masturbating feels better than mathturbating :p love, the divine Mother
Solutions to Triangular and Vandermonde Determinants using methods from Shilov
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n-dimensional Triangular Matrix and its Determinant : :
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We have an nth order determinant of the form,
We expand Dₙ with respect to the first row and first column via factoring a₁₁,
Hence,
We expand Dₙ₋₁ with respect to the first row first column by factoring a₂₂,
Hence,
We have,
Dₙ = a₁₁Dₙ‐₁
Dₙ₋₁ = a₂₂Dₙ₋₂
Therefore,
Dₙ = a₁₁a₂₂Dₙ−₂
We can of course expand Dₙ₋₂ by factoring out a₃₂ and then do the same thing to Dₙ―₃ and so on.
Continuing expansions leads to the result,
D = a₁₁a₂₂a₃₃ . . . aₙₙ
That is the triangular determinant equals the product of the elements appearing along the principal diagonal.
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Vandermonde Determinant : :
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Let the Vandermonde Determinant be denoted as W(x₁, . . ., x₂)
We can use row transformations R to change the determinant while conserving its value,
After simplifying column one we have,
We expand the determinant by with respect to the first row first column by factoring 1,
Factor out all terms of the form xⱼ-x₁,
The product (x₂-x₁)(x₃-x₁)...(xₙ-x₁) = Πⁿⱼ₌₂(xⱼ-x₁) while the determinant is now of order n-1
Πⁿⱼ₌₂(xⱼ-x₁) * Wₙ₋₁(x₂,x₃,...,xₙ)
We can continue to expand Wₙ₋₁(x₂,x₃,...,xₙ) by factoring out consecutive terms and reducing its order, this gives a product similar to that one the case for order n but with reducing values of i.
Wₙ₋₁(x₂,x₃,...,xₙ) = Π₂<ᵢ<ⱼ<ₙ(xⱼ-xᵢ)
As such we now have,
Πⁿⱼ₌₂(xⱼ-x₁) Π₂<ᵢ<ⱼ<ₙ(xⱼ-xᵢ)
Given the first product has i=1,
Πⁿⱼ₌₂(xⱼ-x₁) Π₂<ᵢ<ⱼ<ₙ(xⱼ-xᵢ) = Π₁<ᵢ<ⱼ<ₙ(xⱼ-xᵢ)
Hence,
W(x₁, . . ., xₙ) = Π₁<ᵢ<ⱼ<ₙ(xⱼ-xᵢ)
last edit on
6/6/2022 9:26:33 PM
1.7 Cramer's Rule
Consider a system of the special form,
a₁₁x₁ + a₁₂x₂ + . . . + a₁ₙxₙ = b₁
a₂₁x₁ + a₂₂x₂ + . . . + a₂ₙxₙ = b₂
. . . . . . . . . . . . . . .
aₙ₁x₁ + aₙ₂x₂ + . . . + aₙₙxₙ = bₙ
This system has the same number unknowns as equations.
A system like the one above is always Compatible and Determinant.
The coefficients aᵢⱼ form the coefficient matrix of the system.
We assume that c₁, c₂, . . . , cₙ is a solution, so we have the following system,
a₁₁c₁ + a₁₂c₂ + . . . + a₁ₙcₙ = b₁
a₂₁c₁ + a₂₂c₂ + . . . + a₂ₙcₙ = b₂
. . . . . . . . . . . . . . . .
aₙ₁c₁ + aₙ₂c₂ + . . . + aₙₙcₙ = bₙ
We multiply the first equation by the co-factor by A₁₁ of the elements a₁₁, then the second equation by A₂₁, and the third by A₃₁, and so on until Aₙ₁.
A₁₁(a₁₁c₁ + a₁₂c₂ + . . . + a₁ₙcₙ) = A₁₁b₁
A₂₁(a₂₁c₁ + a₂₂c₂ + . . . + a₂ₙcₙ) = A₂₁b₂
. . . . . .
A₁ₙ(a₁ₙc₁ + a₂ₙc₂ + . . . + aₙₙcₙ) = A₁ₙbₙ
By distributive property in the field K,
a₁₁A₁₁c₁ + a₁₂A₁₁c₂ + . . . + a₁ₙA₁₁cₙ = A₁₁b₁
a₂₁A₂₁c₁ + a₂₂A₂₁c₂ + . . . + a₂ₙA₂₁cₙ = A₂₁b₂
. . . . . .
aₙ₁Aₙ₁c₁ + aₙ₂Aₙ₁c₂ + . . . + aₙₙAₙ₁cₙ = Aₙ₁bₙ
Recall that each solution Cⱼ (j=1,2,...,n) corresponds to a column in the coefficient matrix aᵢⱼ,
and given D = a₁ⱼA₁ⱼ + a₂ⱼA₂ⱼ + . . . + aₙⱼAₙⱼ we have,
a₁₁A₁₁c₁ + a₂₁A₂₁c₁ + . . . + aₙ₁Aₙ₁c₁ = (a₁₁A₁₁ + a₂₁A₂₁ + . . . + aₙ₁Aₙ₁)c₁
a₁₂A₁₁c₂ + a₂₂A₂₁c₂ + . . . + aₙ₂Aₙ₁c₂ = (a₁₂A₁₁ + a₂₂A₂₁ + . . . + aₙ₂A₂₁)c₂
. . . . . . . . . .
aₙ₁A₁₁cₙ + a₂ₙA₂₁cₙ + . . . + aₙₙAₙ₁cₙ = (aₙ₁A₁₁ + a₂ₙA₂₁ + . . . + aₙₙAₙ₁)cₙ
By the linear property of determinants,
(a₁₁A₁₁ + a₂₁A₂₁ + . . . + aₙ₁Aₙ₁)c₁ = A₁₁b₁ + A₂₁b₂ + . . . + Aₙ₁bₙ
(a₁₂A₁₁ + a₂₂A₂₁ + . . . + aₙ₂A₂₁)c₂ = A₁₁b₁ + A₂₁b₂ + . . . + Aₙ₁bₙ
. . . . . . . . .
(aₙ₁A₁₁ + a₂ₙA₂₁ + . . . + aₙₙAₙ₁)cₙ = A₁₁b₁ + A₂₁b₂ + . . . + Aₙ₁bₙ
Given D = aᵢₖAᵢₖ and Dⱼ(bᵢ),
Dc₁ = Dⱼ(bᵢ)
Hence,
c₁ = Dⱼ / D
It follows that if a system exits and can be solved via the method above that solution is unique.
1.3 Determinants of Order n
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.3.1 Matrix and Determinant
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Square matrix is an array of n² numbers aᵢⱼ (i, j = 1, 2, . . . , n) who belong to field K.
Number of rows and columns of a matrix is called Order.
aᵢⱼ are elements of the matrix.
i index indicates row, j index indicates column.
Elements a₁₁, a₂₂, . . . , aₙₙ form the principal diagonal of a matrix.
Definition of Determinant
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For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ===========================================================
1.3.1.1 Fundamentals of Permutations
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Permutations ::
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A permutation is a function α:{1,2,...,n} -> {1,2,...,n}A permutation on some set {1,2,...,n} is an ordering of values from that set, that ordering can be a rearrangement.
A permutation, f : {1, 2, 3, 4, 5} -> {3 5 4 1 2}
Given a permutation is the output of a function α each permutation of n objects corresponds to some function, hence to derive the set of all permutations of n objects is to derive all functions α.
The number of permutations of n objects is always n!.
Ex. For instance, given a set {1,2,3,4,5,6} possible permutations are 1 2 3 4 5 6 or 6 5 4 3 2 1 or 4 1 2 3 6 5 etc.
The function α determines the order of a permutation.
Ex. If {1, 2, 3} and the permutation is 2 3 1 then α(1) = 2, α(2) = 3 and α(3) = 1. α maps a position to a value from the set, hence α(1) maps the position first position in the order to the value 3 from the set {1,2,3}.
We can obtain additional permutations via the group operation composition
Consider f : {1,2,3,4,5} -> {3,5,4,1,2} and g : {1,2,3,4,5} -> {3,1,5,2,5}
Composition,
f(g) : {1, 2, 3, 4, 5} -> {f(g(1)),f(g(2)),f(g(3)),f(g(4)),f(g(5))} = {f(3),f(1),f(5),f(2),f(4)} = {4,3,2,5,1}-----------------------------------------------------------
Symmetric Groups ::
-----------------------------------------------------------A symmetric group is the set of all permutations of n objects.
The symmetric group is denoted as Sₙ where n is the number of objects in that group.
Given a set of n objects has n! permutations, the number of elements of Sₙ is n!.
The group operation of a symmetric group is composition.
Ex. S₄ is a set of 4!=24 permutations where its domain is {1,2,3,4}
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Two-Row Notation ::
-----------------------------------------------------------The functional form α :{1,2,...n} -> {1,2,...n} is known as One-Row notation.
Two row notation stacks the sets so that the mappings between individual values is more explicit.
Two-Row notation,
α:{1,2,...,n}
{1,2,...,n}
Ex. π : {1,2,3,4} -> {4,2,3,1}
π: {1,2,3,4}
{4,2,3,1}-----------------------------------------------------------
Cycle Notation ::
-----------------------------------------------------------
A cycle is a kind of notation for permutations, it is a way of writing a permutation where the only information we care about are the mappings between values in α.The rule behind the mappings is that there is a map between any numbers that reference each other in anyway.
Ex. Consider permutation {1,2,3,4,5}
{3,5,4,1,2},α(1) = 3, 1 -> 3 ;; Now 3 becomes the position of domain,
α(3) = 4, 3 -> 4 ;; Now 4 becomes the position of domain,
α(4) = 1, 4 -> 1Notice the rule is that any value pointed to becomes the position to find the next mapping, this continues until a value maps to a previously referenced value. This is considered all one large mapping in the following way,
1 -> 3 -> 4 -> 1This set of mappings is written as a cycle (1 3 4), a cycle with 3 unique values is called a 3-cycle.
Two values have not been referenced yet, 2 and 5.
α(2) = 5 ;; Now 5 becomes the position of domain
α(5) = 2
2 -> 5 -> 2
(2 5) a cycle with 2 unique values is called a 2-cycle or transpositionAll together, (1 3 4)(2 5)
We can start at different positions, the order of our cycles may change but the cycles themselves (3-cycle, 2-cycle, etc) always remain the same. Under this notation we only care about the mappings and how they relate to cycles.
Ex. {1 2 3 4 5 6}
{3 4 2 6 5 1}
(1 3 2 4 5)(5)Single cycles are called identity cycles.
We can convert cycle notation back to two line notation (or functional) by reversing the process and tracking the maps.
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Products of Cycles ::
-----------------------------------------------------------Finding the product of cycles allows us to deal with cycles that have overlap.
We read through our cycles from right to left matching values to our positions via tracking their mappings, the goal being reconstruct the correct permutation.
Ex. Consider the cycles, (3 4 1 5)(2 3 6 1)(3 1)
We note that we are trying to map the set {1,2,3,4,5,6} to the permutation we want to find.
α(1)
1 -> 3 -> 6 ;; α(1) = 6
α(2)
2 -> 3 -> 4 ;; α(2) = 4
α(3)
3 -> 1 -> 2 ;; α(3) = 2
α(4)
4 -> 1 ;; α(4) = 1
α(5)
5 -> 3 ;; α(5) = 3
α(6)
6 -> 1 -> 5 ;; α(6) = 5
α : {1,2,3,4,5,6} -> {6,4,2,1,3,5}-----------------------------------------------------------
Permutations as Products of Transpositions ::
-----------------------------------------------------------All permutations can be represented as a product of transpositions (2-cycles).
Every cycle can be expressed as a product of transpositions,
(a₁a₂a₃...aₙ₋₁aₙ) = (a₁aₙ)(a₁aₙ₋₁)(a₁aₙ₋₂)...(a₁a₃)(a₁a₂)Ex. Consider, α : {1,2,3,4,5,6}
{6,4,2,1,3,5}
We convert to cycle notation,
(1 6 5 3 2 4)
Express as product of transpositions,
(1 4)(1 2)(1 3)(1 5)(1 6)Ex. Consider π : {1,2,3,4,5}
{3,5,4,1,2},
Convert to cycle notation,
(1 3 4)(2 5)
Express as product of transpositions,
(1 4)(1 3)(2 5)-----------------------------------------------------------
Inversions ::
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An inversion takes place whenever α(x) > α(y) and x < y.Ex. Given the ordered sequence 2 3 1 we say that α(1) = 2, α(2) = 3, and α(3) = 1.
α(1) < α(2) and 1 < 2, hence by definition no inversion.
α(2) > α(3) and 2 < 3, hence by definition an inversion.For each α(n) there is number of terms in a ordered sequence less than α(n), we call this value βₙ
Ex. Given an ordered sequence 3 1 2 where α(1) = 3, α(2) = 1, and α(3) = 2 we know α(1) > α(2) while 1 < 2 and
α(1) > α(3) while 1 < 3, hence β₁ = 2.The number of inversions of a permutation α is defined as N(α(1), α(2), . . ., α(n)) = β₁ + β₂ + . . . + βₙ
Ex.Consider the ordered sequence 3 2 1 where α(1) = 3, α(2) = 2, and α(3) = 1.
α(1) > α(2) and 1 < 2 ; α(1) > α(3) and 1 < 3 ---> β₁ = 2 because two inversions take place in respect to position 1. α(2) > α(3) and 2 < 3 ----> β₂ = 1 because one inversion takes place after position 2.
α(3) has no values that follow it, so β₃ = 0
N(α(1), α(2), α(3)) = β₁ + β₂ + β₃ = 2 + 1 + 0 = 3A parity is even when N(α(1),α(2),...,α(n)) is even and odd when N(α(1),α(2),...,α(n)) is odd.
Ex.The ordered sequence 3 2 1 has a odd parity because N(α(1), α(2), α(3)) = β₁ + β₂ + β₃ = 2 + 1 + 0 = 3
"Definition of Determinant
-----------------------------------------------------------
For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ"
Lets say D = 0 and the matrix is singular. How do I make it not so? Adjust the matrix so D not = 0.
FEAR! FEAR! FEAR! FEAR! FEAR! FEAR!
LiYang said:"Definition of Determinant
-----------------------------------------------------------
For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ"
Lets say D = 0 and the matrix is singular. How do I make it not so? Adjust the matrix so D not = 0.
The question is why would you do so in the first place.
You are no longer working with the same matrix.
LiYang said:"Definition of Determinant
-----------------------------------------------------------
For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ"
Lets say D = 0 and the matrix is singular. How do I make it not so? Adjust the matrix so D not = 0.The question is why would you do so in the first place.
You are no longer working with the same matrix.
The real world is not perfect. Of course it is not the same matrix.
Do you know the answer or not?
How do you know the offending linear equations contained within the set? The ones making the singularity possible.
FEAR! FEAR! FEAR! FEAR! FEAR! FEAR!
LiYang said:"Definition of Determinant
-----------------------------------------------------------
For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ"
Lets say D = 0 and the matrix is singular. How do I make it not so? Adjust the matrix so D not = 0.The question is why would you do so in the first place.
You are no longer working with the same matrix.
The real world is not perfect. Of course it is not the same matrix.
Do you know the answer or not?
How do you know the offending linear equations contained within the set? The ones making the singularity possible.
Okay, we are not talking about a generalization.
The matrix itself is a representation of some real world data set which must have been sampled, a subset of samples has led to a zeroth column/row causing the determinant to vanish.
Maybe samples vary enough that if I resampled there would be no zero columns or rows.
If the solution is purely mathematical I have to change the column or row that causes the determinant to vanish. I assume adding some value into to the zeros is necessary, however this would change all of the terms of the determinant.
I want a solution to the system but any solution I derive purely through mathematical means will be derived from an artificial change of the coefficients of that system. How tolerant I am of such a solution depends on how accurate of approximation I want. I fear in many real world circumstances that such a solution would be as worthless as one that cannot be computed at all given a vanishing determinant.
That was fun to think about, thanks.
What would you?
LiYang said:"Definition of Determinant
-----------------------------------------------------------
For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ"
Lets say D = 0 and the matrix is singular. How do I make it not so? Adjust the matrix so D not = 0.The question is why would you do so in the first place.
You are no longer working with the same matrix.
The real world is not perfect. Of course it is not the same matrix.
Do you know the answer or not?
How do you know the offending linear equations contained within the set? The ones making the singularity possible.
Okay, we are not talking about a generalization.
The matrix itself is a representation of some real world data set which must have been sampled, a subset of samples has led to a zeroth column/row causing the determinant to vanish.
Maybe samples vary enough that if I resampled there would be no zero columns or rows.
If the solution is purely mathematical I have to change the column or row that causes the determinant to vanish. I assume adding some value into to the zeros is necessary, however this would change all of the terms of the determinant.
I want a solution to the system but any solution I derive purely through mathematical means will be derived from an artificial change of the coefficients of that system. How tolerant I am of such a solution depends on how accurate of approximation I want. I fear in many real world circumstances that such a solution would be as worthless as one that cannot be computed at all given a vanishing determinant.
That was fun to think about, thanks.
What would you?
Yes, this is correct. The large data being sampled sometimes causes matrix singularity to machine precision. D ~= 0, And a slight error added or subtracted to one term then causes an adequate solution to be available. Of course this small error removes the linear coefficient dependency. But how to determine the offending coefficients when they are many? Re-sampling ALL the set is time consuming and at times continues, by chance, to be singular. Imagine sampling a very large data set taking several minutes like an MRI. Then having to say, sorry guy, the matrix is singular go back into the MRI.
FEAR! FEAR! FEAR! FEAR! FEAR! FEAR!
1.8 Minors of Arbitrary Order. Laplace's Theorem
The expansion of the determinant D written as D = (-1)¹⁺ʲa₁ⱼM₁ⱼ + (-1)²⁺ʲa₂ⱼM₂ⱼ + . . . + (-1)ⁿ⁺ʲaₙⱼMₙⱼ is a special case of the Laplace Theorem, as we will find out.
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Minor of Order K : :
-----------------------------------------------------------
Suppose for a matrix of order n we specify k<=n different rows and k columns.
The elements appearing at the intersection of these rows and columns form a square matrix of order k.
For instance the matrix A below is of order n,
We could specify the order k rows as 2 and k columns as 2, as such the intersection of these rows and columns is contains the elements a₁₁, a₁₂, a₂₁ a₂₂ which forms the following minor,
Hence, the intersection is a minor of order k.
ᵢ₁,ᵢ₂,...,ᵢₗ
We denote a minor of order k as M = Mⱼ₁,ⱼ₂,...,ⱼₗ where i and j are those columns and rows that make up the intersection.
If in the original matrix we delete the rows and columns that make up the minor M, the remaining elements again form a square matrix of order n-k.
₁,₂
From our example using matrix A, given minor M₁,₂ we have a new matrix of order n - 2,
We call such a matrix the complementary matrix of the minor M and denote it as ⁻M
₁₂ ₃,...,ₙ
Hence, the matrix Aₙ₋₂ is the complementary minor of M₁₂ and is denoted as ⁻M₃,...,ₙ
Note that if the original minor is of order 1, that is just some element aᵢⱼ, then the complementary minor is the minor Mᵢⱼ the expansion definition.
-----------------------------------------------------------
Determining Sign and Properties of Equivalence : :
-----------------------------------------------------------
In the equation ∑sign(α) Πaₐ₍ₙ₎ₙ we group together all terms of the determinant whose first k elements belong to the minor M, thus the remaining elements n-k belong to minor ⁻M.
Consider a matrix of order 3,
₁₂
We specify a minor of order k=2, that is M₁₂
₃
Hence, the complementary minor is ⁻M₃ which has a order of n-k = 3-2 = 1
M₂ = |a₃₃|
We know that for n=3 determinant, D = a₁₁a₂₂a₃₃ + a₂₁a₃₂a₁₃ + a₃₁a₁₂a₂₃ - a₃₁a₂₂a₁₃ - a₂₁a₁₂a₃₃ - a₁₁a₃₂a₂₃
We group all of the terms of the determinant whose first k elements belong to M₁ which is k=2, hence if any term has two elements from M₁ we add them two the group,
a₁₁a₂₂a₃₃ a₂₁a₁₂a₃₃
Let one of the terms be denoted as C, we want to ascribe to c a sing.
c = a₁₁a₂₂a₃₃
The first k elements of M₁ belong to a term c₁ in M₁,
c₁ = a₁₁a₂₂
If we denote by N₁ the number of inversions corresponding to these elements, the sign which must be put in front of c₁ is determined by (-1)ᴺ¹
N₁(1, 2) = 0 -> (-1)⁰ = 1
The remaining n - k elements of c belong to a term c₂ in M₂.
c₂ = a₃₃ ∈ c
The sign which must be put in front of c₂ is determined by (-1)ᴺ²
N₂(3) = 0 -> (-1)⁰ = 1
The sign of c can be determined by computing (-1)ᴺ¹⁺ᴺ²
N₁ = 0 and N₂ = 0 -> 0 + 0 = 0 -> (-1)⁰ = 0
Hence, c has a positive sign.
Note that the product of any term of the minor M₁ and any term of the minor M₂ give us one of the terms of the determinant D that we have grouped together,
Let the term from M₁ be a₁₁a₂₂ and M₂ be a₃₃, the product is a₁₁a₂₂a₃₃ which is equivalent to the chosen c.
It follows that the sum of all the terms that we have grouped together from the expression fro the determinant is equal to the product of minors M₁ and M₂
The sum of our grouped terms, including their respective signs,
a₁₁a₂₂a₃₃ - a₂₁a₁₂a₃₃ = M₁M₂ = (a₁₁a₂₂ - a₂₁a₁₂)a₃₃ = a₁₁a₂₂a₃₃ - a₂₁a₁₂a₃₃
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Analogous Problem : :
-----------------------------------------------------------
Consider the same matrix A₃
Hence the complementary minor M₂ = |a₁₁|
We interchange adjacent rows and columns until M₁ is in the upper left hand corner of the determinant D,
To determine the number of interchange we use (i₁-1)+(i₂-2)+...+(iₖ-k)+(j₁-1)+(j₂-2)+...+(jₖ-k)
₂₃
M₁ = M₂₃ therefore i₁=2, i₂=3, j₁=2, and j₂=3
As such we have (2-1)+(3-2)+(2-1)+(3-2) = 1 + 1 + 1 + 1 = 4 interchanges.
After the interchanges we obtain a determinant D₁, that is the determinant where M₁ occupies the upper left most position of the determinant.
Determinant D₁ has the same terms as determinant D, as such it will have the same value, however the sign of D₁ may very and is determined by (-1)ⁱ⁺ʲ where i= i₁ + i₂ + . . . + iₖ and j = j₁ + j₂ + . . . + jₖ.
Given i = 3 + 2 and j = 3 + 2, i + j = 10 and (-1)¹⁰ = 1, hence D₁ has a positive sign in this case.
It follows from the above that the sum of all terms in the determinant D₁ whose first k elements appear in the minor M₁ is equal to the product M₁M₂.
a₁₁a₂₂a₃₃ - a₁₁a₃₂a₂₃ = M₁M₂ = (a₂₂a₃₃ - a₃₂a₂₃)a₁₁ = a₁₁a₂₂a₃₃ - a₁₁a₃₂a₂₃
Hence, the cum of the corresponding terms of D is equal to the product (-1)ⁱ⁺ʲM₁M₂ = M₁A₂ where A₂ is called the cofactor of the minor M₁ in determinant D.
ᵢ₁,ᵢ₂,...,ᵢₖ
A₂ = ⁻Aⱼ₁,ⱼ₂,...,ⱼₖ
In this case A₂ = a₁₁(-1)¹⁰ = a₁₁
Hence, M₁A₂ = a₁₁M₁
Let the rows of the determinant D with indices i₁,i₂,...,iₖ be fixed; some elements from these rows appear in every term of D.
Group together all the terms of D such that the elements from the fixed rows i₁,i₂,...,iₖ belong to the columns with indices j₁,j₂,...,jₖ.
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Laplace's Theorem : :
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All terms of D can be divided into groups, each of which is characterized by specifying k columns.
The sum of the terms in each group is equal to the product of the corresponding minor and its cofactor.
Therefore, the determinant can be written as the sum D = ∑MA whose indices i₁,i₂,...,iₖ are fixed and the sum is over all possible values of the column indices j₁,j₂,...,jₖ.