I'm kind of lost on this problem: "Prove or give a counterexample to the following statement: If the coefficient matrix of a system of $m$ linear equations in $n$ unknowns has rank $m$, then the system has a solution".
My initial thought on it is that yes, it has a solution. But I am horrid at proving things. So I figured if I talk through my thoughts, someone could assist in carrying them over to some sort of proof.
I found this site (http://mathhelpforum.com/advanced-algebra/29134-linear-equations.html) but I believe the answer is incorrect: the problem states the rank of $m$ must equal the number of equations. The "answer" gives two equations in both examples, but the rank of the examples listed still equals one, so the number of equations being must be limited to 1.
Additionally, any inconsistent matrices would be eliminated, since the rank $m$ would be less than the number of equations (also $m$). I suspect that is the real "answer", but how would I go about proving this?
For a system $AX=B$ where $A$ is $m×n$ ($m\leq n$)and $B$ is $m×1$ matrix, construct an augmented matric $C=[A:B]$. Reducing $C$ into its echelon form, $2$ cases arise:
$1$. $Rank(A)=Rank(C)$, then the system is consistent and has at least one solution.
$2$. $Rank(A)\neq Rank(C)$, then the system is inconsistent and has no solution.
In your question, case $2$ cannot happen as $Rank(A)=m$ necessitates $Rank(C)\geq m$ but for a system of $m$ equations $Rank(C)>m$ is ruled out, so you are left with $Rank(C)=m$ i.e case $1$.