# Big formula for determinants

## Main Question or Discussion Point

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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Fredrik
Staff Emeritus
Gold Member
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}

• HallsofIvy
HallsofIvy
Homework Helper
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 $$a_{1i}a_{2j}\cdot\cdot\cdot a_{nk}$$ 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, $$\left|\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right|$$ 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 $a_{22}$ from "second row second column" so have product $a_{11}a_{22}$. The second indices in each number "12" is obviously an even permutation of "12" so we have $+a_{11}a_{22}$.
Now, choose $a_{12}$ from the first row. That is from the second column so we must also choose $a_{21}$ from the second row, second column. The product is $a_{12}a_{21}[/tex] and the second indices, "21" are an odd permutation of "12" so this is negative: [itex]-a_{12}a_{21}$. Those are all possible such choices so the determinant is $$a_{11}a_{22}- a_{12}a_{21}$$.

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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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