Intuition behind determinant of a matrix with $2$ equal rows In my linear algebra course, we have just proved that if a matrix $A$ contains $2$ equal rows, then $\det(A)=0$. 
I understand how the proof works, but could somebody offer a more intuitive explanation of why this is the case?
 A: Apologies as this is my first StackExchange post.
As Robert Israel says, but in less mathematical terms.
The 2D determinant is the area of the parallelogram formed by the two vectors
of the rows (or columns).
See:
Why determinant of a 2 by 2 matrix is the area of a parallelogram?
The 3D determinant is the volume of the parallelepiped formed by the three vectors of the rows or columns and so on into higher dimensions.
If any of the two vectors point in the same direction, this area/volume/hypervolume becomes zero.
I hope this is an intuitive answer.
A: If 2 rows are equal then they are linearly dependant so you'll have a zero eigen value and so the determinant (which is the product of the eigen values) will be zero.
A: If you subtract the two rows the new matrix you get has the same determinant and has a row with only zeros in it. If you apply Laplace's expansion with that row you get $0$
A: It depends on what properties of determinant you consider "intuitive".
One possibility: as a mapping of $\mathbb R^n$ to itself, $A$ multiplies $n$-dimensional volumes by $|\det(A)|$.  If rows $i$ and $j$ of $A$ are equal, $A$ maps $\mathbb R^n$ to vectors whose $i$'th and $j$'th entries are equal.  This set has $n$-dimensional measure $0$, so the determinant must be $0$.
A: If you swap the two rows the determinant is multiplied by $-1$ as the determinant is an alternating multi linear function of its rows. Hence
$$
\det(A)=-\det(A)
$$ 
whence
$$
\det(A)=0
$$
A: If the rows of the matrix are vectors and the determinant is not zero, then the row vectors form a basis for $\mathbb R^n$. If two rows of a matrix are equal then there are at most $n-1$ distinct row vectors and they can not form a basis which requires $n$ vectors.
A: Please correct me if I'm wrong. A determinant of n*n matrix can be consider the "volume" of the shape enclosed by the n column/row vectors in the space of R^n.
In particular, suppose we have a 3*3 matrix but with 2 row vectors equal/linear dependent and the 3rd vector linear independent with them. The "volume" in this case is just an area in R^2 since it's only enclosed by 2 linearly independent vector, which has 0 volume in R^3. Similar argument applies to higher dimension. 
This is just my intuitions on determinant, sorry it's very loose.
Also, I found 3Blue1Brown's video on linear algebra gives very good intuitions on the object:
https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab
A: Consider the following system of equations:
$
2x + 6y + z = A \\
4x + 12y + 2z = B \\
3x + 7y + 2z = C
$
Geometrically, this system of equations represents planes (or lines, etc in other dimensions) and 
 the solution represents their point of intersection. If the coefficients of any two equations are linearly related, it implies the two planes are parallel to eachother. If two of the planes are parallel, the system does not have a unique solution.
This system can be represented using matrices by putting the coefficients into a matrix:
$
\begin{bmatrix} 2 & 6 & 1 \\ 4 & 12 & 2 \\ 3 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} A \\ B \\
C\end{bmatrix}
$
The solution of the system can be found using inverse matrices:
$
 \begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 2 & 6 & 1 \\ 4 & 12 & 2 \\ 3 & 7 & 2 \end{bmatrix}^{-1} \begin{bmatrix} A \\ B \\
C\end{bmatrix}
$
Observe that the second row is linearly related to the first (R2 = R1 * 2). This implies there is not a unique solution to the system (parallel planes). This means that the inverse of the matrix is undefined: the determinant must be 0. This fact holds for any linear relationship between the rows.Therefore, if two rows of a matrix are equal, they are linearly related, which means the determinant is 0. 
