What is the simplest way to calculate this determinant using properties of determinants?
$begingroup$
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
asked Jan 6 at 8:03
mBartmBart
82
$endgroup$
add a comment |
$begingroup$
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
asked Jan 6 at 8:03
mBartmBart
82
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 6 at 8:03
mBartmBart
82
$endgroup$
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
asked Jan 6 at 8:03
mBartmBart
82
asked Jan 6 at 8:03
mBartmBart
82
asked Jan 6 at 8:03
mBartmBart
82
asked Jan 6 at 8:03
mBartmBart
82
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $(a-b)(b-c)(c-a)$.
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
$endgroup$
add a comment |
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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 $(a-b)(b-c)(c-a)$.
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
$endgroup$
add a comment |
$begingroup$
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 $(a-b)(b-c)(c-a)$.
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
$endgroup$
add a comment |
$begingroup$
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 $(a-b)(b-c)(c-a)$.
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
$endgroup$
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 $(a-b)(b-c)(c-a)$.
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
edited Jan 6 at 8:28
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
answered Jan 6 at 8:21
Shubham JohriShubham Johri
4,689717
4,689717
add a comment |
add a comment |
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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