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Relationship between determinant of matrix and determinant of adjoint?
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We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it.
linear-algebra determinant adjoint-operators
asked Nov 17 '17 at 17:32
CluelessCoder
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up vote
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We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it.
linear-algebra determinant adjoint-operators
asked Nov 17 '17 at 17:32
CluelessCoder
325
For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants).
– Prasun Biswas
Nov 17 '17 at 17:35
1
@CluelessCoder: see here: math.stackexchange.com/questions/516127/…
– Hector Blandin
Nov 17 '17 at 17:49
Possible duplicate of The determinant of adjugate matrix
– Math Lover
Nov 17 '17 at 17:51
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it.
linear-algebra determinant adjoint-operators
asked Nov 17 '17 at 17:32
CluelessCoder
325
We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it.
linear-algebra determinant adjoint-operators
asked Nov 17 '17 at 17:32
CluelessCoder
325
asked Nov 17 '17 at 17:32
CluelessCoder
325
asked Nov 17 '17 at 17:32
CluelessCoder
325
asked Nov 17 '17 at 17:32
CluelessCoder
325
325
For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants).
– Prasun Biswas
Nov 17 '17 at 17:35
1
@CluelessCoder: see here: math.stackexchange.com/questions/516127/…
– Hector Blandin
Nov 17 '17 at 17:49
Possible duplicate of The determinant of adjugate matrix
– Math Lover
Nov 17 '17 at 17:51
add a comment |Â
For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants).
– Prasun Biswas
Nov 17 '17 at 17:35
1
@CluelessCoder: see here: math.stackexchange.com/questions/516127/…
– Hector Blandin
Nov 17 '17 at 17:49
Possible duplicate of The determinant of adjugate matrix
– Math Lover
Nov 17 '17 at 17:51
For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants).
– Prasun Biswas
Nov 17 '17 at 17:35
For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants).
– Prasun Biswas
Nov 17 '17 at 17:35
1
1
@CluelessCoder: see here: math.stackexchange.com/questions/516127/…
– Hector Blandin
Nov 17 '17 at 17:49
@CluelessCoder: see here: math.stackexchange.com/questions/516127/…
– Hector Blandin
Nov 17 '17 at 17:49
Possible duplicate of The determinant of adjugate matrix
– Math Lover
Nov 17 '17 at 17:51
Possible duplicate of The determinant of adjugate matrix
– Math Lover
Nov 17 '17 at 17:51
add a comment |Â
1 Answer
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oldest
votes
up vote
2
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accepted
If $A$ is of size $ntimes n$, then
$$det(adj(A))=det(A)^n-1$$
also, you can verify that:
$$ Acdot adj(A) = det(A), I_ntimes n $$
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $A$ is of size $ntimes n$, then
$$det(adj(A))=det(A)^n-1$$
also, you can verify that:
$$ Acdot adj(A) = det(A), I_ntimes n $$
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
add a comment |Â
up vote
2
down vote
accepted
If $A$ is of size $ntimes n$, then
$$det(adj(A))=det(A)^n-1$$
also, you can verify that:
$$ Acdot adj(A) = det(A), I_ntimes n $$
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $A$ is of size $ntimes n$, then
$$det(adj(A))=det(A)^n-1$$
also, you can verify that:
$$ Acdot adj(A) = det(A), I_ntimes n $$
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
If $A$ is of size $ntimes n$, then
$$det(adj(A))=det(A)^n-1$$
also, you can verify that:
$$ Acdot adj(A) = det(A), I_ntimes n $$
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
answered Nov 17 '17 at 17:48
Hector Blandin
1,741816
1,741816
add a comment |Â
add a comment |Â
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For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants).
– Prasun Biswas
Nov 17 '17 at 17:35
1
@CluelessCoder: see here: math.stackexchange.com/questions/516127/…
– Hector Blandin
Nov 17 '17 at 17:49
Possible duplicate of The determinant of adjugate matrix
– Math Lover
Nov 17 '17 at 17:51