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Solving a linear homogeneous system setting the determinant equal to zero - theorem?
On solution(s) to nonhomogeneous and corresponding homogeneous system of linear equationshomogeneous linear systemWhy does $det A=0$ imply that there exists a non-zero solution to the homogeneous linear equations determined by $A$?Relation between homogeneous and non-homogeneous system of linear equationsCreating a homogeneous system of linear equations that has no solution?Why do homogenous linear systems have determinant zero?Multiplying a homogeneous linear system with a matrixWhat is the relationship between linear, non-homogeneous system of differential equations and linear, non-homogeneous system of equations?Solutions to homogeneous linear system of equations over the integers.$4 times 4$ matrix and homogeneous system of equations.
$begingroup$
I often face a matrix problem of this form: AB=CB. To solve it they write
(A-C)B=0 and then to find the solutions det(A-C)=0. So is there a theorem about linear systems that tells us how to find the solution in this way?
If I imagine a 2X2 matrix I would have an homogeneous system like this:
(a b)(x)=0
(c f)(y)=0
so why should I find the det = 0? I will just get ad - bc = 0 and from there I can find x and y?
linear-algebra systems-of-equations determinant
edited Mar 11 at 21:12
Adam
1033
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
$endgroup$
add a comment |
$begingroup$
I often face a matrix problem of this form: AB=CB. To solve it they write
(A-C)B=0 and then to find the solutions det(A-C)=0. So is there a theorem about linear systems that tells us how to find the solution in this way?
If I imagine a 2X2 matrix I would have an homogeneous system like this:
(a b)(x)=0
(c f)(y)=0
so why should I find the det = 0? I will just get ad - bc = 0 and from there I can find x and y?
linear-algebra systems-of-equations determinant
edited Mar 11 at 21:12
Adam
1033
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
$endgroup$
add a comment |
$begingroup$
I often face a matrix problem of this form: AB=CB. To solve it they write
(A-C)B=0 and then to find the solutions det(A-C)=0. So is there a theorem about linear systems that tells us how to find the solution in this way?
If I imagine a 2X2 matrix I would have an homogeneous system like this:
(a b)(x)=0
(c f)(y)=0
so why should I find the det = 0? I will just get ad - bc = 0 and from there I can find x and y?
linear-algebra systems-of-equations determinant
edited Mar 11 at 21:12
Adam
1033
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
$endgroup$
I often face a matrix problem of this form: AB=CB. To solve it they write
(A-C)B=0 and then to find the solutions det(A-C)=0. So is there a theorem about linear systems that tells us how to find the solution in this way?
If I imagine a 2X2 matrix I would have an homogeneous system like this:
(a b)(x)=0
(c f)(y)=0
so why should I find the det = 0? I will just get ad - bc = 0 and from there I can find x and y?
linear-algebra systems-of-equations determinant
linear-algebra systems-of-equations determinant
edited Mar 11 at 21:12
Adam
1033
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
edited Mar 11 at 21:12
Adam
1033
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
edited Mar 11 at 21:12
Adam
1033
edited Mar 11 at 21:12
Adam
1033
edited Mar 11 at 21:12
Adam
1033
1033
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
asked Mar 11 at 20:26
Andrea AngelettiAndrea Angeletti
11
11
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
I guess A and C depend on some variable here?
If $det(A-C) neq 0$ then $A - C$ is invertible, so the only solution is
$$B = (A-C)^-10 = 0.$$
Conversely if $det(A-C) = 0$ then $A-C$ has a nontrivial kernel so there's some nonzero solution for B.
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
add a comment |
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$begingroup$
I guess A and C depend on some variable here?
If $det(A-C) neq 0$ then $A - C$ is invertible, so the only solution is
$$B = (A-C)^-10 = 0.$$
Conversely if $det(A-C) = 0$ then $A-C$ has a nontrivial kernel so there's some nonzero solution for B.
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
add a comment |
$begingroup$
I guess A and C depend on some variable here?
If $det(A-C) neq 0$ then $A - C$ is invertible, so the only solution is
$$B = (A-C)^-10 = 0.$$
Conversely if $det(A-C) = 0$ then $A-C$ has a nontrivial kernel so there's some nonzero solution for B.
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
add a comment |
$begingroup$
I guess A and C depend on some variable here?
If $det(A-C) neq 0$ then $A - C$ is invertible, so the only solution is
$$B = (A-C)^-10 = 0.$$
Conversely if $det(A-C) = 0$ then $A-C$ has a nontrivial kernel so there's some nonzero solution for B.
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
$endgroup$
I guess A and C depend on some variable here?
If $det(A-C) neq 0$ then $A - C$ is invertible, so the only solution is
$$B = (A-C)^-10 = 0.$$
Conversely if $det(A-C) = 0$ then $A-C$ has a nontrivial kernel so there's some nonzero solution for B.
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
answered Mar 11 at 21:18
Daniel McLauryDaniel McLaury
15.9k32981
15.9k32981
add a comment |
add a comment |
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