x -  y + 3z =  -8      (1)
2x + 3y -  z =   5      (2)
3x + 2y + 2kz = -3k     (3)


A) for which values of K will the system have infinitely many solutions?
B) For which values of K will the system have exactly one solution?
C) For which values of K will the system have no solution?

Add equations (1) and (2)  (both sides). You will get

3x + 2y + 2z  = -3      (4)

3x + 2y + 2kz = -3k     (3)


I placed eq(3) of the original system below equation (4). 


Now subtract eq(4) from eq(3). You will get

2z*(k-1) = -3(k-1).     (5)


Now from (5), it should be clear to you, that the value k = 1 is a SPECIAL value.



a)  If k= 1, then both sides of (5) are equal each other and are equal to zero.

    At k= 1, equation (3) is simply the sum of equations (1) and (3), and, therefore, does not bring any new information or any new 

    restrictions to the unknowns x, y and z, comparing with the system of two equations (1) and (2).


    In other words, at k= 1 the system (1),(2),(3) is EQUIVALENT to the system of two equations (1) and (2).

    This last system has, OBVIOUSLY, infinitely many solutions.


    Thus the answer to question a) is k= 1.



b)  If, on the contrary, k=/=1,  then from (5) you have the unique solution z=  for z.

    You can substitute it into equations (1) and (2), and then you will get the system of two equations in two unknowns x and y.

     The 2x2 coefficient matrix of this system has a non-zero determinant.

     It provides the unique solution in x and y for this reduced system.

     Thus the answer to question b) is k=/= 1.


c)   From the solution above we got that

         - at k = 1 the system has INFINITELY MANY solution;

         - at k =/= 1 the system has a unique solution.

     It exhausts all possible options for k; hence, the set of those {k} the system has no solution is EMPTY.