correct Answer: 1