how to prove if a rank(A)<n, then det(A)=0. (A is a matrix)

hi, all,
Does anybody know how to prove that for the nxn matrix, if rank(A)<n, then det(A)=0;
any help will be highly appreciated.
Chris
chrisguoAsked:
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ozoConnect With a Mentor Commented:
If rank(A) < n then one of the rows or columns of the matrix is a linear combination of the other rows or columns
if one of the rows or columns of the matrix is a linear combination of the other rows or columns then the determinate of the matrix is 0
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chrisguoAuthor Commented:
ozo,
thanks for the answer, I think I got the answer already.
Since the determinant of a matrix has such properties as follows.
1. any row or column of a matrix added to one row or column, the determinant will not change.
2. if matrix B is one row or column of a matrix A multiply a constant k, then the det(B)=k*det(A);
3. if any row or column of a matrix A is 0, then det(A)=0;
And if rank(A)<n, then we must have at least two rows or columns are linearly dependant, it means k*Ai+Aj=0(Ai and Aj are the vectors of matrix A).
So if we do this operation on matrix A to get a matrix B in which one vector is 0, then we can have det(A)=k*det(B)=0. Then we got it proved.
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