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**Last_Singularity** Prove or provide counterexample: if the coefficient matrix of a system of $\displaystyle m$ linear equations in $\displaystyle n$ unknowns has rank $\displaystyle m$, then the system has a solution.

I have an idea for the proof but I am not sure how to formalize it. Some advice, please?

Here's the proof: let $\displaystyle A$ be the coefficient matrix in the system $\displaystyle Ax=b$. If $\displaystyle A$ has rank of $\displaystyle m$, then $\displaystyle n \geq m$. Say that's not true - then $\displaystyle n < m$, implying that the maximum rank of $\displaystyle A$ which equivalent to the max number of linearly independent columns, has a maximum of $\displaystyle n$, which is less than $\displaystyle m$ by assumption, meaning that the rank will never equal $\displaystyle m$. So we must have $\displaystyle n \geq m$. Then if the rank of $\displaystyle A$ is $\displaystyle m$, we can reduce the matrix [A|b] using row operations to a form [A'|b'] where all $\displaystyle x_i,i=1,...,m$ have coefficients of $\displaystyle 1$. Then a solution becomes $\displaystyle x_i=b'_i$ for $\displaystyle i=1,...,m$ and for $\displaystyle i=m+1,...,n$, we have $\displaystyle x_i = 0$.