The equations

Question

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

Answer 1

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.