Determinant/Leibniz formula/Fact
< Determinant < Leibniz formula
Leibniz formula for the determinant
For the
determinant
of an
n
×
n
{\displaystyle {}n\times n}
-
matrix
M
=
(
a
i
j
)
i
j
,
{\displaystyle {}M={\left(a_{ij}\right)}_{ij}\,,}
the formula
det
M
=
∑
π
∈
S
n
sgn
(
π
)
a
1
π
(
1
)
⋯
a
n
π
(
n
)
{\displaystyle {}\det M=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )a_{1\pi (1)}\cdots a_{n\pi (n)}\,}
holds.
Proof
,
Write another proof
-matrix
