Can you explain the determinant formula using permutation notation?

In summary, the formula for calculating the determinant of a matrix A is det(A)=∑ det(P)a1p(1)a2p(2)...a nP(n)/P, where P is a permutation on the set {1,2,...,n}. The notation for the signature of the permutation is sgn P or (-1)^P. The permutation can be written in the form (a b c) where the numbers represent the order of the elements in the set. The determinant is found by choosing one number from each row and column and multiplying them together, with a positive or negative sign depending on whether the permutation is even or odd. This process is repeated for all possible choices and added together to get the determinant.
  • #1
Unusualskill
35
1
Can anyone explain to me this formula?

det(A)=∑ det(P)a1p(1)a2p(2)...a nP(n)
----------P

I understand the reasoning behind the formula, but i don't understand this notation...
 
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  • #2
P is a permutation on the set {1,2,...,n}. A permutation is a bijection. So the sets {1,2,...,n}={P(1),P(2),...,P(n)} are the same. That shouldn't be "det P" after the summation sigma. It should be "sgn P" or some other notation for the signature of the permutation. A permutation is said to be even if it's equivalent to an even number of swaps of two elements. For example, the permutation P defined by P(1)=2, P(2)=3, P(3)=1 is even, because you can rearrange (1,2,3) to (2,3,1) by first swapping 1 and 2 to get (2,1,3) and then swapping 1 and 3 to get (2,3,1). Odd permutations are defined similarly. Every permutation is either even or odd. The signature of a permutation is defined to be +1 if the permutation is even, and -1 if the permutation is odd. The most common notation for the signature of P is sgn P. I have also seen the notation ##(-1)^P##.

The specific permutation I used as an example can be written as (2 3 1), i.e. you simply list the numbers that 1,2,3 are taken to, in the appropriate order. The even permutations on the set {1,2,3} are (1 2 3), (2 3 1) and (3 1 2). The odd ones are (2 1 3), (3 2 1) and (1 3 2). So for a 3x3 matrix A,
\begin{align}
&=\det A =\sum_P(\operatorname{sgn}P) A_{1,P(n)}\dots,A_{n,P(n)}=\\
&=A_{11}A_{22}A_{33}+ A_{12}A_{23}A_{31}+A_{13}A_{21}A_{32} -A_{13}A_{22}A_{31} - A_{12}A_{21}A_{33}- A_{11}A_{23}A_{32}
\end{align}
 
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  • #3
What that says is "choose one number from each row and column of the array and multiply them together"
That will be a product of the form [tex]a_{1i}a_{2j}\cdot\cdot\cdot a_{nk}[/tex] where the "1, 2, ..., n" are the row numbers and "i, j, ..., k" are the column numbers. Since there is one number from each row, we do have "1, 2, ..., n". Since there is one number from each column, "i, j, ..., k" is a permutation of "1, 2, ..., n". Multiply the product by 1 if it is an even permutation and by -1 if an odd permutation. Do that for all possible choices of "one number from each row and each column" and add them all together.

For example, for the 2 by 2 determinant, [tex]\left|\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right|[/tex] if we choose "itex]a_{11}[/itex] from the "first row first column" since we already have a number from the first row, we must next choose from the second row. Since we already have a number from the first column, we must choose from the second column- we must choose [itex]a_{22}[/itex] from "second row second column" so have product [itex]a_{11}a_{22}[/itex]. The second indices in each number "12" is obviously an even permutation of "12" so we have [itex]+a_{11}a_{22}[/itex].
Now, choose [itex]a_{12}[/itex] from the first row. That is from the second column so we must also choose [itex]a_{21}[/itex] from the second row, second column. The product is [itex]a_{12}a_{21}[/tex] and the second indices, "21" are an odd permutation of "12" so this is negative: [itex]-a_{12}a_{21}[/itex]. Those are all possible such choices so the determinant is [tex]a_{11}a_{22}- a_{12}a_{21}[/tex].

For a 3 by 3 array, there are 3 choices for a number from the first row, then two choices for a number from the second rwo, NOT in the same column, then 1 choice for a number from the third row not in either of the first two columns chosen. Thus, there area 3!= 6 such choices because there are 3! permutations of "123", half of them even and half of them odd. So we would have the sum and differences of 6 terms just as Fredrik shows.

In general, there are n! such choices for an n by n array. Half will be positive and half negative. Of course, that would be an insane way of actually calculating determinants which is why we use other things like "expansion by minors".
 
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  • #4
thanks guys...anyway i understand the ideas behind already...it was just the notation that confused me...I know how to compute using big formula anyway.thx
 
  • #5


The notation used in this formula is called the permutation notation. It is used to represent the different ways in which a set of numbers or variables can be arranged. In this case, the numbers a1, a2, ..., an represent the elements of the matrix A, and P is a permutation of the numbers 1 to n.

The formula itself is used to calculate the determinant of a square matrix A. The determinant is a special value that is associated with a square matrix and is used in various mathematical and scientific applications. The formula essentially involves taking the sum of the products of the elements of the matrix A, multiplied by the corresponding elements of a permutation matrix P. The permutation matrix P is essentially a matrix that represents the different arrangements of the elements of A.

In summary, the notation in this formula may seem complicated, but it is simply a way to represent the different arrangements of elements in a matrix and the formula itself is used to calculate the determinant of a square matrix. I hope this helps to clarify the notation and the purpose of the formula.
 

1. What is the big formula for determinants?

The big formula for determinants is the method used to calculate the value of a determinant for a square matrix. It involves finding the product of the elements in the main diagonal and subtracting the product of the elements in the opposite diagonal. This process is repeated for each row or column until the determinant is found.

2. How is the big formula for determinants derived?

The big formula for determinants is derived from the properties of determinants, such as linearity and the effect of elementary row operations on the determinant. By using these properties, the formula can be simplified and applied to any square matrix.

3. Why is the big formula for determinants important?

The big formula for determinants is important because it allows us to find the value of a determinant, which is a useful tool in various applications, such as solving systems of linear equations and calculating areas and volumes. It also helps us understand the properties and behavior of matrices.

4. Are there any limitations to the big formula for determinants?

Yes, the big formula for determinants can only be used for square matrices. It also becomes increasingly complex and time-consuming to use for larger matrices, so other methods, such as row reduction, may be more efficient.

5. How can I apply the big formula for determinants in real-life situations?

The big formula for determinants can be applied in various fields, such as engineering, physics, and economics. For example, it can be used to calculate the stability of a structure or to determine the optimal production levels for a company.

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