# 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 $det(C) = -2$?

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Made a typo in C. "4, 5, 6" should be "4, 6, 8". Fixed above. Thanks for the quick answer, @copper.hat. –  Brian Curran May 3 '12 at 0:02
copper.hat's method still applies. You just have to figure out how to write $(4,6,8)$ as a linear combination of $(1,1,1)$ and $(1,2,3)$. –  Michael Joyce May 3 '12 at 0:05

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 $-2$.

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Apologies for making your result wrong because of a silly typo :[. –  Brian Curran May 3 '12 at 0:06
No problem, fixed! –  copper.hat May 3 '12 at 1:05
$$\begin{pmatrix} 1 & 1 & 1 & 3 \\ 2 & 3 & 1 & -4 \\ 5 & 6 & 4 & x \end{pmatrix}$$
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.
Thanks for catching my typo! C should actually look like: $$C = \begin{pmatrix} a & b & c \\ 4 & 6 & 8 \\ d & e & f \end{pmatrix}$$ I've edited the original question above. –  Brian Curran May 3 '12 at 0:00