$AC^T = det(A)I$












1














Let A = $begin{bmatrix} a_{11} & a_{12} & dots & a_{1n} \ a_{21} & a_{22} & dots & a_{2n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
dots & a_{nn} end{bmatrix}$
and the matrix of cofactors of $A$ is
$$C=begin{bmatrix} C_{11} & C_{12} & dots & C_{1n} \ C_{21} & C_{22} & dots & C_{2n} \ vdots & vdots & ddots & vdots \ C_{n1} & C_{n2} &
dots & C_{nn} end{bmatrix}.
$$
I try to understand why $AC^T = det(A)I$ necessarily.



Why is it that $a_{11}C_{21} + a_{12}C_{22} + dots + a_{1n}C_{2n} = 0$?






linear-algebra determinant laplace-expansion







share|cite|improve this question





























    1














    Let A = $begin{bmatrix} a_{11} & a_{12} & dots & a_{1n} \ a_{21} & a_{22} & dots & a_{2n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
    dots & a_{nn} end{bmatrix}$
    and the matrix of cofactors of $A$ is
    $$C=begin{bmatrix} C_{11} & C_{12} & dots & C_{1n} \ C_{21} & C_{22} & dots & C_{2n} \ vdots & vdots & ddots & vdots \ C_{n1} & C_{n2} &
    dots & C_{nn} end{bmatrix}.
    $$
    I try to understand why $AC^T = det(A)I$ necessarily.



    Why is it that $a_{11}C_{21} + a_{12}C_{22} + dots + a_{1n}C_{2n} = 0$?






    linear-algebra determinant laplace-expansion







    share|cite|improve this question



























      1












      1








      1







      Let A = $begin{bmatrix} a_{11} & a_{12} & dots & a_{1n} \ a_{21} & a_{22} & dots & a_{2n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
      dots & a_{nn} end{bmatrix}$
      and the matrix of cofactors of $A$ is
      $$C=begin{bmatrix} C_{11} & C_{12} & dots & C_{1n} \ C_{21} & C_{22} & dots & C_{2n} \ vdots & vdots & ddots & vdots \ C_{n1} & C_{n2} &
      dots & C_{nn} end{bmatrix}.
      $$
      I try to understand why $AC^T = det(A)I$ necessarily.



      Why is it that $a_{11}C_{21} + a_{12}C_{22} + dots + a_{1n}C_{2n} = 0$?






      linear-algebra determinant laplace-expansion







      share|cite|improve this question















      Let A = $begin{bmatrix} a_{11} & a_{12} & dots & a_{1n} \ a_{21} & a_{22} & dots & a_{2n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
      dots & a_{nn} end{bmatrix}$
      and the matrix of cofactors of $A$ is
      $$C=begin{bmatrix} C_{11} & C_{12} & dots & C_{1n} \ C_{21} & C_{22} & dots & C_{2n} \ vdots & vdots & ddots & vdots \ C_{n1} & C_{n2} &
      dots & C_{nn} end{bmatrix}.
      $$
      I try to understand why $AC^T = det(A)I$ necessarily.



      Why is it that $a_{11}C_{21} + a_{12}C_{22} + dots + a_{1n}C_{2n} = 0$?






      linear-algebra determinant laplace-expansion




      linear-algebra determinant laplace-expansion






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 15:04









      A.Γ.

      22.6k32656




      22.6k32656










      asked Jan 4 at 14:24









      Kid CudiKid Cudi

      456




      456






















          1 Answer
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          1














          The expression
          $$
          color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}
          $$
          is the Laplace expansion along the second row of
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{red}{a_{11}} & color{red}{a_{12}} & color{red}dots & color{red}{a_{1n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
          dots & a_{nn}
          end{vmatrix}=color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}=0.
          $$



          Edit: take the matrix $A$ and do the determinant expansion along the second row
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{blue}{a_{21}} & color{blue}{a_{22}} & color{blue}dots & color{blue}{a_{2n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} & dots & a_{nn}
          end{vmatrix}=color{blue}{a_{21}}C_{21} + color{blue}{a_{22}}C_{22} + dots + color{blue}{a_{2n}}C_{2n}.
          $$
          Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.






          share|cite|improve this answer























          • Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
            – Kid Cudi
            Jan 4 at 14:54








          • 1




            @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
            – A.Γ.
            Jan 4 at 15:00








          • 1




            @KidCudi I have edited the answer, hopefully it is clearer now.
            – A.Γ.
            Jan 4 at 15:13






          • 1




            @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
            – A.Γ.
            Jan 4 at 15:17






          • 1




            I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
            – darij grinberg
            Jan 4 at 20:26











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          The expression
          $$
          color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}
          $$
          is the Laplace expansion along the second row of
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{red}{a_{11}} & color{red}{a_{12}} & color{red}dots & color{red}{a_{1n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
          dots & a_{nn}
          end{vmatrix}=color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}=0.
          $$



          Edit: take the matrix $A$ and do the determinant expansion along the second row
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{blue}{a_{21}} & color{blue}{a_{22}} & color{blue}dots & color{blue}{a_{2n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} & dots & a_{nn}
          end{vmatrix}=color{blue}{a_{21}}C_{21} + color{blue}{a_{22}}C_{22} + dots + color{blue}{a_{2n}}C_{2n}.
          $$
          Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.






          share|cite|improve this answer























          • Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
            – Kid Cudi
            Jan 4 at 14:54








          • 1




            @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
            – A.Γ.
            Jan 4 at 15:00








          • 1




            @KidCudi I have edited the answer, hopefully it is clearer now.
            – A.Γ.
            Jan 4 at 15:13






          • 1




            @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
            – A.Γ.
            Jan 4 at 15:17






          • 1




            I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
            – darij grinberg
            Jan 4 at 20:26
















          1














          The expression
          $$
          color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}
          $$
          is the Laplace expansion along the second row of
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{red}{a_{11}} & color{red}{a_{12}} & color{red}dots & color{red}{a_{1n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
          dots & a_{nn}
          end{vmatrix}=color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}=0.
          $$



          Edit: take the matrix $A$ and do the determinant expansion along the second row
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{blue}{a_{21}} & color{blue}{a_{22}} & color{blue}dots & color{blue}{a_{2n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} & dots & a_{nn}
          end{vmatrix}=color{blue}{a_{21}}C_{21} + color{blue}{a_{22}}C_{22} + dots + color{blue}{a_{2n}}C_{2n}.
          $$
          Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.






          share|cite|improve this answer























          • Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
            – Kid Cudi
            Jan 4 at 14:54








          • 1




            @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
            – A.Γ.
            Jan 4 at 15:00








          • 1




            @KidCudi I have edited the answer, hopefully it is clearer now.
            – A.Γ.
            Jan 4 at 15:13






          • 1




            @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
            – A.Γ.
            Jan 4 at 15:17






          • 1




            I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
            – darij grinberg
            Jan 4 at 20:26














          1












          1








          1






          The expression
          $$
          color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}
          $$
          is the Laplace expansion along the second row of
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{red}{a_{11}} & color{red}{a_{12}} & color{red}dots & color{red}{a_{1n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
          dots & a_{nn}
          end{vmatrix}=color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}=0.
          $$



          Edit: take the matrix $A$ and do the determinant expansion along the second row
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{blue}{a_{21}} & color{blue}{a_{22}} & color{blue}dots & color{blue}{a_{2n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} & dots & a_{nn}
          end{vmatrix}=color{blue}{a_{21}}C_{21} + color{blue}{a_{22}}C_{22} + dots + color{blue}{a_{2n}}C_{2n}.
          $$
          Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.






          share|cite|improve this answer














          The expression
          $$
          color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}
          $$
          is the Laplace expansion along the second row of
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{red}{a_{11}} & color{red}{a_{12}} & color{red}dots & color{red}{a_{1n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} &
          dots & a_{nn}
          end{vmatrix}=color{red}{a_{11}}C_{21} + color{red}{a_{12}}C_{22} + dots + color{red}{a_{1n}}C_{2n}=0.
          $$



          Edit: take the matrix $A$ and do the determinant expansion along the second row
          $$
          begin{vmatrix}
          a_{11} & a_{12} & dots & a_{1n} \ color{blue}{a_{21}} & color{blue}{a_{22}} & color{blue}dots & color{blue}{a_{2n}} \ a_{31} & a_{32} & dots & a_{3n} \ vdots & vdots & ddots & vdots \ a_{n1} & a_{n2} & dots & a_{nn}
          end{vmatrix}=color{blue}{a_{21}}C_{21} + color{blue}{a_{22}}C_{22} + dots + color{blue}{a_{2n}}C_{2n}.
          $$
          Observe that the blue elements are not used to build $C_{2j}$. If we replace the blue elements with the red elements above we get exactly what we need.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 20:26









          darij grinberg

          10.2k33062




          10.2k33062










          answered Jan 4 at 14:31









          A.Γ.A.Γ.

          22.6k32656




          22.6k32656












          • Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
            – Kid Cudi
            Jan 4 at 14:54








          • 1




            @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
            – A.Γ.
            Jan 4 at 15:00








          • 1




            @KidCudi I have edited the answer, hopefully it is clearer now.
            – A.Γ.
            Jan 4 at 15:13






          • 1




            @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
            – A.Γ.
            Jan 4 at 15:17






          • 1




            I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
            – darij grinberg
            Jan 4 at 20:26


















          • Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
            – Kid Cudi
            Jan 4 at 14:54








          • 1




            @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
            – A.Γ.
            Jan 4 at 15:00








          • 1




            @KidCudi I have edited the answer, hopefully it is clearer now.
            – A.Γ.
            Jan 4 at 15:13






          • 1




            @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
            – A.Γ.
            Jan 4 at 15:17






          • 1




            I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
            – darij grinberg
            Jan 4 at 20:26
















          Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
          – Kid Cudi
          Jan 4 at 14:54






          Where did that determinant come from? Why do the first two rows suddenly become the same? Sorry I'm slow, don't see the jump. I got the same answer from the book.
          – Kid Cudi
          Jan 4 at 14:54






          1




          1




          @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
          – A.Γ.
          Jan 4 at 15:00






          @KidCudi Yes, it may take time to understand. The determinant expansion of a matrix is the sum of the element${}_{ij}cdot$the cofactor${}_{ij}$. The expression has the same structure as the determinant expansion along the second row. The red elements must be the elements of the row and the cofactors are the same as for the matrix $A$ for the same row. Observe that the second row is not used when building the cofactors. Try to do the expanstion for this matrix yourself.
          – A.Γ.
          Jan 4 at 15:00






          1




          1




          @KidCudi I have edited the answer, hopefully it is clearer now.
          – A.Γ.
          Jan 4 at 15:13




          @KidCudi I have edited the answer, hopefully it is clearer now.
          – A.Γ.
          Jan 4 at 15:13




          1




          1




          @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
          – A.Γ.
          Jan 4 at 15:17




          @KidCudi I am saying that the expression in question $a_{11}C_{21}+ldots$ is the determinant expansion of $B$, which is zero. Then the expression is also zero.
          – A.Γ.
          Jan 4 at 15:17




          1




          1




          I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
          – darij grinberg
          Jan 4 at 20:26




          I have added a $3$-rd row to your two matrices to make the pattern behind the vdots clearer.
          – darij grinberg
          Jan 4 at 20:26


















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