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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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    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 $(1,2,3)$.2012-05-03

2 Answers 2

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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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    No problem, fixed!2012-05-03
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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.

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    @BrianCurran, well then your method and solution are perfectly valid. Still note the alternative (and equivalent) way of describing it via copper.hat's explanation.2012-05-03