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.
- 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.
- 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.
Best Answer
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.