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The Leibniz formula gives $\det (\begin{array}{cc} a & b \\ c & d \end{array}) = a d - b c$

Example

For every commutative ring, $\det (\begin{array}{cc} a & b \\ c & d \end{array}) = a d - b c .$

Facts & Assumptions

Given: A $2 \times 2$ matrix $A = (\begin{array}{cc} a & b \\ c & d \end{array})$ over a commutative ring.

[L1]

Determinant is the signed Leibniz sum over $S_{2}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L2]

Sign is $(- 1)$ to the inversion number (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

Verification

technique · direct

1.1

The two permutations of $2 = {0, 1}$ are the identity, with sign $1$ , and the swap, with one inversion and sign $- 1$ .

L1L2

2.1

Their Leibniz monomials are $a_{00} a_{11} = a d$ and $a_{10} a_{01} = c b = b c$ , so the signed sum is $a d - b c$ .

step 1.1algebra∎

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Example 4.27 (standard reference, not scraped)
D. Margalit and J. Rabinoff, Interactive Linear Algebra, §4.1 (standard reference, not scraped)

The Leibniz formula gives det⁡(abcd)=ad−bcdet(ac​bd​)=ad−bc