Determinant of a $ntimes n$ matrix in terms of $n-1times n-1$ matrix












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Suppose we know the determinant of matrix $A=(a_{ij})_{i,j=1}^{n-1}$. Can we express determinant of matrix $A'=(a_{ij})_{i,j=1}^{n}$ in terms of the determinant of matrix $A$?



We see that the matrix $A'$ differs from matrix $A$ only in the extra right-most column and bottom row. As a starting point, we can think of case $n=3$.






matrices determinant determinant-functions







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  • $begingroup$
    No. Take any $A$ with nonzero determinant and have last row of $A'$ be all $0$s.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:07












  • $begingroup$
    Actually, yes. $det(A') = 0cdot det(A)+det(A')$.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:08
















2












$begingroup$


Suppose we know the determinant of matrix $A=(a_{ij})_{i,j=1}^{n-1}$. Can we express determinant of matrix $A'=(a_{ij})_{i,j=1}^{n}$ in terms of the determinant of matrix $A$?



We see that the matrix $A'$ differs from matrix $A$ only in the extra right-most column and bottom row. As a starting point, we can think of case $n=3$.






matrices determinant determinant-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    No. Take any $A$ with nonzero determinant and have last row of $A'$ be all $0$s.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:07












  • $begingroup$
    Actually, yes. $det(A') = 0cdot det(A)+det(A')$.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:08














2












2








2





$begingroup$


Suppose we know the determinant of matrix $A=(a_{ij})_{i,j=1}^{n-1}$. Can we express determinant of matrix $A'=(a_{ij})_{i,j=1}^{n}$ in terms of the determinant of matrix $A$?



We see that the matrix $A'$ differs from matrix $A$ only in the extra right-most column and bottom row. As a starting point, we can think of case $n=3$.






matrices determinant determinant-functions







share|cite|improve this question









$endgroup$




Suppose we know the determinant of matrix $A=(a_{ij})_{i,j=1}^{n-1}$. Can we express determinant of matrix $A'=(a_{ij})_{i,j=1}^{n}$ in terms of the determinant of matrix $A$?



We see that the matrix $A'$ differs from matrix $A$ only in the extra right-most column and bottom row. As a starting point, we can think of case $n=3$.






matrices determinant determinant-functions




matrices determinant determinant-functions






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asked Dec 20 '18 at 23:54









ershersh

438113




438113












  • $begingroup$
    No. Take any $A$ with nonzero determinant and have last row of $A'$ be all $0$s.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:07












  • $begingroup$
    Actually, yes. $det(A') = 0cdot det(A)+det(A')$.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:08


















  • $begingroup$
    No. Take any $A$ with nonzero determinant and have last row of $A'$ be all $0$s.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:07












  • $begingroup$
    Actually, yes. $det(A') = 0cdot det(A)+det(A')$.
    $endgroup$
    – mathworker21
    Dec 21 '18 at 0:08
















$begingroup$
No. Take any $A$ with nonzero determinant and have last row of $A'$ be all $0$s.
$endgroup$
– mathworker21
Dec 21 '18 at 0:07






$begingroup$
No. Take any $A$ with nonzero determinant and have last row of $A'$ be all $0$s.
$endgroup$
– mathworker21
Dec 21 '18 at 0:07














$begingroup$
Actually, yes. $det(A') = 0cdot det(A)+det(A')$.
$endgroup$
– mathworker21
Dec 21 '18 at 0:08




$begingroup$
Actually, yes. $det(A') = 0cdot det(A)+det(A')$.
$endgroup$
– mathworker21
Dec 21 '18 at 0:08










4 Answers
4






active

oldest

votes


















1












$begingroup$

No, e.g., consider:
$$begin{array}{ccc}
A = left(matrix{0 & 0 \ 0 & 1}right) & quadquad & B = left(matrix{1 & 0 \ 0 & 0 }right) \
A' = left(matrix{0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1}right) & quadquad & B' = left(matrix{1 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 1}right)
end{array}$$



$A$ and $B$ have the same determinant (namely $0$) and $A'$ and $B'$ are obtained from them by adding the same third row and column, but $|A'| = -1 neq 0 = |B'|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
    $endgroup$
    – ersh
    Dec 21 '18 at 0:50






  • 1




    $begingroup$
    No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
    $endgroup$
    – Rob Arthan
    Dec 21 '18 at 0:56










  • $begingroup$
    Yes That is right!
    $endgroup$
    – ersh
    Dec 21 '18 at 1:23



















4












$begingroup$

In general no. You need to know the values of the entries of the extra row and column. Note that if it were possible, one could iterate the process and express the determinant of any matrix in terms of the value of its corner entry (say first row, first column).



Edit. As Rob pointed out in the comments, you need to know the internals of $A$ too.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
    $endgroup$
    – Rob Arthan
    Dec 21 '18 at 0:21












  • $begingroup$
    You are right about the internals of $A$ being needed too.
    $endgroup$
    – positrón0802
    Dec 21 '18 at 0:33



















1












$begingroup$

No-ish, in that there exists a formula for computing the determinant perturbatively
$$detleft(mathbf{A} + mathbf{uv}^textsf{T}right) = detleft(mathbf{A}right) + mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$$
but the perturbing term $mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$ is not solely a function of the determinant $detleft(mathbf{A}right)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    What connects the determinant of a matrix and its leading principal minor is Schur complement. In general, if a leading principal submatrix $A$ of a partitioned matrix $M$ is nonsingular, we have
    $$
    detunderbrace{pmatrix{A&B\ C&D}}_M=det(A)det(D-CA^{-1}B),
    $$
    where $S=D-CA^{-1}B$ is called the Schur complement of $A$ in the partitioned matrix $M$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Closely related
      $endgroup$
      – ersh
      Dec 21 '18 at 16:10











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    No, e.g., consider:
    $$begin{array}{ccc}
    A = left(matrix{0 & 0 \ 0 & 1}right) & quadquad & B = left(matrix{1 & 0 \ 0 & 0 }right) \
    A' = left(matrix{0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1}right) & quadquad & B' = left(matrix{1 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 1}right)
    end{array}$$



    $A$ and $B$ have the same determinant (namely $0$) and $A'$ and $B'$ are obtained from them by adding the same third row and column, but $|A'| = -1 neq 0 = |B'|$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
      $endgroup$
      – ersh
      Dec 21 '18 at 0:50






    • 1




      $begingroup$
      No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:56










    • $begingroup$
      Yes That is right!
      $endgroup$
      – ersh
      Dec 21 '18 at 1:23
















    1












    $begingroup$

    No, e.g., consider:
    $$begin{array}{ccc}
    A = left(matrix{0 & 0 \ 0 & 1}right) & quadquad & B = left(matrix{1 & 0 \ 0 & 0 }right) \
    A' = left(matrix{0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1}right) & quadquad & B' = left(matrix{1 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 1}right)
    end{array}$$



    $A$ and $B$ have the same determinant (namely $0$) and $A'$ and $B'$ are obtained from them by adding the same third row and column, but $|A'| = -1 neq 0 = |B'|$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
      $endgroup$
      – ersh
      Dec 21 '18 at 0:50






    • 1




      $begingroup$
      No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:56










    • $begingroup$
      Yes That is right!
      $endgroup$
      – ersh
      Dec 21 '18 at 1:23














    1












    1








    1





    $begingroup$

    No, e.g., consider:
    $$begin{array}{ccc}
    A = left(matrix{0 & 0 \ 0 & 1}right) & quadquad & B = left(matrix{1 & 0 \ 0 & 0 }right) \
    A' = left(matrix{0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1}right) & quadquad & B' = left(matrix{1 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 1}right)
    end{array}$$



    $A$ and $B$ have the same determinant (namely $0$) and $A'$ and $B'$ are obtained from them by adding the same third row and column, but $|A'| = -1 neq 0 = |B'|$.






    share|cite|improve this answer









    $endgroup$



    No, e.g., consider:
    $$begin{array}{ccc}
    A = left(matrix{0 & 0 \ 0 & 1}right) & quadquad & B = left(matrix{1 & 0 \ 0 & 0 }right) \
    A' = left(matrix{0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1}right) & quadquad & B' = left(matrix{1 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 1}right)
    end{array}$$



    $A$ and $B$ have the same determinant (namely $0$) and $A'$ and $B'$ are obtained from them by adding the same third row and column, but $|A'| = -1 neq 0 = |B'|$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 0:17









    Rob ArthanRob Arthan

    29.5k42967




    29.5k42967












    • $begingroup$
      I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
      $endgroup$
      – ersh
      Dec 21 '18 at 0:50






    • 1




      $begingroup$
      No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:56










    • $begingroup$
      Yes That is right!
      $endgroup$
      – ersh
      Dec 21 '18 at 1:23


















    • $begingroup$
      I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
      $endgroup$
      – ersh
      Dec 21 '18 at 0:50






    • 1




      $begingroup$
      No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:56










    • $begingroup$
      Yes That is right!
      $endgroup$
      – ersh
      Dec 21 '18 at 1:23
















    $begingroup$
    I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
    $endgroup$
    – ersh
    Dec 21 '18 at 0:50




    $begingroup$
    I am already given the fixed entries of matrices $A$ and $A'$. For example, the matrices $A$ and $A'$ you have provided. Can we split the determinant of $A'$ as a function of the determinant of $A$. Anyway, I was already convinced that the question is too much general.
    $endgroup$
    – ersh
    Dec 21 '18 at 0:50




    1




    1




    $begingroup$
    No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
    $endgroup$
    – Rob Arthan
    Dec 21 '18 at 0:56




    $begingroup$
    No: think of my $B$ and $B'$ as what your $A$ and $A'$ might have been in an alternative universe. You can't give $|A'|$ as a function of $|A|$ and the entries in the extra row and column, because as $|A| = |B|$ that would give you the same answer for $|B'|$, but $|A'| neq |B'|$.
    $endgroup$
    – Rob Arthan
    Dec 21 '18 at 0:56












    $begingroup$
    Yes That is right!
    $endgroup$
    – ersh
    Dec 21 '18 at 1:23




    $begingroup$
    Yes That is right!
    $endgroup$
    – ersh
    Dec 21 '18 at 1:23











    4












    $begingroup$

    In general no. You need to know the values of the entries of the extra row and column. Note that if it were possible, one could iterate the process and express the determinant of any matrix in terms of the value of its corner entry (say first row, first column).



    Edit. As Rob pointed out in the comments, you need to know the internals of $A$ too.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:21












    • $begingroup$
      You are right about the internals of $A$ being needed too.
      $endgroup$
      – positrón0802
      Dec 21 '18 at 0:33
















    4












    $begingroup$

    In general no. You need to know the values of the entries of the extra row and column. Note that if it were possible, one could iterate the process and express the determinant of any matrix in terms of the value of its corner entry (say first row, first column).



    Edit. As Rob pointed out in the comments, you need to know the internals of $A$ too.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:21












    • $begingroup$
      You are right about the internals of $A$ being needed too.
      $endgroup$
      – positrón0802
      Dec 21 '18 at 0:33














    4












    4








    4





    $begingroup$

    In general no. You need to know the values of the entries of the extra row and column. Note that if it were possible, one could iterate the process and express the determinant of any matrix in terms of the value of its corner entry (say first row, first column).



    Edit. As Rob pointed out in the comments, you need to know the internals of $A$ too.






    share|cite|improve this answer











    $endgroup$



    In general no. You need to know the values of the entries of the extra row and column. Note that if it were possible, one could iterate the process and express the determinant of any matrix in terms of the value of its corner entry (say first row, first column).



    Edit. As Rob pointed out in the comments, you need to know the internals of $A$ too.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 0:30

























    answered Dec 21 '18 at 0:03









    positrón0802positrón0802

    4,488520




    4,488520












    • $begingroup$
      You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:21












    • $begingroup$
      You are right about the internals of $A$ being needed too.
      $endgroup$
      – positrón0802
      Dec 21 '18 at 0:33


















    • $begingroup$
      You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
      $endgroup$
      – Rob Arthan
      Dec 21 '18 at 0:21












    • $begingroup$
      You are right about the internals of $A$ being needed too.
      $endgroup$
      – positrón0802
      Dec 21 '18 at 0:33
















    $begingroup$
    You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
    $endgroup$
    – Rob Arthan
    Dec 21 '18 at 0:21






    $begingroup$
    You need to know the internals of $A$ too. See my answer. I think the OP was hoping to be able to calculate the determinant as a function of the determinant of $A$ and of the values added in the last row and column in $A'$.
    $endgroup$
    – Rob Arthan
    Dec 21 '18 at 0:21














    $begingroup$
    You are right about the internals of $A$ being needed too.
    $endgroup$
    – positrón0802
    Dec 21 '18 at 0:33




    $begingroup$
    You are right about the internals of $A$ being needed too.
    $endgroup$
    – positrón0802
    Dec 21 '18 at 0:33











    1












    $begingroup$

    No-ish, in that there exists a formula for computing the determinant perturbatively
    $$detleft(mathbf{A} + mathbf{uv}^textsf{T}right) = detleft(mathbf{A}right) + mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$$
    but the perturbing term $mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$ is not solely a function of the determinant $detleft(mathbf{A}right)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      No-ish, in that there exists a formula for computing the determinant perturbatively
      $$detleft(mathbf{A} + mathbf{uv}^textsf{T}right) = detleft(mathbf{A}right) + mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$$
      but the perturbing term $mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$ is not solely a function of the determinant $detleft(mathbf{A}right)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        No-ish, in that there exists a formula for computing the determinant perturbatively
        $$detleft(mathbf{A} + mathbf{uv}^textsf{T}right) = detleft(mathbf{A}right) + mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$$
        but the perturbing term $mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$ is not solely a function of the determinant $detleft(mathbf{A}right)$.






        share|cite|improve this answer









        $endgroup$



        No-ish, in that there exists a formula for computing the determinant perturbatively
        $$detleft(mathbf{A} + mathbf{uv}^textsf{T}right) = detleft(mathbf{A}right) + mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$$
        but the perturbing term $mathbf{v}^textsf{T}mathrm{adj}left(mathbf{A}right)mathbf{u}$ is not solely a function of the determinant $detleft(mathbf{A}right)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 0:08









        K B DaveK B Dave

        3,677317




        3,677317























            1












            $begingroup$

            What connects the determinant of a matrix and its leading principal minor is Schur complement. In general, if a leading principal submatrix $A$ of a partitioned matrix $M$ is nonsingular, we have
            $$
            detunderbrace{pmatrix{A&B\ C&D}}_M=det(A)det(D-CA^{-1}B),
            $$
            where $S=D-CA^{-1}B$ is called the Schur complement of $A$ in the partitioned matrix $M$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Closely related
              $endgroup$
              – ersh
              Dec 21 '18 at 16:10
















            1












            $begingroup$

            What connects the determinant of a matrix and its leading principal minor is Schur complement. In general, if a leading principal submatrix $A$ of a partitioned matrix $M$ is nonsingular, we have
            $$
            detunderbrace{pmatrix{A&B\ C&D}}_M=det(A)det(D-CA^{-1}B),
            $$
            where $S=D-CA^{-1}B$ is called the Schur complement of $A$ in the partitioned matrix $M$.






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            $endgroup$













            • $begingroup$
              Closely related
              $endgroup$
              – ersh
              Dec 21 '18 at 16:10














            1












            1








            1





            $begingroup$

            What connects the determinant of a matrix and its leading principal minor is Schur complement. In general, if a leading principal submatrix $A$ of a partitioned matrix $M$ is nonsingular, we have
            $$
            detunderbrace{pmatrix{A&B\ C&D}}_M=det(A)det(D-CA^{-1}B),
            $$
            where $S=D-CA^{-1}B$ is called the Schur complement of $A$ in the partitioned matrix $M$.






            share|cite|improve this answer









            $endgroup$



            What connects the determinant of a matrix and its leading principal minor is Schur complement. In general, if a leading principal submatrix $A$ of a partitioned matrix $M$ is nonsingular, we have
            $$
            detunderbrace{pmatrix{A&B\ C&D}}_M=det(A)det(D-CA^{-1}B),
            $$
            where $S=D-CA^{-1}B$ is called the Schur complement of $A$ in the partitioned matrix $M$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 21 '18 at 5:41









            user1551user1551

            73.8k566129




            73.8k566129












            • $begingroup$
              Closely related
              $endgroup$
              – ersh
              Dec 21 '18 at 16:10


















            • $begingroup$
              Closely related
              $endgroup$
              – ersh
              Dec 21 '18 at 16:10
















            $begingroup$
            Closely related
            $endgroup$
            – ersh
            Dec 21 '18 at 16:10




            $begingroup$
            Closely related
            $endgroup$
            – ersh
            Dec 21 '18 at 16:10


















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