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 $det(C) = -2$?

The determinant is a multilinear function of the rows (or columns). Since $(4,6,8) = 2(1,1,1)+2(1,2,3)$, we have $\det C = 2 \det A + 2 \det B$. Hence the answer is $-2$.
