If an nxn matrix has rank n how do you know it's invertible?

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An nxn matrix A is invertible if its rank is n, as established by the Fundamental Theorem of Invertible Matrices. This theorem confirms that a full rank matrix (rank n) implies linear independence of its rows and columns, leading to a non-zero determinant. Consequently, the nullity of the matrix, dim(Nul(A)), is zero, indicating that there are no non-trivial solutions to the equation Ax = 0. Thus, the matrix A is guaranteed to be invertible.

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Homework Statement
If an nxn matrix has rank n how do you know it's invertible?

The attempt at a solution
I know that by the Fundamental Theorem of Invertible Matrices if Rank(A) = n then A is invertible. However, I don't know if that is enough of an answer so it kind of seems like I'm supposed to prove it but I'm not sure where to start with that.

Any nudge in the right direction to get my brain cranking is appreciated.
 
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If rank(A)=n, what does that say about dim(Nul(A))?
 
If an nxn matrix has rank n, then all the rows are linearly independent from one another, so what can you say about it's determinant?
 
Let A be that matrix

Col(A) = Span(A)

dim(Col(A)) = dim(Span(A))

What dose that mean? When can the above happen, ever?
 

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