Determinant equal to zero I would like a demonstration of the following theorem:
Original question:  If a square matrix of order $n\gt 1$ has a row (column) that is linear combination of rows (columns) parallel to it, then the determinant of this matrix is null.
Actual intended question: If the determinant of a square matrix of order $n\gt 1$ is zero, then this matrix has some row (column) that is linear combination of rows (columns) parallel to it.
 A: If you know the determinant is multilinear (linear in each row/linear in each column), then express the the determinant as a sum of determinants of matrices that have repeated rows/columns. Then show that the determinant of a matrix with two identical rows/columns is always zero.
If you do not know the determinant is multilinear in the rows/columns, and you know the definition in terms of sums of products over all permutations, then the same argument works to show the determinant is a sum of determinants of matrices with repeated rows/columns.
If your only definition of determinant is inductively (by minors), then prove it by induction, expanding along a row or column that is not a linear combination of other rows/columns. 
Edit: You've edited your question, and now the question is the precise converse of what you originally posted. This is a bit of bad form, since any answers already posted would automatically look completely wrong-headed. In the future, consider editing to add the new question to the old, indicating this is an edit or addition.
So, now you want to show that if the determinant is zero, then a row/column is a linear combination of other rows/columns. Looking at the tranpose if necessary it suffices to consider the case of rows. Note that applying elementary row operations does not change whether the determinant is zero or nonzero (an elementary row operation will either multiply the determinant by a nonzero constant, or will not change the determinant). So the determinant of the matrix is zero if and only if the determinant of its row-echelon form is zero; the determinant of the row-echelon form is zero if and only if there is a row of zeros. Tracing back the elementary row operations will express the original row as a linar combination of the other rows. 
A: Think of this square n x n matrix as a transformation T on a vector space V. This transformation sends $V^{n} \rightarrow V^{n}$. Now, if the determinant of a matrix is equal to 0, we know it isn't invertible. 
Since our vector space is finite-dimensional, a transformation that has a non-invertible matrix representation is neither onto nor one-to-one. However, if all our columns were linearly independent vectors in $V^n$, they would span $V^n$ and T would be onto. One of the columns must therefore be a linear combination of the others. 
