Determinant from matrix entirely composed of variables I don't want the answer, but I'd love to kick in the right direction. I'm really not sure how to approach this question.
$$\begin{align}
& -6 = det\begin{bmatrix}
a & b & c \\ 
d & e & f \\ 
g & h & i \\
\end{bmatrix} \\
& x = det\begin{bmatrix}
a & b & c \\
2d & 2e & 2f \\
g+3a & h+3b & i+3c \\
\end{bmatrix}
& \text{Solve for }x
\end{align}$$
I believe if I set $a=1$, $e=2$, and $i=3$ (all other variables $0$), the determinant of the first matrix is $6$, and then for the second matrix is $12$. These were arbitrary variable initializations and can be any number. The relationship between the two (a scalar multiple of 2) will be the same irrespective of what I set the variables to.
I can then infer that the determinant of the second matrix is $2*det[A]$ or $-12$ because I'm investigating the relationship between the two matrices rather than actually calculating anything. I imagine this is a cheap way out, though. But, from this method I do get $x = -12$, which I believe is the correct answer.
What is the proper way of solving this? I don't want the answer, but I'd like to know the process.
 A: Your answer will be closely related to the $-6$ that was given. You are NOT supposed to find values of $a,b,c,d,...$ that "work". You are supposed to use pure theory. 
A) when you add a multiple of one row to another row, the determinant stays the same.
B) multiplying a row by a constant multiplies the determinant by __.
A: I just solved this by expanding the determinants, and then comparing the two equations. You may be shocked when you do the same. One determinant turns out to be an integer multiple of the other, and so you can very quickly solve for $x$. One way: Actually expand the two determinants and compare. You are clearly supposed to be solving for $x$. Expand and have faith. ;))
A: $$\begin{align}
 det\begin{bmatrix}
a & b & c \\
2d & 2e & 2f \\
g+3a & h+3b & i+3c \\
\end{bmatrix}
=\det \begin{bmatrix}
a & b & c \\
2d & 2e & 2f \\
g & h & i \\
\end{bmatrix}+\det \begin{bmatrix}
a & b & c \\
2d & 2e & 2f \\
3a & 3b & 3c \\
\end{bmatrix}=\det \begin{bmatrix}
a & b & c \\
2d & 2e & 2f \\
g & h & i \\
\end{bmatrix}=2\det \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{bmatrix}
\end{align}$$
It would be useful for you in future  if you can prove what i have done is valid always.
(splitting up into two matrices is valid and taking out a multiple from a row as well)
