The equations
x+ ky + 2z =0
x + (2k-1)y + 3z =0
x + ky + (k+3)z = 2k-1
Find the values of k such that
a) the system has a unique solution
b) the system has no solutions
c) the system has infinitely many solutions
Thanks
To find the values of k that satisfy each scenario, we need to consider the determinant of the coefficient matrix. The coefficient matrix for this system is:
| 1 k 2 |
| 1 2k-1 3 |
| 1 k k+3 |
Let's go through each scenario and explain the process to determine the values of k:
a) To have a unique solution, the determinant of the coefficient matrix must be non-zero. So, we need to calculate the determinant of the matrix and identify the values of k that make it non-zero. If the determinant is non-zero, the system will have a unique solution.
b) To have no solutions, the determinant of the coefficient matrix must be zero, and the system must be inconsistent. We'll calculate the determinant and look for the values of k that make it equal to zero. If the determinant is zero, it implies that the system has no solutions.
c) To have infinitely many solutions, the determinant of the coefficient matrix must be zero, and there should be dependent equations in the system. We'll calculate the determinant and check if it's zero and if there are any dependent equations. If the determinant is zero and there are dependent equations, it indicates that the system has infinitely many solutions.
Let's calculate the determinant of the coefficient matrix and analyze the values of k for each scenario:
a) Unique Solution:
For a unique solution, the determinant of the coefficient matrix must be non-zero:
Determinant = (1)((2k-1)(k+3) - (k)((2k-1)(1)) - (2)((k)(k+3))
= (2k^2 + k - 3 - 2k^2 + k + 1 - 2k^2 - 2k)
= k - 2
For the system to have a unique solution, we need k ≠ 2.
b) No Solution:
For no solution, the determinant of the coefficient matrix must be zero:
Determinant = (1)((2k-1)(k+3) - (k)((2k-1)(1)) - (2)((k)(k+3))
= (2k^2 + k - 3 - 2k^2 + k + 1 - 2k^2 - 2k)
= k - 2
For the system to have no solution, we require k = 2.
c) Infinitely Many Solutions:
For infinitely many solutions, the determinant of the coefficient matrix must be zero and there should be dependent equations:
Determinant = (1)((2k-1)(k+3) - (k)((2k-1)(1)) - (2)((k)(k+3))
= (2k^2 + k - 3 - 2k^2 + k + 1 - 2k^2 - 2k)
= k - 2
For the system to have infinitely many solutions, we need k = 2.
To summarize:
a) The system has a unique solution when k ≠ 2.
b) The system has no solution when k = 2.
c) The system has infinitely many solutions when k = 2.
These are the values of k that satisfy each scenario.