6. 三阶行列式
6.1 三阶行列式的定义
对三阶方阵
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\begin{pmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} &c_{3} \end{pmatrix}
a1b1c1a2b2c2a3b3c3
，在直角坐标系中取向量
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\pmb{\alpha}=(a_{1},a_{2},a_{3}),\pmb{\beta}=(b_{1},b_{2},b_{3}),\pmb{\gamma}=(c_{1},c_{2},c_{3})
α=(a1,a2,a3),β=(b1,b2,b3),γ=(c1,c2,c3)
定义三阶行列式
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V
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\begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} &c_{3} \end{vmatrix}=V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma})
a1b1c1a2b2c2a3b3c3
=V(α,β,γ)，其中
V
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V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma})
V(α,β,γ)是以
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\pmb{\alpha},\pmb{\beta},\pmb{\gamma}
α,β,γ为相邻边的平行六面体的有向体积。
6.2 三阶行列式的性质
和二阶行列式的性质类似。
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V ( α , α , γ ) = 0 V(\pmb{\alpha},\pmb{\alpha},\pmb{\gamma})=0 V(α,α,γ)=0
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V ( α , β , γ ) = − V ( β , α , γ ) = V ( α , γ ) , β V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma})=-V(\pmb{\beta},\pmb{\alpha},\pmb{\gamma})=V(\pmb{\alpha},\pmb{\gamma}),\pmb{\beta} V(α,β,γ)=−V(β,α,γ)=V(α,γ),β
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V ( k α , β , γ ) = k V ( α , β , γ ) V(k\pmb{\alpha},\pmb{\beta},\pmb{\gamma})=kV(\pmb{\alpha},\pmb{\beta},\pmb{\gamma}) V(kα,β,γ)=kV(α,β,γ)
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V ( α , β , γ + δ ) = V ( α , β , γ ) + V ( α , β , δ ) V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma}+\pmb{\delta})=V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma})+V(\pmb{\alpha},\pmb{\beta},\pmb{\delta}) V(α,β,γ+δ)=V(α,β,γ)+V(α,β,δ)
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V ( α , β , k α + γ ) = V ( α , β , γ ) V(\pmb{\alpha},\pmb{\beta},k\pmb{\alpha}+\pmb{\gamma})=V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma}) V(α,β,kα+γ)=V(α,β,γ)
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∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ = V ( α , β , γ ) = a 1 b 2 c 3 + a 2 b 3 c 1 + a 3 b 1 c 2 − a 1 b 3 c 2 − a 2 b 1 c 3 − a 3 b 2 c 1 \begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} &c_{3} \end{vmatrix}=V(\pmb{\alpha},\pmb{\beta},\pmb{\gamma})=a_{1}b_{2}c_{3}+a_{2}b_{3}c_{1}+a_{3}b_{1}c_{2}-a_{1}b_{3}c_{2}-a_{2}b_{1}c_{3}-a_{3}b_{2}c_{1} a1b1c1a2b2c2a3b3c3 =V(α,β,γ)=a1b2c3+a2b3c1+a3b1c2−a1b3c2−a2b1c3−a3b2c1（可以用二阶行列式的拆0的方法证明，我这里不证明了，主对角线和减去反对角线和）
n n n阶方阵也有类似的性质，线性代数（或者高等代数）学过，不记了。
【例】证明：在空间仿射坐标系中有
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\pmb{\alpha}=(x_{1},x_{2},x_{3}),\pmb{\beta}=(b_{1},b_{2},b_{3})
α=(x1,x2,x3),β=(b1,b2,b3)，若
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⇔
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\pmb{\alpha}//\pmb{\beta}\Leftrightarrow\begin{vmatrix} x_{1} & x_{2}\\ y_{1} & y_{2} \end{vmatrix}=\begin{vmatrix} x_{1} & x_{3}\\ y_{1} & y_{3} \end{vmatrix}=\begin{vmatrix} x_{2} & x_{3}\\ y_{2} & y_{3} \end{vmatrix}=0
α//β⇔
x1y1x2y2
=
x1y1x3y3
=
x2y2x3y3
=0
【证】先证必要性：由于
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\pmb{\alpha}//\pmb{\beta}
α//β，则
∃
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\exists\lambda\in\mathbb{R}
∃λ∈R使得
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y_{i}=\lambda x_{i}(i=1,2,3)
yi=λxi(i=1,2,3)，所以
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\begin{vmatrix} x_{1} & x_{2}\\ y_{1} & y_{2} \end{vmatrix}=\begin{vmatrix} x_{1} & x_{3}\\ y_{1} & y_{3} \end{vmatrix}=\begin{vmatrix} x_{2} & x_{3}\\ y_{2} & y_{3} \end{vmatrix}=0
x1y1x2y2
=
x1y1x3y3
=
x2y2x3y3
=0
再证充分性，
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\begin{vmatrix} x_{1} & x_{2}\\ y_{1} & y_{2} \end{vmatrix}=0
x1y1x2y2
=0，则
∃
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\exists\lambda
∃λ使得
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y_{1}=\lambda x_{1},y_{2}=\lambda x_{2}
y1=λx1,y2=λx2，
再由
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\begin{vmatrix} x_{2} & x_{3}\\ y_{2} & y_{3} \end{vmatrix}=0
x2y2x3y3
=0可知
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y_{2}=\lambda x_{2},y_{3}=\lambda x_{3}
y2=λx2,y3=λx3
故
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\pmb{\alpha}//\pmb{\beta}
α//β
【定理】构造平面仿射坐标系
[
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[0:\pmb{e}_{1},\pmb{e}_{2}]
[0:e1,e2]，有三个点
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A=(a_{1},a_{2}),B=(b_{1},b_{2}),C=(c_{1},c_{2})
A=(a1,a2),B=(b1,b2),C=(c1,c2)，则
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A,B,C
A,B,C三点共线
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\Leftrightarrow\begin{vmatrix} a_{1} & a_{2} & 1\\ b_{1} & b_{2}& 1\\ c_{1} & c_{2} & 1 \end{vmatrix}=0
⇔
a1b1c1a2b2c2111
=0
【证】若要
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\begin{vmatrix} a_{1} & a_{2} & 1\\ b_{1} & b_{2}& 1\\ c_{1} & c_{2} & 1 \end{vmatrix}=\begin{vmatrix} a_{1}-c_{1} & a_{2}-c_{2} & 0\\ b_{1}-c_{1} & b_{2}-c_{2}& 0\\ c_{1} & c_{2} & 1 \end{vmatrix}=\begin{vmatrix} a_{1}-c_{1} & a_{2}-c_{2} \\ b_{1}-c_{1} & b_{2}-c_{2} \end{vmatrix}=\begin{vmatrix} \vec{CA} \\ \vec{CB} \end{vmatrix}=0
a1b1c1a2b2c2111
=
a1−c1b1−c1c1a2−c2b2−c2c2001
=
a1−c1b1−c1a2−c2b2−c2
=
CACB
=0
当且仅当A,B,C三点共线
