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 $det(C) = -2$?