Let $A$ a square matrix with the size of $n \times n$.
I know that if the rank of the matrix is $<n$, then there must be a "zeroes-line", therefore $\det(A)=0$.
What about $\text{rank}(A)=n$? Why does it imply $\det(A)\ne0$?
Of course, there is no "zeroes-line", but that doesn't prove it yet.
I've seen a proof in a book which does this conclusion immediately, but IMHO this alone, doesn't prove it.
What's the missing part?
Best Answer
Let $A$ be an $n\times n$ matrix.
Note that $\det(A) \neq 0$ iff the rows are linearly independent iff $rank(A)=n$.