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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.

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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.

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  • $\begingroup$ Yes, it was a typo, thank you for noticing. Fixed it. Unfortunately, I do not clearly understand where does x= 4y- 4z come from. Following your advice, I eliminated X in first two equations by multiplying first one by -2. Then, with a result, if did [y-(5+k)z=2] + [y-2z=-1], so Z was -3/(5-k). Is it possible to say that if k+5, there are no solutions? Probably not, if you have come with 22... $\endgroup$
    – Daniil
    Apr 11 '20 at 13:07
  • $\begingroup$ That was MY mistake- a typo that I didn't catch! From x- 4y+ 3z= 0 we have x= 4y- 3z, not "4z". $\endgroup$
    – user247327
    Apr 11 '20 at 17:09

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