Relation between determinant and matrix rank

The rank of $A$ can be viewed as $m$ where $m$ is the size of the largest non-zero $m\times m$ submatrix with non-zero determinant.

Alternatively, you can row reduce the matrix to give you an upper triangular matrix using row interchanges and adding scalar multiples of a row to another row. This will only affect the sign of the determinant. If an $n \times n$ matrix has rank $n$ then it has $n$ pivot columns (and therefore $n$ pivot rows). This means you will be able to row reduce it to an upper triangular form with pivots along the diagonal. The determinant is the product of these elements along the diagonal. Can you prove that? Pivots are necessarily non-zero and therefore their product is non-zero, regardless of sign.

Let $A$ be an $n\times n$ matrix.

Note that $\det(A) \neq 0$ iff the rows are linearly independent iff $rank(A)=n$.