Linear algebra terminology: unique, trivial, non-trivial, inconsistent and consistent Let me get this straight:


*

*Unique solution- has an exact solution (such as a POI of 2 intersecting lines)

*Trivial solution- when 0 needs to equal the zero vector in $Ax=0 vector$ 

*Non-Trivial-? Perhaps when 0 does not equal the zero vector in $Ax=0$. This is due to the presence of free variables in Matrix A.

*Consistent- Where there is a unique solution?

*Inconsistent? When there is an inequality such as 4 can't equal 3 meaning the solution can not be solved. This is due to parallel lines.


Am I correct on these  technical terms? 
I put inequality in italics because I'm not sure if I used it correctly.
 A: Your formulations/phrasings are not very precise and should be modified:


*

*Unique solution: Say you are given a $\mathbf{b}$ for which $A\mathbf{x}=\mathbf{b}$; then there is only one $\mathbf{x}$ (i.e., $\mathbf{x}$ is unique) for which the system is consistent. In the case of two lines in $\mathbb{R^2}$, this may be thought of as one and only one point of intersection. 

*Trivial solution: The only solution to $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$. 

*Non-trivial solution: There exists $\mathbf{x}$ for which $A\mathbf{x}=\mathbf{0}$ where $\mathbf{x}\neq\mathbf{0}$.

*Consistent: A system of linear equations is said to be consistent when there exists one or more solutions that makes this system true. For example, the simple system $x+y=2$ is consistent when $x=y=1$, when $x=0$ and $y=2$, etc.

*Inconsistent: This is the opposite of a consistent system and is simply when a system of linear equations has no solution for which the system is true. A simple example $x-x=5$. This is the same as saying $0=5$, and we know this is not true regardless of the value for $x$. Thus, the simple system $x-x=5$ is inconsistent. 

A: Inconsistent is when the system has NO solution. Consistent is the negation of this, e.g., we might have just one solution or infinitely many solutions. And a nontrivial solution might be a solution that's not the zero vector. So say $Ax = \begin{bmatrix}1 & 2 & 3\end{bmatrix}^T$, where $x=(1, 2, 3)^T$ and $A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$.
