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
Think of Cramer's rule. The determinant of the coefficients is
D = k^2-1
So, for any k≠1 or -1 there is a unique solution
If k=1, we have
x + y + 2z =0
x + y + 3z =0
x + y + 4z = 1
since z=0 ans x+y=0 are the only solutions to the first two, but fail on the 3rd, no solution.
If k = -1, we have
x+ -y + 2z =0
x + -3y + 3z =0
x - y + 2z = -3
again, no solution.
I can't find any k such that the three equations describe the same plane.
To find the values of k that satisfy each condition, we need to analyze the system of equations and determine the relationship between its equations.
a) For the system to have a unique solution, it means that the equations are non-parallel and non-coincident. In other words, the determinant of the coefficient matrix should be non-zero.
Let's write the coefficients into a matrix A:
A = [1, k, 2;
1, 2k-1, 3;
1, k, k+3]
Now, calculate the determinant of A:
det(A) = (1)((2k-1)(k+3) - 3k) - (0) - (1)((1)(k+3) - (3k)) = (2k^2 - 5k + 3) - (k^2 - 2k + 3) = k^2 - 5k + 3 - k^2 + 2k - 3 = -3k
For the system to have a unique solution, det(A) ≠ 0. Therefore, we need to solve the equation -3k ≠ 0.
Setting -3k ≠ 0, we obtain k ≠ 0.
So, for the system to have a unique solution, k cannot equal 0.
b) For the system to have no solutions, it means that the equations are parallel and non-intersecting. In other words, the determinant of the coefficient matrix should be zero.
To find the values of k, we need to solve the equation -3k = 0.
Setting -3k = 0, we get k = 0.
Therefore, for the system to have no solutions, k must equal 0.
c) For the system to have infinitely many solutions, it means that the equations are parallel and coincident. In other words, the determinant of the coefficient matrix is zero, and the right-hand side of the equations are proportional.
To find the values of k, we need to solve the equation -3k = 0 and check if the right-hand sides of the equations are proportional.
Setting -3k = 0, we get k = 0.
Next, compare the right-hand sides of the equations:
2k - 1 = k^2 - 2k + 3
Simplifying, we have:
k^2 - 4k + 4 = 0
(k - 2)^2 = 0
k - 2 = 0
k = 2
Therefore, for the system to have infinitely many solutions, k must equal 0 and 2.
To summarize:
a) The system has a unique solution if k is any real number except 0.
b) The system has no solutions if k = 0.
c) The system has infinitely many solutions if k = 0 or k = 2.