$Ax = b$ and $Ax = 0$ solutions I'm having trouble classifying the solution set of the systems $Ax = b$ and $Ax = 0$.
Let A be an ($m \times n$) matrix.
Case I:
For $Ax = 0$, the system has a unique solution (the trivial one) when A is invertible, and infinitely many solutions when A is not. 


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*We can scratch off the "no solution" case because there is always the zero matrix solution correct?


Case II: 
For $Ax = b$, if A is invertible, then for all ($n \times 1$) vector b, the matrix equation has a unique solution given by $x = A^{-1}b$.
Else, there are two remaining cases: infinitely many solutions or no solutions. 


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*How would I know which it is? Can we not tell until we have row reduced the augmented matrix?


Finally is the following true or false?
If y and z are solutions of the system Ax = b then any linear combination of y and z is also a solution.
My thoughts are that it is correct but I am not too sure.
 A: In case II, note  that in general, $A$ will not be invertible, as this requires $m=n$.
The condition for having solutions is the rank of the augmented matrix is equal to $\operatorname{rank}A$ (in all cases it is $\ge\operatorname{rank}A$). It is the case if $A$ has rank $m$, which implies $n\ge m$ and $A$ is right-invertible – in particular if $A$ is invertible.
Concerning the last question, the solution is not a subspace, but it is an affine subspace, which means any weighted mean of solutions is again a solution.
A: 1) yes, there is always the zero - vector solution $x = 0$
2) Yes, you need to compute the rank of the augmented matrix.
3) No. If $Ay = b$ and $Az=b$ then $$A(y+z) = Ay + Az = b + b = 2b \neq b$$
except the case when $b= 0$. If $b=0$ then it is true and that is why $Ker(A)$ is a vector space.
Is there a "certain" combination that works? Let's see:
$$A(\alpha y + \beta z) = \alpha Ay + \beta Az = (\alpha + \beta)b = b$$ and again, with $b \neq 0$ this implies $\alpha + \beta = 1$.
So you still have a solution provided that the "weights" sum up to $1$
