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A=$\begin{bmatrix} \frac ab & \frac bc & \frac ca\\a & b & c\\ab & bc & ca\end{bmatrix}$

Or maby there is no shorcut to calculate the det(A)?

mBart
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    It seems that Sarrus rule is good enough – Hagen von Eitzen Jan 06 '19 at 08:08
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    You can divide the first column by $a/b$, the second one by $b/c$, the third one by $c/a$, since the product of these numbers is $1$, the determinant is unchanged but the matrix has a better-known form. – Aphelli Jan 06 '19 at 08:14

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Perform $C_3\to C_3-(c/b)C_2,C_2\to C_2-(b/a)C_1$, $$\sim\begin{vmatrix} \frac ab & \frac bc-1& \frac ca-1\\a &0 & 0\\ab & b(c-b) & c(a-c)\end{vmatrix}$$Now, factor out $a,(b-c),(c-a)$ from the first three columns respectively,$$\sim a(c-a)(b-c) \begin{vmatrix} \frac 1b & \frac 1c& \frac 1a\\1 &0 & 0\\b &-b& -c\end{vmatrix}$$and expand along the second row to get the answer $(a-b)(b-c)(c-a)$.

Shubham Johri
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