# For what value of a will the system have no solution?

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.

• Do you know linear space? I would like to answer based on your reply. Oct 2 '17 at 17:49
• No, I only learn the method of row operations Oct 2 '17 at 17:51
• I am studying the math for economics ,therefore i do not know much about the method Oct 2 '17 at 17:52
• Before I answer, there is something make me confused. What is your problem? It seems like that you have had an answer to your question. "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." Oct 2 '17 at 17:55
• I do not know much about the system of linear equations. Is it possible that no matter the value of a is, the system is always have solution? Under the above system, is it true that the system will never have no solution , no matter what the value of a is? Thank You. Oct 2 '17 at 17:58

Counterexample: \begin{align} x&+&y&+&z&=&1 \\ x&+&2y&+&3z&=&10 \\ x&+&3y&+&4z&=&11+\epsilon \end{align} for $\epsilon \neq 0$.