Library / Algebra / Chapter III / § 8
Example
Tag 064A, permanent. Cite it as [Bourbaki, Tag 064A].
Algebra, chapter III, § 8. Determinants
(1) The determinant of the empty matrix is equal to 1; the determinant of a square matrix of order 1 is equal to the unique element of this matrix. For a matrix of order 2
( ξ 11 ξ 12 ξ 21 ξ 22 )
\begin{pmatrix}
\xi_{11} & \xi_{12} \\
\xi_{21} & \xi_{22}
\end{pmatrix}
( ξ 11 ξ 21 ξ 12 ξ 22 ) then, in the above notation,
x 1 ∧ x 2 = ( ξ 11 e 1 + ξ 21 e 2 ) ∧ ( ξ 12 e 1 + ξ 22 e 2 ) = ξ 11 ξ 22 e 1 ′ ∧ e 2 + ξ 21 ξ 12 e 2 ∧ e 1
x_1 \wedge x_2 = (\xi_{11} e_1 + \xi_{21} e_2) \wedge (\xi_{12} e_1 + \xi_{22} e_2) = \xi_{11} \xi_{22} e_1' \wedge e_2 + \xi_{21} \xi_{12} e_2 \wedge e_1
x 1 ∧ x 2 = ( ξ 11 e 1 + ξ 21 e 2 ) ∧ ( ξ 12 e 1 + ξ 22 e 2 ) = ξ 11 ξ 22 e 1 ′ ∧ e 2 + ξ 21 ξ 12 e 2 ∧ e 1 whence
∣ ξ 11 ξ 12 ξ 21 ξ 22 ∣ = ξ 11 ξ 22 − ξ 12 ξ 21 .
\begin{vmatrix}
\xi_{11} & \xi_{12} \\
\xi_{21} & \xi_{22}
\end{vmatrix} = \xi_{11} \xi_{22} - \xi_{12} \xi_{21}.
ξ 11 ξ 21 ξ 12 ξ 22 = ξ 11 ξ 22 − ξ 12 ξ 21 . We translate into the language of matrices some of the results of nos. 1 and 2:
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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Transcribed from the printed edition, 1998, Springer. Not translated.