What is the simplest way to calculate this determinant using properties of determinants?












0












$begingroup$


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



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






determinant







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$endgroup$








  • 3




    $begingroup$
    It seems that Sarrus rule is good enough
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 8:08






  • 3




    $begingroup$
    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.
    $endgroup$
    – Mindlack
    Jan 6 at 8:14
















0












$begingroup$


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



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






determinant







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    It seems that Sarrus rule is good enough
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 8:08






  • 3




    $begingroup$
    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.
    $endgroup$
    – Mindlack
    Jan 6 at 8:14














0












0








0





$begingroup$


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



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






determinant







share|cite|improve this question









$endgroup$




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



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






determinant




determinant






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 6 at 8:03









mBartmBart

82




82








  • 3




    $begingroup$
    It seems that Sarrus rule is good enough
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 8:08






  • 3




    $begingroup$
    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.
    $endgroup$
    – Mindlack
    Jan 6 at 8:14














  • 3




    $begingroup$
    It seems that Sarrus rule is good enough
    $endgroup$
    – Hagen von Eitzen
    Jan 6 at 8:08






  • 3




    $begingroup$
    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.
    $endgroup$
    – Mindlack
    Jan 6 at 8:14








3




3




$begingroup$
It seems that Sarrus rule is good enough
$endgroup$
– Hagen von Eitzen
Jan 6 at 8:08




$begingroup$
It seems that Sarrus rule is good enough
$endgroup$
– Hagen von Eitzen
Jan 6 at 8:08




3




3




$begingroup$
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.
$endgroup$
– Mindlack
Jan 6 at 8:14




$begingroup$
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.
$endgroup$
– Mindlack
Jan 6 at 8:14










1 Answer
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$begingroup$

Perform $C_3to C_3-(c/b)C_2,C_2to C_2-(b/a)C_1$,
$$simbegin{vmatrix} frac ab & frac bc-1& frac ca-1\a &0 & 0\ab & b(c-b) & c(a-c)end{vmatrix}$$Now, factor out $a,(b-c),(c-a)$ from the first three columns respectively,$$sim a(c-a)(b-c)
begin{vmatrix} frac 1b & frac 1c& frac 1a\1 &0 & 0\b &-b& -cend{vmatrix}$$and expand along the second row to get the answer $(a-b)(b-c)(c-a)$.






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    $begingroup$

    Perform $C_3to C_3-(c/b)C_2,C_2to C_2-(b/a)C_1$,
    $$simbegin{vmatrix} frac ab & frac bc-1& frac ca-1\a &0 & 0\ab & b(c-b) & c(a-c)end{vmatrix}$$Now, factor out $a,(b-c),(c-a)$ from the first three columns respectively,$$sim a(c-a)(b-c)
    begin{vmatrix} frac 1b & frac 1c& frac 1a\1 &0 & 0\b &-b& -cend{vmatrix}$$    and expand along the second row to get the answer $(a-b)(b-c)(c-a)$.






    share|cite|improve this answer











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      0












      $begingroup$

      Perform $C_3to C_3-(c/b)C_2,C_2to C_2-(b/a)C_1$,
      $$simbegin{vmatrix} frac ab & frac bc-1& frac ca-1\a &0 & 0\ab & b(c-b) & c(a-c)end{vmatrix}$$Now, factor out $a,(b-c),(c-a)$ from the first three columns respectively,$$sim a(c-a)(b-c)
      begin{vmatrix} frac 1b & frac 1c& frac 1a\1 &0 & 0\b &-b& -cend{vmatrix}$$      and expand along the second row to get the answer $(a-b)(b-c)(c-a)$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Perform $C_3to C_3-(c/b)C_2,C_2to C_2-(b/a)C_1$,
        $$simbegin{vmatrix} frac ab & frac bc-1& frac ca-1\a &0 & 0\ab & b(c-b) & c(a-c)end{vmatrix}$$Now, factor out $a,(b-c),(c-a)$ from the first three columns respectively,$$sim a(c-a)(b-c)
        begin{vmatrix} frac 1b & frac 1c& frac 1a\1 &0 & 0\b &-b& -cend{vmatrix}$$        and expand along the second row to get the answer $(a-b)(b-c)(c-a)$.






        share|cite|improve this answer











        $endgroup$



        Perform $C_3to C_3-(c/b)C_2,C_2to C_2-(b/a)C_1$,
        $$simbegin{vmatrix} frac ab & frac bc-1& frac ca-1\a &0 & 0\ab & b(c-b) & c(a-c)end{vmatrix}$$Now, factor out $a,(b-c),(c-a)$ from the first three columns respectively,$$sim a(c-a)(b-c)
        begin{vmatrix} frac 1b & frac 1c& frac 1a\1 &0 & 0\b &-b& -cend{vmatrix}$$        and expand along the second row to get the answer $(a-b)(b-c)(c-a)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 8:28

























        answered Jan 6 at 8:21









        Shubham JohriShubham Johri

        5,204718




        5,204718






























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