Proving determinant The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How do you show that $det(A)$ is the product of a linear form in $x_1, x_2, ldots, x_n$ and a linear form in $y_1, y_2, ldots, y_n$?How to find k given determinant?Vandermonde determinant for order 4Proving determinant of Vandermonde matrixCalculate the determinant of the matrix using cofactor expansion along the first rowHow to compute the given determinant?Find the determinant of the following $4 times 4$ matrixDeterminant of the vandermonde matrix times the sum of its base terms?Why is this matrix-transformation true? (determinant)Determinant = product of the difference of 2 rows
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Proving determinant
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How do you show that $det(A)$ is the product of a linear form in $x_1, x_2, ldots, x_n$ and a linear form in $y_1, y_2, ldots, y_n$?How to find k given determinant?Vandermonde determinant for order 4Proving determinant of Vandermonde matrixCalculate the determinant of the matrix using cofactor expansion along the first rowHow to compute the given determinant?Find the determinant of the following $4 times 4$ matrixDeterminant of the vandermonde matrix times the sum of its base terms?Why is this matrix-transformation true? (determinant)Determinant = product of the difference of 2 rows
$begingroup$
$$det beginpmatrix
0&x_1&x_2&ldots&x_n\
x_1&a_11&a_12&ldots&a_1n\
x_2&a_21&a_22&ldots&a_2n\
ldots&ldots&ldots&ldots&ldots\
x_n&a_n1&a_n2&ldots&a_nn\
endpmatrix = -sum_i=1^n sum_j=1^n A_ijx_ix_j.$$
where $A_ij textis the cofactor of a_ij text and n>1.$
My question is proving this problem.
First, I tried to take the cofactor expansion along the first row.
But I don't know what I do next.
Please give me some hints for proving this problem
matrices determinant
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
$endgroup$
|
show 1 more comment
$begingroup$
$$det beginpmatrix
0&x_1&x_2&ldots&x_n\
x_1&a_11&a_12&ldots&a_1n\
x_2&a_21&a_22&ldots&a_2n\
ldots&ldots&ldots&ldots&ldots\
x_n&a_n1&a_n2&ldots&a_nn\
endpmatrix = -sum_i=1^n sum_j=1^n A_ijx_ix_j.$$
where $A_ij textis the cofactor of a_ij text and n>1.$
My question is proving this problem.
First, I tried to take the cofactor expansion along the first row.
But I don't know what I do next.
Please give me some hints for proving this problem
matrices determinant
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
$endgroup$
$begingroup$
The next thing to do is a cofactor expansion along the first column of each of the cofactors of your first expansion.
$endgroup$
– Andreas Caranti
Apr 8 at 14:10
$begingroup$
What exactly is a cofactor, may I ask?
$endgroup$
– user477343
Apr 8 at 14:23
$begingroup$
@Andreas Caranti I'll try it!
$endgroup$
– Prater Darrius
Apr 8 at 14:30
$begingroup$
@user477343 Cofactor is A_ij=(-1)^(i+j) det(M_ij) det(M_ij) is submatrix deleting the i row and j column of A
$endgroup$
– Prater Darrius
Apr 8 at 14:33
$begingroup$
I need more context than that, hahah. I just learned about using matrices yesterday to solve for simultaneous equations (though I have been familiar with their determinants). I'd nevertheless hate to go off-topic.
$endgroup$
– user477343
Apr 8 at 14:35
|
show 1 more comment
$begingroup$
$$det beginpmatrix
0&x_1&x_2&ldots&x_n\
x_1&a_11&a_12&ldots&a_1n\
x_2&a_21&a_22&ldots&a_2n\
ldots&ldots&ldots&ldots&ldots\
x_n&a_n1&a_n2&ldots&a_nn\
endpmatrix = -sum_i=1^n sum_j=1^n A_ijx_ix_j.$$
where $A_ij textis the cofactor of a_ij text and n>1.$
My question is proving this problem.
First, I tried to take the cofactor expansion along the first row.
But I don't know what I do next.
Please give me some hints for proving this problem
matrices determinant
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
$endgroup$
$$det beginpmatrix
0&x_1&x_2&ldots&x_n\
x_1&a_11&a_12&ldots&a_1n\
x_2&a_21&a_22&ldots&a_2n\
ldots&ldots&ldots&ldots&ldots\
x_n&a_n1&a_n2&ldots&a_nn\
endpmatrix = -sum_i=1^n sum_j=1^n A_ijx_ix_j.$$
where $A_ij textis the cofactor of a_ij text and n>1.$
My question is proving this problem.
First, I tried to take the cofactor expansion along the first row.
But I don't know what I do next.
Please give me some hints for proving this problem
matrices determinant
matrices determinant
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
asked Apr 8 at 14:05
Prater DarriusPrater Darrius
65
65
$begingroup$
The next thing to do is a cofactor expansion along the first column of each of the cofactors of your first expansion.
$endgroup$
– Andreas Caranti
Apr 8 at 14:10
$begingroup$
What exactly is a cofactor, may I ask?
$endgroup$
– user477343
Apr 8 at 14:23
$begingroup$
@Andreas Caranti I'll try it!
$endgroup$
– Prater Darrius
Apr 8 at 14:30
$begingroup$
@user477343 Cofactor is A_ij=(-1)^(i+j) det(M_ij) det(M_ij) is submatrix deleting the i row and j column of A
$endgroup$
– Prater Darrius
Apr 8 at 14:33
$begingroup$
I need more context than that, hahah. I just learned about using matrices yesterday to solve for simultaneous equations (though I have been familiar with their determinants). I'd nevertheless hate to go off-topic.
$endgroup$
– user477343
Apr 8 at 14:35
|
show 1 more comment
$begingroup$
The next thing to do is a cofactor expansion along the first column of each of the cofactors of your first expansion.
$endgroup$
– Andreas Caranti
Apr 8 at 14:10
$begingroup$
What exactly is a cofactor, may I ask?
$endgroup$
– user477343
Apr 8 at 14:23
$begingroup$
@Andreas Caranti I'll try it!
$endgroup$
– Prater Darrius
Apr 8 at 14:30
$begingroup$
@user477343 Cofactor is A_ij=(-1)^(i+j) det(M_ij) det(M_ij) is submatrix deleting the i row and j column of A
$endgroup$
– Prater Darrius
Apr 8 at 14:33
$begingroup$
I need more context than that, hahah. I just learned about using matrices yesterday to solve for simultaneous equations (though I have been familiar with their determinants). I'd nevertheless hate to go off-topic.
$endgroup$
– user477343
Apr 8 at 14:35
$begingroup$
The next thing to do is a cofactor expansion along the first column of each of the cofactors of your first expansion.
$endgroup$
– Andreas Caranti
Apr 8 at 14:10
$begingroup$
The next thing to do is a cofactor expansion along the first column of each of the cofactors of your first expansion.
$endgroup$
– Andreas Caranti
Apr 8 at 14:10
$begingroup$
What exactly is a cofactor, may I ask?
$endgroup$
– user477343
Apr 8 at 14:23
$begingroup$
What exactly is a cofactor, may I ask?
$endgroup$
– user477343
Apr 8 at 14:23
$begingroup$
@Andreas Caranti I'll try it!
$endgroup$
– Prater Darrius
Apr 8 at 14:30
$begingroup$
@Andreas Caranti I'll try it!
$endgroup$
– Prater Darrius
Apr 8 at 14:30
$begingroup$
@user477343 Cofactor is A_ij=(-1)^(i+j) det(M_ij) det(M_ij) is submatrix deleting the i row and j column of A
$endgroup$
– Prater Darrius
Apr 8 at 14:33
$begingroup$
@user477343 Cofactor is A_ij=(-1)^(i+j) det(M_ij) det(M_ij) is submatrix deleting the i row and j column of A
$endgroup$
– Prater Darrius
Apr 8 at 14:33
$begingroup$
I need more context than that, hahah. I just learned about using matrices yesterday to solve for simultaneous equations (though I have been familiar with their determinants). I'd nevertheless hate to go off-topic.
$endgroup$
– user477343
Apr 8 at 14:35
$begingroup$
I need more context than that, hahah. I just learned about using matrices yesterday to solve for simultaneous equations (though I have been familiar with their determinants). I'd nevertheless hate to go off-topic.
$endgroup$
– user477343
Apr 8 at 14:35
|
show 1 more comment
0
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$begingroup$
The next thing to do is a cofactor expansion along the first column of each of the cofactors of your first expansion.
$endgroup$
– Andreas Caranti
Apr 8 at 14:10
$begingroup$
What exactly is a cofactor, may I ask?
$endgroup$
– user477343
Apr 8 at 14:23
$begingroup$
@Andreas Caranti I'll try it!
$endgroup$
– Prater Darrius
Apr 8 at 14:30
$begingroup$
@user477343 Cofactor is A_ij=(-1)^(i+j) det(M_ij) det(M_ij) is submatrix deleting the i row and j column of A
$endgroup$
– Prater Darrius
Apr 8 at 14:33
$begingroup$
I need more context than that, hahah. I just learned about using matrices yesterday to solve for simultaneous equations (though I have been familiar with their determinants). I'd nevertheless hate to go off-topic.
$endgroup$
– user477343
Apr 8 at 14:35