Apply Definition 6.1.3 to a 2 × 2 matrix, and verify that the result agrees with the formula given in Theorem 2.4.9a.
There are two patterns in the 2 × 2 matrix A=\begin{bmatrix}a&b\\c&d\end{bmatrix}:
Therefore, det A = (−1)^0ad + (−1)^1bc = ad − bc.