proof of all integer entries in an inverse matrix The Next CEO of Stack OverflowCofactor matrix 4x4, evaluated by handProve that if $A$ is regular then $operatornameadj(operatornameadj(A)) = (det A)^n-2 A$Adjoint of Complex Matrix$A^-1$ has integer entries if and only if the $rm det (A) =pm 1$Integer solutions of a quadratic equation with combined variablesFind the inverse of the $ntimes n$ matrix whose entries are given by $a_ij = max (i,j)$The difference between adjoint in linear algebra and adjoint of operator?proving that the area of a 2016 sided polygon is an even integerrandomized approximate matrix inverse or adjoint of a square matrixAll multiplicative sets of the integers
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proof of all integer entries in an inverse matrix
The Next CEO of Stack OverflowCofactor matrix 4x4, evaluated by handProve that if $A$ is regular then $operatornameadj(operatornameadj(A)) = (det A)^n-2 A$Adjoint of Complex Matrix$A^-1$ has integer entries if and only if the $rm det (A) =pm 1$Integer solutions of a quadratic equation with combined variablesFind the inverse of the $ntimes n$ matrix whose entries are given by $a_ij = max (i,j)$The difference between adjoint in linear algebra and adjoint of operator?proving that the area of a 2016 sided polygon is an even integerrandomized approximate matrix inverse or adjoint of a square matrixAll multiplicative sets of the integers
$begingroup$
"Prove that if $det(A)=1$ and all the entries in $A$ are integers, then all the entries in $A^-1$ are integers."
I began by setting up the adjoint method for finding the inverse.
$A^-1 = cfrac 1det(A) adj(A)$
given that $det(A)=1$,
$A^-1=adj(A)$
from here I'm stuck. I think I need to work with an example containing integers, but I'm not quite sure.
adjoint-operators integers
asked May 3 '15 at 22:26
shadeptshadept
4526
$endgroup$
add a comment |
$begingroup$
"Prove that if $det(A)=1$ and all the entries in $A$ are integers, then all the entries in $A^-1$ are integers."
I began by setting up the adjoint method for finding the inverse.
$A^-1 = cfrac 1det(A) adj(A)$
given that $det(A)=1$,
$A^-1=adj(A)$
from here I'm stuck. I think I need to work with an example containing integers, but I'm not quite sure.
adjoint-operators integers
asked May 3 '15 at 22:26
shadeptshadept
4526
$endgroup$
$begingroup$
So you need to show that the adjoint has integer entries. What were the entries of the adjoint again?
$endgroup$
– Daniel Fischer
May 3 '15 at 22:28
$begingroup$
The coefficients in the adjugate are determinants of matrices with integer coefficients, being the algebraic complements.
$endgroup$
– egreg
May 3 '15 at 22:29
$begingroup$
They don't give one.
$endgroup$
– shadept
May 3 '15 at 22:29
add a comment |
$begingroup$
"Prove that if $det(A)=1$ and all the entries in $A$ are integers, then all the entries in $A^-1$ are integers."
I began by setting up the adjoint method for finding the inverse.
$A^-1 = cfrac 1det(A) adj(A)$
given that $det(A)=1$,
$A^-1=adj(A)$
from here I'm stuck. I think I need to work with an example containing integers, but I'm not quite sure.
adjoint-operators integers
asked May 3 '15 at 22:26
shadeptshadept
4526
$endgroup$
"Prove that if $det(A)=1$ and all the entries in $A$ are integers, then all the entries in $A^-1$ are integers."
I began by setting up the adjoint method for finding the inverse.
$A^-1 = cfrac 1det(A) adj(A)$
given that $det(A)=1$,
$A^-1=adj(A)$
from here I'm stuck. I think I need to work with an example containing integers, but I'm not quite sure.
adjoint-operators integers
adjoint-operators integers
asked May 3 '15 at 22:26
shadeptshadept
4526
asked May 3 '15 at 22:26
shadeptshadept
4526
asked May 3 '15 at 22:26
shadeptshadept
4526
asked May 3 '15 at 22:26
shadeptshadept
4526
asked May 3 '15 at 22:26
shadeptshadept
4526
4526
$begingroup$
So you need to show that the adjoint has integer entries. What were the entries of the adjoint again?
$endgroup$
– Daniel Fischer
May 3 '15 at 22:28
$begingroup$
The coefficients in the adjugate are determinants of matrices with integer coefficients, being the algebraic complements.
$endgroup$
– egreg
May 3 '15 at 22:29
$begingroup$
They don't give one.
$endgroup$
– shadept
May 3 '15 at 22:29
add a comment |
$begingroup$
So you need to show that the adjoint has integer entries. What were the entries of the adjoint again?
$endgroup$
– Daniel Fischer
May 3 '15 at 22:28
$begingroup$
The coefficients in the adjugate are determinants of matrices with integer coefficients, being the algebraic complements.
$endgroup$
– egreg
May 3 '15 at 22:29
$begingroup$
They don't give one.
$endgroup$
– shadept
May 3 '15 at 22:29
$begingroup$
So you need to show that the adjoint has integer entries. What were the entries of the adjoint again?
$endgroup$
– Daniel Fischer
May 3 '15 at 22:28
$begingroup$
So you need to show that the adjoint has integer entries. What were the entries of the adjoint again?
$endgroup$
– Daniel Fischer
May 3 '15 at 22:28
$begingroup$
The coefficients in the adjugate are determinants of matrices with integer coefficients, being the algebraic complements.
$endgroup$
– egreg
May 3 '15 at 22:29
$begingroup$
The coefficients in the adjugate are determinants of matrices with integer coefficients, being the algebraic complements.
$endgroup$
– egreg
May 3 '15 at 22:29
$begingroup$
They don't give one.
$endgroup$
– shadept
May 3 '15 at 22:29
$begingroup$
They don't give one.
$endgroup$
– shadept
May 3 '15 at 22:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The coefficient at place $(i,j)$ of $operatornameadjA$ is the determinant of the matrix obtained by removing row $j$ and column $i$ from $A$, multiplied by $(-1)^i+j$.
The determinant of a matrix with integer coefficients is an integer.
answered May 3 '15 at 22:31
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
For a $2 times 2$ matrix $A$, the adjoint $Adj(A)$ coincides with its matrix of cofactors, with (egreg refreshed my memory) a change of sign for the even entries, meaning those entries $a_ij$ with $i+j$ even.
And you can go further: an integer-valued matrix, i.e., a matrix with integer entries is invertible iff its determinant is $ pm 1$: You know the determinant will be Integer-valued, and you need $Det(A)Det(A^-1)=1$. So the determinant DetA will have to be a unit in the ring of integers , so that it must be $pm 1$
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
$endgroup$
1
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The coefficient at place $(i,j)$ of $operatornameadjA$ is the determinant of the matrix obtained by removing row $j$ and column $i$ from $A$, multiplied by $(-1)^i+j$.
The determinant of a matrix with integer coefficients is an integer.
answered May 3 '15 at 22:31
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
The coefficient at place $(i,j)$ of $operatornameadjA$ is the determinant of the matrix obtained by removing row $j$ and column $i$ from $A$, multiplied by $(-1)^i+j$.
The determinant of a matrix with integer coefficients is an integer.
answered May 3 '15 at 22:31
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
The coefficient at place $(i,j)$ of $operatornameadjA$ is the determinant of the matrix obtained by removing row $j$ and column $i$ from $A$, multiplied by $(-1)^i+j$.
The determinant of a matrix with integer coefficients is an integer.
answered May 3 '15 at 22:31
egregegreg
185k1486206
$endgroup$
The coefficient at place $(i,j)$ of $operatornameadjA$ is the determinant of the matrix obtained by removing row $j$ and column $i$ from $A$, multiplied by $(-1)^i+j$.
The determinant of a matrix with integer coefficients is an integer.
answered May 3 '15 at 22:31
egregegreg
185k1486206
answered May 3 '15 at 22:31
egregegreg
185k1486206
answered May 3 '15 at 22:31
egregegreg
185k1486206
answered May 3 '15 at 22:31
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
$begingroup$
For a $2 times 2$ matrix $A$, the adjoint $Adj(A)$ coincides with its matrix of cofactors, with (egreg refreshed my memory) a change of sign for the even entries, meaning those entries $a_ij$ with $i+j$ even.
And you can go further: an integer-valued matrix, i.e., a matrix with integer entries is invertible iff its determinant is $ pm 1$: You know the determinant will be Integer-valued, and you need $Det(A)Det(A^-1)=1$. So the determinant DetA will have to be a unit in the ring of integers , so that it must be $pm 1$
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
$endgroup$
1
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
add a comment |
$begingroup$
For a $2 times 2$ matrix $A$, the adjoint $Adj(A)$ coincides with its matrix of cofactors, with (egreg refreshed my memory) a change of sign for the even entries, meaning those entries $a_ij$ with $i+j$ even.
And you can go further: an integer-valued matrix, i.e., a matrix with integer entries is invertible iff its determinant is $ pm 1$: You know the determinant will be Integer-valued, and you need $Det(A)Det(A^-1)=1$. So the determinant DetA will have to be a unit in the ring of integers , so that it must be $pm 1$
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
$endgroup$
1
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
add a comment |
$begingroup$
For a $2 times 2$ matrix $A$, the adjoint $Adj(A)$ coincides with its matrix of cofactors, with (egreg refreshed my memory) a change of sign for the even entries, meaning those entries $a_ij$ with $i+j$ even.
And you can go further: an integer-valued matrix, i.e., a matrix with integer entries is invertible iff its determinant is $ pm 1$: You know the determinant will be Integer-valued, and you need $Det(A)Det(A^-1)=1$. So the determinant DetA will have to be a unit in the ring of integers , so that it must be $pm 1$
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
$endgroup$
For a $2 times 2$ matrix $A$, the adjoint $Adj(A)$ coincides with its matrix of cofactors, with (egreg refreshed my memory) a change of sign for the even entries, meaning those entries $a_ij$ with $i+j$ even.
And you can go further: an integer-valued matrix, i.e., a matrix with integer entries is invertible iff its determinant is $ pm 1$: You know the determinant will be Integer-valued, and you need $Det(A)Det(A^-1)=1$. So the determinant DetA will have to be a unit in the ring of integers , so that it must be $pm 1$
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
edited May 3 '15 at 22:45
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
answered May 3 '15 at 22:32
Gary.Gary.
1,96659
1,96659
1
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
add a comment |
1
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
1
1
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
$$Adjbeginpmatrixa&b\c&dendpmatrix=beginpmatrixd&-b\-c&aendpmatrix$$
$endgroup$
– user228113
May 3 '15 at 22:41
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
$begingroup$
Yea, I mixed up the cofactor entries with the transpose. Just edited.
$endgroup$
– Gary.
May 3 '15 at 22:46
add a comment |
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$begingroup$
So you need to show that the adjoint has integer entries. What were the entries of the adjoint again?
$endgroup$
– Daniel Fischer
May 3 '15 at 22:28
$begingroup$
The coefficients in the adjugate are determinants of matrices with integer coefficients, being the algebraic complements.
$endgroup$
– egreg
May 3 '15 at 22:29
$begingroup$
They don't give one.
$endgroup$
– shadept
May 3 '15 at 22:29