Determinant of a 2 by 2 matrix

Statement

Let $K$ be a commutative ring and let $a, b, c, d \in K$. Then $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc,$$ where $\det$ is the determinant.

Remarks

This is the Leibniz formula for $n = 2$, with the identity and one transposition. Over $\R$, $\abs{ad - bc}$ is the area of the parallelogram spanned by the columns $(a, c)$ and $(b, d)$. For example, $\det \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = 4 - 6 = -2$.

Explanations

L1 human

The determinant of a $2 \times 2$ matrix is the product of the main diagonal minus the product of the other diagonal. It is zero exactly when one column is a multiple of the other, over a field. Then the matrix squeezes the plane onto a line or a point.