Answer is : (D) inconsistent having no solution
A system of equations is considered overdetermined if there are more equations than unknowns.
Different cases of a overdetermined system , which are inconsistent and have no solution:
case 1: Consider the system of 3 equations and 2 unknowns (X and Y), which is overdetermined because 3>2,
There is one solution for each pair of linear equations: for the first and second equations (0.2, −1.4), for the first and third (−2/3, 1/3), and for the second and third (1.5, 2.5). However, there is no solution that satisfies all three simultaneously.
case 2: A system of three linearly independent equations, three lines (two parallel), no solutions.
case 3: A system of three linearly independent equations, three lines (all parallel), no solution
In systems of linear equations, Li=ci for 1 ≤ i ≤ M, in variables X1, X2, ..., XN the equations are sometimes linearly dependent; in fact the number of linearly independent equations cannot exceed N+1. We have the following possible cases for an overdetermined system with N unknowns and M equations (M>N).
- M = N+1 and all M equations are linearly independent. This hull yields no solution. Example: x = 1, x = 2.
- M > N but only K equations (K < M and K ≤ N+1) are linearly independent. There exist three possible sub-cases of this:
- K = N+1. This hull yields no solutions. Example: 2x = 2, x = 1, x = 2.
- K = N. This hull yields either a single solution or no solution, the latter occurring when the coefficient vector of one equation can be replicated by a weighted sum of the coefficient vectors of the other equations but that weighted sum applied to the constant terms of the other equations does not replicate the one equation's constant term. Example with one solution: 2x = 2, x = 1. Example with no solution: 2x + 2y = 2, x + y = 1, x + y = 3.
- K < N. This hull yields either infinitely many solutions or no solution, the latter occurring as in the previous sub-. Example with infinitely many solutions: 3x + 3y = 3, 2x + 2y = 2, x + y = 1. Example with no solution: 3x + 3y + 3z = 3, 2x + 2y + 2z = 2, x + y + z = 1, x + y + z = 4.