Determinants - adding one row to another If we think of the determinant of a matrix as the magnitude of the space enclosed by its columns (NOT by its rows), then what's the geometric interpretation of this property:
"Adding one row of a matrix to another doesn't change its determinant"
I know that if we were to consider the ROWS of the matrix as vectors, then its just Cavalieri's principle - all we'd be doing is skewing the parallelogram, which wouldn't change the area enclosed within it.
But keeping the interpretation of the columns being the vectors...
Thanks!
 A: Here's an attempt.  Let's work with this matrix.
$$
    A = \begin{bmatrix}
    a & d & g \\
    b & e & h \\
    c & f & i \\
    \end{bmatrix}
$$
Without loss of generality, let's assume we're going to add the 1st row to the 3rd row.  Also, let's assume $a$ is nonzero.  At least one of the elements in the 1st row must be nonzero otherwise the determinant is zero.
Before we add one row to another, let's use some column operations to find the determinant of the original matrix.
Let's use two column operations (sheering/skewing of the parallelepiped, picture Cavalieri) to "knock out" $d$ and $g$ with $a$ without changing the volume.
$$
    \begin{bmatrix}
    a & (d - \frac{d}{a} a) & (g - \frac{g}{a} a) \\
    b & (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    c & (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
which becomes  
$$
    \begin{bmatrix}
    a & 0 & 0 \\
    b & (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    c & (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
What we've done is "projected" (not orthogonally) the column vectors $(d\;e\;f)^T$ and $(g\;h\;i)^T$ into the $y,z$ plane (the $x=0$ plane).  By shoving them over there with $(a\;b\;c)^T$.
So now, the two column vectors
$$
    \begin{bmatrix}
     0 & 0 \\
    (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
form a parallelogram in the $x=0$ plane.  A 2d parallelogram embedded in 3d.
Hopefully you can see that the area of that parallelogram is the determinant of the matrix you get by removing their (0-valued) x-coordinates, namely
$$
    \begin{bmatrix}
    (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
Consider this 2-dimensional parallelogram to be the base of our reshaped 3 dimensional parallelepiped.  Remember, the volume of our reshaped 3d parallelepiped is the same as that of our original matrix.  We've just sheered it to make one of its faces lie in the $x=0$ plane.  And we're deciding to consider that face the base.
If that's the base, then what's the height (to get base times height)?  It's $a$.  It doesn't matter whether whether the edge extending out of the base is $(a\;b\;c)^T$ or $(a\;0\;0)^T$.  Again, this is Cavalieri's principle.  All that matters is how far this edge goes in the direction orthogonal to the plane that the base exists in.  Since the base is conveniently in the $x=0$ plane, that would be simply the x-coordinate of this edge.
Therefore the following matrix still has the same volume:
$$
    \begin{bmatrix}
    a & 0 & 0 \\
    0 & (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    0 & (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
You can also get this by realizing we could have used the 2nd and 3rd columns to "knock out" the $y$ and $z$ coordinates of $(a\;b\;c)^T$ with column operations, none of which would change the volume.
Now, to get to the question, let's do the same process, but instead start with the matrix you get by adding the 1st row to the 3rd row
$$
    B = \begin{bmatrix}
    a & d & g \\
    b & e & h \\
    (c + a) & (f + d) & (i + g) \\
    \end{bmatrix}
$$
Try to immediately picture this as something we've done to columns again.  We've added the column vector $(0\;0\;a)^T$ to our first vector, and $(0\;\;0\;d)^T$ to the second vector, etc.
When we do the exact same column operations we first did to knock out $d$ and $g$ with $a$, notice that the $+ a$ (the $a$ we've just added on) in position (3, 1) will likewise knock out the $+ d$ and $+ g$ that we've added in (3, 2) and (3, 3).
This means we'll end up with 
$$
    \begin{bmatrix}
    a & 0 & 0 \\
    b & (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    (c + a) & (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
which by similar arguments to above has the same determinant as
$$
    \begin{bmatrix}
    a & 0 & 0 \\
    0 & (e - \frac{d}{a} b) & (h - \frac{g}{a} b) \\
    0 & (f - \frac{d}{a} c) & (i - \frac{g}{a} c) \\
    \end{bmatrix}
$$
again.
The big challenge is in trying to visualize this happening.
The two columns
$$
    \begin{bmatrix}
    d & g \\
    e & h \\
    (f + d) & (i + g) \\
    \end{bmatrix}
$$
still end up shoved into the same place in the $x=0$ plane because the extra bits we added on to the 2nd and 3rd columns (adding $(0\;0\;d)^T$ and $(0\;0\;g)^T$ respectively) are just right to get disappeared away by the extra bit we added to the 1st column when we do the shoving.
For instance, when knocking out $d$ in $A$, we replaced the 2nd column, $(d\;e\;f)^T$, with the following combination of the 1st and 2nd columns:
$$\begin{bmatrix}
d \\
e \\
f
\end{bmatrix} - \frac{d}{a}\begin{bmatrix}
a \\
b \\
c
\end{bmatrix} = \begin{bmatrix}
\color{red}{d - \frac{d}{a}a} \\
e - \frac{d}{a}b \\
f - \frac{d}{a}c
\end{bmatrix} = \begin{bmatrix}
0 \\
e - \frac{d}{a}b \\
f - \frac{d}{a}c
\end{bmatrix}$$
And when knocking out $d$ in $B$, we replaced the 2nd column, $(d\;e\;(f + d))^T$, with the following combination of the 1st and 2nd columns:
$$\begin{bmatrix}
d \\
e \\
(f + d)
\end{bmatrix} - \frac{d}{a}\begin{bmatrix}
a \\
b \\
c + a
\end{bmatrix} = \begin{bmatrix}
\color{red}{d - \frac{d}{a}a} \\
e - \frac{d}{a}b \\
(f + \color{red}{d}) - \color{red}{\frac{d}{a}}(c + \color{red}{a})
\end{bmatrix} = \begin{bmatrix}
0 \\
e - \frac{d}{a}b \\
f - \frac{d}{a}c
\end{bmatrix}$$
which ended up the same.
To emphasize the fact that we had added column vectors $(0\;0\;a)^T$ and $(0\;0\;d)^T$ to the 1st and 2nd columns of $A$ to get the 1st and 2nd columns of $B$ when doing the row operation of interest, you can also view the above combination more explicitly as:
$$
\begin{equation}
\left( \begin{bmatrix}
\color{red}{d} \\
e \\
f
\end{bmatrix} + \begin{bmatrix}
0 \\
0 \\
\color{red}{d}
\end{bmatrix} \right) - \color{red}{\frac{d}{a}} \left( \begin{bmatrix}
\color{red}{a} \\
b \\
c
\end{bmatrix} + \begin{bmatrix}
0 \\
0 \\
\color{red}{a}
\end{bmatrix} \right) = \left( \begin{bmatrix}
\color{red}{d} \\
e \\
f
\end{bmatrix} - \color{red}{\frac{d}{a}} \begin{bmatrix}
\color{red}{a} \\
b \\
c
\end{bmatrix} \right) + \left( \begin{bmatrix}
0 \\
0 \\
\color{red}{d}
\end{bmatrix} - \color{red}{\frac{d}{a}} \begin{bmatrix}
0 \\
0 \\
\color{red}{a}
\end{bmatrix} \right) \\
= \left( \begin{bmatrix}
\color{red}{d} \\
e \\
f
\end{bmatrix} - \color{red}{\frac{d}{a}} \begin{bmatrix}
\color{red}{a} \\
b \\
c
\end{bmatrix} \right) + 0 \end{equation} $$
and so on
A: As the transposed matrice has the same determinant, the geometric intuition with the columns can be identified by that with the rows by taking $M^T$ instead of $M$.
