I am solving some calculus exercises, and encountered one which put me into a confusion. It requires to:
determine how many solutions the system has
provide geometrical interpretation of each solution as plane in space for any k
determine number of solutions for system if k is such that second plane passes through point Q=(1, 1, 1).
The system itself is:
x-4y+3z=0
2x-7y+kz=2
3y-6z=-3
By applying Cramer's rule I have found:
dX/D = (12k-93)/(-3k+12) dY/D = (-30+3k)/(-3k+12) dZ/D = (-9)/(-3k+12)
And after that I don't really understand how to proceed. I have tried to find value of k for every equation and then perform division (e.g. for dZ/D is -9/4=2,25), but I think this is a wrong strategy, which doesn't bring me to anything.
I would be grateful for couple of hints on how to proceed with solution.
P.S. sorry for messiness, didn't figure out how to use MathJax yet.
The first thing I note is that the third equation, 3y- 6z= -3 doesn't involve x (and can be more simply written as y- 2z= -1) so I would eliminate x from the other two equations, x- 4y+ 3z= 0 and 2x+ 7y+ kz= 2 (you have "+kx" but I assume that was a typo). Of course x= 4y- 4z so 2(4y- 4z)+ 7y+ kz= 15y- (8- k)z= 2. Since y= 2z- 1, that is 30z- 15- (8- k)z= (22- k)z- 15= 2 and then z= 17/(22- k). If k is not 22, there is a single solution. If k= 22 there are NO solutions.