SOLUTION: By solving these equation Px+y+z=6 3x-y+11z=6 2x+y+4z=q Find the condition of (i) a unique solution (ii) no solution (iii) infinitely many solutions

Px + y +   z = 6    (1)
3x - y + 11z = 6    (2)
2x + y +  4z = q    (3)

Let me exclude "y" from the system. In other words, I want to reduce the given 3x3-system to 2x2-system for unknowns "x" and "z".
For it, I will add eqns (1) and (2) to get the equation (4)

(P+3)x + 12z = 12.      (4)

Also, add eqns (2) and (3) to get the equation (5)

5x + 15z = 6+q.         (5)


So, instead of (1),(2) and (3) I have now the system of 2 linear eqns in 2 unknowns

(P+3)x + 12z = 12.      (4)
    5x + 15z = 6+q.     (5)

The determinant of the coefficient matrix M is

det(M) = (P+3)*15 - 5*12 = 15P + 45 - 60 = 15P - 15.

The determinant is zero, 15P - 15 = 0, if and only if P = 1.


Having this, we can make our FIRST conclusion:


     If P is different from 1, P =/= 1, then the system has a unique solution for any value of the parameter q.


Now consider the case P = 1.

In this case the system (4),(5) takes the form

4x + 12z = 12,      
5x + 15z =  6 + q.   

or, simplifying,

 x +  3z =  3,        (6)
5x + 15z =  6 + q.    (7)

Express x from (6):  x = 3 - 3z,  and substitute it into (7). You will get

5*(3-3z) + 15z = 6 + q.

Simplify:

15 - 15z + 15z = 6 + q,

15 = 6 + q,

q = 9.


Having this, we can make two next conclusions.


    For P = 1,  the system (6),(7) has INFINITELY MANY solutions, if q = 9.

    For P = 1, the system (6),(7) has NO solutions if q =/= 9.


It gives the FINAL solution for the original system (1),(2),(3):

   It has a unique solution if P =/=1.

   If P = 1, the system (1)-(3) has INFINITELY MANY solutions at q = 9.

   If P = 1, the system (1)-(3) has NO              solutions at q =/= 9.