Understanding Ax = 0 in Linear Algebra If we have 3 systems of equations, that intersect at the point (1,2,3) does it have trivial or nontrivial solutions?
Let's assume it is this system of equations which intersects at (1,2,3) and row reduces to the Identity matrix:
$x+y+2z=9$
$2x + 4y - 3z = 1$
$3x + 6y -5z = 0$
I know that Trivial solutions must be at (0,0,0). Since the 3 systems of equations can be reduced to I, the identity matrix, they must be invertible, but since they are not trivial they are not invertible. 
Why the contradiction, or am I just misunderstanding? 
 A: I am sorry to say things are a bit off. It's hard to say what's a misunderstanding and what are typos/terminological/language issues. But, I'll just dive right in :)

If we have 3 systems of equations, 

But here, we have a single system of three equations, not three systems.

that intersect at the point (1,2,3) does it have trivial or nontrivial solutions?

One of the motivations for the study of linear algebra is determining when a system of linear equations has a solution and beyond that, describing the solution(s). Only systems of the form $A\mathbf{x} = \mathbf{0}$ (we call them homogeneous when the right side is the zero vector) "obviously" have a solution (apply $A$ to $\mathbf{0}$, get $\mathbf{0}$ back), and it's only in this case that we even use the term trivial solution, and we only call the "obvious" solution $\mathbf{x} = \mathbf{0} = (0, 0, 0)$ trivial.
Since your right side isn't the zero vector, I wouldn't even use the terms trivial or nontrivial; a solution may or may not exist, we are not guaranteed one, unlike in the homogeneous case. I would just call $(1, 2, 3)$ a solution, I wouldn't call it nontrivial (or trivial).

Let's assume it is this system of equations which intersects at (1,2,3) and row reduces to the Identity matrix:
\begin{align}x+y+2z&=9\\
2x + 4y - 3z &= 1 \\
3x + 6y -5z &= 0
\end{align}
I know that Trivial solutions must be at (0,0,0). 

That's true, only the zero vector would be a trivial solution. The zero vector isn't a solution, but if it were, we'd call it trivial.

Since the $\overset{\text{left hand side of the}}{\wedge}$ 3 systems system of $\overset{\text{three}}{\wedge}$ equations can be reduced to $\mathbf{I}$, the identity matrix, they must be invertible, 

If the left hand side of an augmented matrix can be row reduced to the identity matrix, it means that particular $3 \times 3$ matrix (of coefficients of the left hand sides of the equations)/the linear transformation it represents is invertible, not the system of equations. The implication for the system of equation is that there is a unique solution.

but since they are not trivial they are not invertible. 
Why the contradiction, or am I just misunderstanding? 

When we have a homogeneous system $A\mathbf{x} = \mathbf{0}$, and we know that there are nontrivial solutions, then we can say the matrix/transformation must not be invertible. 
More generally, a system of equations has exactly one solution if and only if the matrix of coefficients is invertible. When the right side is the zero vector, we know $\mathbf{x} = \mathbf{0}$ is a solution. Any other solutions (which we would call nontrivial) would signal that the matrix isn't invertible, since we would have more than one solution.
