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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?
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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.
Solved.
