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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A: I see what you're doing now.  You are using the properties of determinants.  You look fine, but the only time you change signs is when you swap rows.  I think there is a theorem that states that elementary row operations do not change the value of the det of a matrix.  You were right to factor that the coefficient but I don't think you change the sign unless you swap the rows.
A: I'm not sure I understand. The determinant 
\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}=
$a e i-a f h-b d i+b f g+c d h-c e g$=-6.
While
\begin{vmatrix} a+d & b+e & c+f \\ -d & -e & -f \\ g & h & i \end{vmatrix}=
$-a e i+a f h+b d i-b f g-c d h+c e g$
And
\begin{vmatrix} a & b & c \\ 2d & 2e & 2f \\ g+3a & h+3b & i+3c \end{vmatrix}=
$2 a e i-2 a f h-2 b d i+2 b f g+2 c d h-2 c e g$
In order to compute the determinant of a "x" matrix you can use the Sarrus rule (http://en.wikipedia.org/wiki/Rule_of_Sarrus)
