Consider the following system of linear equations
2x + 2y + 3z = 0 (1) 4x + 8y + 12z = -4 (2) 6x + 2y + az = 4 (3)
Is the following system always consistent no matter what the value of a is ? By applying the row operation, I find that R3 of the matrix is
0x + 0y + (a-3)z = 0
Therefore, if the value of a is 3, the system have infinitely many solutions. And if the value of a != 3, the system have a unique solution.
Does it mean that this system must have solution no matter what the value of $a$ is?
If no, how can I find the value of $a$ such that the system has no solution? Thank You.