Determinant of a 3x3 matrix with 6 unknowns given the determinants of two 3x3 matrices with same unknowns? Given:
$$
det(A) = 3 \\ det(B) = -4
$$
$$
A =
\begin{pmatrix}
a & b & c \\
1 & 1 & 1\\
d & e & f
\end{pmatrix} \\
B =
\begin{pmatrix}
a & b & c \\
1 & 2 & 3 \\
d & e & f
\end{pmatrix} \\
C =
\begin{pmatrix}
a & b & c \\
4 & 6 & 8 \\
d & e & f
\end{pmatrix}
$$
Find $det(C)$.
$$
det(A) = (af-cd)+(bd-ae)+(ce-bf) = 3 \\
det(B) = 2(af-cd)+3(bd-ae)+(ce-bf) = -4 \\
det(C) = 6(af-cd)+8(bd-ae)+4(ce-bf) = x
$$
I've written this as an augmented matrix with $(af-cd), (bd-ae), (ce-bf)$ as the unknowns and found the reduced row echelon form to be:
$$
\begin{pmatrix}
1 & 0 & 2 & 3 \\
0 & 1 & -1 & -10 \\
0 & 0 & 0 & x+2
\end{pmatrix}
$$
Can I then conclude that $det(C) = -2$?

copper.hat's answer is a lovely answer, which uses very fundamental attributes of the determinant. Notice that your answer is an algebraic way of saying the same thing. They are really equivalent, just that you have an algebraic error in your working; your augmented matrix should look like (I tried to preserve the column order you used, and can't seem to get the nice vertical line...):
$$
\begin{pmatrix}
1 & 1 & 1 & 3 \\
2 & 3 & 1 & -4 \\
5 & 6 & 4 & x
\end{pmatrix}
$$
Solving this indeed yields $x=5$. The important point here is that once you've extracted your equations and put them in your system, the row reduction you perform is exactly what cooper.hat did in his breakdown.
