$AC^T = det(A)I$
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
add a comment |
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
add a comment |
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
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
edited Jan 4 at 15:04
A.Γ.
22.6k32656
22.6k32656
asked Jan 4 at 14:24
Kid CudiKid Cudi
456
456
add a comment |
add a comment |
1 Answer
1
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oldest
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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.
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
|
show 2 more comments
Your Answer
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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.
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
|
show 2 more comments
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.
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
|
show 2 more comments
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.
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.
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
|
show 2 more comments
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
|
show 2 more comments
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