Q:

Adding rows to calculate the determinant.

Evaluate the determinants given that $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}=-6.$

$\begin{vmatrix} a+d & b+e & c+f \\ -d & -e & -f \\ g & h & i \end{vmatrix}$ 
$\begin{vmatrix} a & b & c \\ 2d & 2e & 2f \\ g+3a & h+3b & i+3c \end{vmatrix}$

Here is what I have tried:
1.
$\begin{vmatrix} a+d & b+e & c+f \\ -d & -e & -f \\ g & h & i \end{vmatrix}\stackrel{\text{add row 2 to row 1}}=\begin{vmatrix} a & b & c \\ -d & -e & -f \\ g & h & i \end{vmatrix}\stackrel{\text{factor out $-1$}}=-\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}=-(-6)=6.$ 
2.
$\begin{vmatrix} a & b & c \\ 2d & 2e & 2f \\ g+3a & h+3b & i+3c \end{vmatrix}\stackrel{\text{row 1 times -3, add to row 3}}{=}\begin{vmatrix} a & b & c \\ 2d & 2e & 2f \\ g & h & i \end{vmatrix}\stackrel{\text{factor out 2}}{=}2\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}=2(-6)=-12.$
Did I do these correctly? 
I've tried some cases with numbers where adding a multiple of one row to another and found that it doesn't not change the value of the determinant. But I can't seem to grasp the intuition as to why this is so from numeric calculations. 
Why is this so? 

A:

Geometrically, the fact that you can add multiples of rows to each other while keeping the determinant the same, is a reflection of the fact that the determinant can be seen as the volume of the parallelepiped with the rows or columns as it's vectors.
Adding a row to a row has the effect of simply skewing the parallelepiped. Much like skewing a parallelogram does not change it's area (since neither the height of base changes length) the same holds for a parallelepiped, hence the determinant stays the same.
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