Consider the following linear system
$$\pmatrix{1&4&7\\2&5&8\\3&6&9} \pmatrix{x_1\\x_2\\x_3} = \pmatrix{0\\0\\0}$$
A solution is $x_1 = 1$, $x_ 2=-2$ and $x_3=1$. However, matrix $A$ is singular, which should make this equation unsolvable, but clearly
$$x= \pmatrix{1\\-2\\3}$$
is a solution. How is it possible for there to exist solution when the inverse of matrix $A$ cannot even exist?
On
If it singular, why do you think that it has no solution ? It will have infinitely many solutions. All element of $$\{(x,-2x,x)\mid x\in\mathbb R\}=\text{Span}\{(1,-2,1)\},$$ is a solution. By the way, the vector $(1,-2,3)^T$ is not a solution.
On
Recall, or correct the idea you have, maybe:
The system of equations will have non-trivial solutions (that is not all $x_1,x_2,x_3$ are zeros) if the coefficient matrix is singular.
In this case, the coefficient matrix is singular, and therefore it has a non-trivial solution, that is $x_1=1, x_2=-2, x_3=3$.
Quite on the contrary, the fact that $A$ is singular implies that such a solution exists.
Let $M$ be another matrix, and let's suppose that $M$ is non-singular, i.e. $M$ admits an inverse, $M^{-1}$. Suppose now you want to solve the same equation with $M$. $$Mx=0\iff x=M^{-1}0\iff x=0$$
You'll find nothing but $\left(\begin{array}{c} 0\\0\\0\\\end{array}\right)$.
EDIT
I put here a solution of the system. $$\pmatrix{1&4&7\\2&5&8\\3&6&9} \pmatrix{x_1\\x_2\\x_3} = \pmatrix{0\\0\\0}$$ which means $$\pmatrix{x_1+4x_2+7x_3\\2x_1+5x_2+8x_3\\3x_1+6x_2+9x_3} = \pmatrix{0\\0\\0}$$ (I just perform the matrix product on the left). Substract the first line from the second line we get $x_1+x_2+x_3=0$. Substitute $x_3=x_1+x_2$ into the three equations we get $x_2=-2x_1$. Then we verify that all vectors of the form $(x_1,-2x_1,x_1)$ are solutions of this system.