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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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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
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2 Answers 2

up vote 4 down vote accepted

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
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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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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
    
@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. –  davin May 3 '12 at 0:08
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