# Linear Algebra: #14 Leibniz Formula

**Deﬁnition**

A permutation of the numbers {1, . . . , n} is a

*bijection*

σ : {1, . . . , n} → {1, . . . , n}.

The set of all permutations of the numbers {1, . . . , n} is denoted S

_{n}. In fact, S

_{n}is a group: the symmetric group of order n. Given a permutation σ ∈ S , we will say that a pair of numbers (i, j), with i, j ∈ {1, . . . , n} is a “

*reversed pair*” if i < j, yet σ(i) > σ(j). Let s(σ) be the total number of reversed pairs in σ. Then the sign of sigma is deﬁned to be the number

sign(σ) = (−1)

^{s(σ)}

**Theorem 37 (Leibniz)**Let the elements in the matrix A be a

_{ij}, for i, j between 1 and n. Then we have

As a consequence of this formula, the following theorems can be proved:

**Theorem 38**Let A be a diagonal matrix

Then det(A) = λ

_{1}λ

_{2}· · · λ

_{n}.

**Theorem 39**Let A be a triangular matrix

Then det(A) = a

_{11}a

_{22}· · · a

_{nn}.

Leibniz formula also gives:

**Deﬁnition**

Let A ∈ M(n × n, F). The transpose A

^{t}of A is the matrix consisting of elements a

_{ij}

^{t}such that for all i and j we have a

_{ij}

^{t}= a

_{ji}, where a

_{ji}are the elements of the original matrix A.

**Theorem 40**det(A

^{t}) = det(A).

**14.1 Special rules for 2 × 2 and 3 × 3 matrices**

For the 2 × 2 matrix, the Leibniz formula reduces to the simple formula

det(A) = a

_{11}a_{22 }- a_{12}a_{21}For the 3 × 3 matrix, the formula is a little more complicated.

det(A) = a

_{11}a_{22}a_{33 }+ a_{12}a_{23}a_{33}+ a_{13}a_{21}a_{32}- a_{11}a_{23}a_{32}- a_{12}a_{21}a_{33}- a_{11}a_{23}a_{32}

**14.2 A proof of Leibniz Formula**Let the rows of the n × n identity matrix be

**ε**, . . . ,

_{1}**ε**. Thus

_{n}**ε**= (1 0 0 · · · 0),

_{1}**ε**= (0 1 0 · · · 0), . . . ,

_{2}**ε**= (0 0 0 · · · 1).

_{n}Therefore, given that the i-th row in a matrix is

ξ

_{i}= (a_{i1}a_{i2}· · · a_{in}),then we have

So let the matrix A be represented by its rows,

It was an exercise to show that the determinant function is additive. That is, if B and C are n × n matrices, then we have det(B + C) = det(B) + det(C). Therefore we can write

To begin with, observe that if

**ε**=

_{jk}**ε**for some j

_{jl}_{k}≠ j

*, then two rows are identical, and therefore the determinant is zero. Thus we need only the sum over all possible*

_{l}*permutations*(j

_{1}, j

_{2}, . . . , j

_{n}) of the numbers (1, 2, . . . , n). Then, given such a permutation, we have the matrix

This can be transformed back into the identity matrix

by means of successively exchanging pairs of rows.

Each time this is done, the determinant changes sign (from +1 to -1, or from -1 to +1). Finally, of course, we know that the determinant of the identity matrix is 1.

Therefore we obtain the Leibniz formula

## No comments:

## Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.

Comments will only be published after moderation