**Note: This isn't a direct answer to the question. I've covered all the possible cases I could. Don't read this if you want a straight answer to the given question.**

The number of equations is $m$.

Hence, there can be a $m*m$ matrix.

Hence the maximum rank can be $m$.

Hence, $m$ is the upper bound of $r$.

We conclude that $r \leq m$ ALWAYS.

### Statement I

For homogeneous system of equations; if $m < n$, then $r < n$. Because $m$ is the upper bound of $r$. We'll have infinite solutions.

For non-homogeneous system of equations; maybe $r(A) \neq r(A|B)$. We'll have no solutions in this case.

So, all such systems have a solution? No. **This statement is false**.

### Statement II

$m > n$. The only thing we know is $r$ lies below $m$. It can be between $m$ and $n$. Or it can be lower than $n$.

ie

- $m > r > n$ is possible.
- $m > n > r$ is possible.

For case 2, $r<n$. It's good.

For case 1, $r > n$. Which is inconsistent and we'll have no solutions in case of non-homogeneous equations.

We always want $r \leq n$ be it any system of equations.

So, **this statement is false**, because we have a favourable case.

### Statement III

When $m = n$

$r$ could be $= m$, and hence, $r = n$.

$r$ could be $< m$, and hence $r < n$.

So, yeah, there exists favourable cases. **This statement is correct**.

Suppose $r$ is such that $r(A) \neq r(A|B)$. Then, we'll have no solution. But we need just one favourable case to make this statement True, and we did it.

Option C