Determinant if Two Rows Are Equivalent What is the reasoning that the determinant will be zero if two rows are exactly the same? I see a lot of pseudo proofs out there for this one and would like to know the reasoning behind it. 
 A: Theorem: for two $n\times n$ matrices of the same size, $A$ and $B$, one has $\det(AB)=\det(A)\det(B)$ (not proven here)
Theorem: Elementary row operations can be described via left multiplication by elementary matrices.  More specifically, the row operation $r_i+kr_j\mapsto r_i$ is described by the matrix with ones along the diagonal and $k$ in the $i^{th}$ row $j^{th}$ column position.  For example, in a $4\times 4$ matrix, doing the operation $r_3-r_4\mapsto r_3$ is accomplished by $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&-1\\0&0&0&1\end{bmatrix}$
Theorem: The previously described elementary matrix for row addition has determinant one.  This is seen immediately by the fact that there is only ever one nonzero pattern in the matrix, namely the main diagonal.

Result: If you have a square matrix $A$ and you wish to compute its determinant, you may add or subtract multiples of rows from other rows and it will not change its determinant since $\det(A)=\det(E)\det(A)=\det(EA)$ for elementary matrices that apply this row summation operation.  If this happens to cause a row of all zeroes to appear, it immediately implies that the determinant will be zero as every pattern in the matrix will have to take an entry from that row and the product will necessarily be zero in that case.
Note: other elementary row operations effect a change to the determinant.  Row swaps will flip the sign of the determinant from positive to negative or vice versa.  Row multiplication by a constant $k$ will multiply the resulting determinant by the same constant $k$.

Another quick way to see this is via the invertible matrix theorem.  Among the many items on the list for the invertible matrix theorem is:
An $n\times n$ matrix has nonzero determinant iff the matrix is invertible iff its rows form a linearly independent set.
If two rows are equal, they are linearly dependent, implying they do not form a linearly independent set, implying the matrix is not invertible, implying the determinant is zero.
A: One way to prove this is by the Rank-Nullity Theorem. This states that for any $n$ by $n$ matrix $A$ $$ \mathrm{dim} (\mathrm{null}(A)) + \mathrm{rank}(A) = n$$
If two rows of A are linearly dependent, A does not have full rank $n$. Therefore, $$\mathrm{dim} (\mathrm{null}(A)) > 0$$
This in turn implies that $\mathrm{det}(A) = 0$.
