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










0












$begingroup$


"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.










share|cite|improve this question









$endgroup$











  • $begingroup$
    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















0












$begingroup$


"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.










share|cite|improve this question









$endgroup$











  • $begingroup$
    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













0












0








0





$begingroup$


"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.










share|cite|improve this question









$endgroup$




"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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    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 $pm 1$






    share|cite|improve this answer











    $endgroup$








    • 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












    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 3 '15 at 22:31









        egregegreg

        185k1486206




        185k1486206





















            0












            $begingroup$

            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 $pm 1$






            share|cite|improve this answer











            $endgroup$








            • 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
















            0












            $begingroup$

            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 $pm 1$






            share|cite|improve this answer











            $endgroup$








            • 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














            0












            0








            0





            $begingroup$

            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 $pm 1$






            share|cite|improve this answer











            $endgroup$



            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 $pm 1$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 3 '15 at 22:45

























            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













            • 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


















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