设 $a,b,c,d,x,y,u,v\in\mathbb{R}$. 则有
\[
\begin{vmatrix}
ax+cu & bx+du\\
ay+cv & by+dv
\end{vmatrix}=
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}\cdot
\begin{vmatrix}
x & y\\
u & v
\end{vmatrix}\tag{*}
\]
应用.
设 $r(u,v)=(x(u,v),y(u,v),z(u,v))$ 是曲面 $\Sigma$ 的参数方程. $(u,v)=\varphi(s,t)$ 是曲面 $\Sigma$ 的重新参数化. 若仍记 $\tilde{r}(s,t)=r(\varphi(s,t))$ 为 $r(s,t)$. 则有
\[
r_s\times r_t=r_u\times r_v\cdot\frac{\partial(u,v)}{\partial(s,t)}.
\]
1
(法一) 直接验证.
\[
\begin{split}
\begin{vmatrix}
ax+cu & bx+du\\
ay+cv & by+dv
\end{vmatrix}&=(ax+cu)(by+dv)-(bx+du)(ay+cv)\\
&=(abxy+adxv+bcuy+cduv)-(abxy+bcxv+aduy+cduv)\\
&=(ad-bc)xv+(bc-ad)uy=(ad-bc)(xv-uy)\\
&=\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}\cdot\begin{vmatrix}
x & y\\
u & v
\end{vmatrix}.
\end{split}
\]
2
(法二) 利用行列式的性质
\[
\begin{split}
\begin{vmatrix}
ax+cu & bx+du\\
ay+cv & by+dv
\end{vmatrix}&=\begin{vmatrix}
ax & bx+du\\
ay & by+dv
\end{vmatrix}+\begin{vmatrix}
cu & bx+du\\
cv & by+dv
\end{vmatrix}\\
&=\begin{vmatrix}
ax & bx\\
ay & by
\end{vmatrix}+\begin{vmatrix}
ax & du\\
ay & dv
\end{vmatrix}+\begin{vmatrix}
cu & bx\\
cv & by
\end{vmatrix}+\begin{vmatrix}
cu & du\\
cv & dv
\end{vmatrix}\\
&=0+ad(xv-uy)+bc(uy-xv)+0\\
&=(ad-bc)(xv-uy).
\end{split}
\]
3
参数曲面
\[
\begin{array}{rcl}
r:\ \Omega&\rightarrow&\mathbb{R}^3\\
(u,v) &\mapsto&(x(u,v),y(u,v),z(u,v))
\end{array}
\]
设 $\varphi:\ \Omega'\rightarrow\Omega$, $(u,v)=\varphi(s,t)$ 是 $r$ 的重新参数化, 即 $\tilde{r}(s,t)=(r\circ\varphi)(s,t)$, 下面仍记 $r(s,t)$ 为 $\tilde{r}(s,t)$.
\[
\begin{aligned}
r_s&=(x_s,y_s,z_s)=(x'_u\cdot u'_s+x'_v\cdot v'_s,\ y'_u\cdot u'_s+y'_v\cdot v'_s,\ z'_u\cdot u'_s+z'_v\cdot v'_s),\\
r_t&=(x_t,y_t,z_t)=(x'_u\cdot u'_t+x'_v\cdot v'_t,\ y'_u\cdot u'_t+y'_v\cdot v'_t,\ z'_u\cdot u'_t+z'_v\cdot v'_t),
\end{aligned}
\]
为简洁, 偏导数 $x'_u$ 为 $x_u$, 其余类似. 于是
\[
\begin{split}
r_s\times r_t&=\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
x_u\cdot u_s+x_v\cdot v_s & y_u\cdot u_s+y_v\cdot v_s & z_u\cdot u_s+z_v\cdot v_s\\
x_u\cdot u_t+x_v\cdot v_t & y_u\cdot u_t+y_v\cdot v_t & z_u\cdot u_t+z_v\cdot v_t
\end{vmatrix}\\
&=\begin{vmatrix}
y_u\cdot u_s+y_v\cdot v_s & z_u\cdot u_s+z_v\cdot v_s\\
y_u\cdot u_t+y_v\cdot v_t & z_u\cdot u_t+z_v\cdot v_t
\end{vmatrix}\vec{i}-
\begin{vmatrix}
x_u\cdot u_s+x_v\cdot v_s & z_u\cdot u_s+z_v\cdot v_s\\
x_u\cdot u_t+x_v\cdot v_t & z_u\cdot u_t+z_v\cdot v_t
\end{vmatrix}\vec{j}\\
&\quad+
\begin{vmatrix}
x_u\cdot u_s+x_v\cdot v_s & y_u\cdot u_s+y_v\cdot v_s \\
x_u\cdot u_t+x_v\cdot v_t & y_u\cdot u_t+y_v\cdot v_t
\end{vmatrix}\vec{k}
\end{split}
\]
对其中每个分量应用公式 (*),
\[
\begin{vmatrix}
y_u\cdot u_s+y_v\cdot v_s & z_u\cdot u_s+z_v\cdot v_s\\
y_u\cdot u_t+y_v\cdot v_t & z_u\cdot u_t+z_v\cdot v_t
\end{vmatrix}=\begin{vmatrix}
y_u & z_u\\
y_v & z_v
\end{vmatrix}\cdot\begin{vmatrix}
u_s & u_t\\
v_s & v_t
\end{vmatrix}
\]
\[
\begin{vmatrix}
x_u\cdot u_s+x_v\cdot v_s & z_u\cdot u_s+z_v\cdot v_s\\
x_u\cdot u_t+x_v\cdot v_t & z_u\cdot u_t+z_v\cdot v_t
\end{vmatrix}=\begin{vmatrix}
x_u & z_u\\
x_v & z_v
\end{vmatrix}\cdot\begin{vmatrix}
u_s & u_t\\
v_s & v_t
\end{vmatrix}
\]
\[
\begin{vmatrix}
x_u\cdot u_s+x_v\cdot v_s & y_u\cdot u_s+y_v\cdot v_s \\
x_u\cdot u_t+x_v\cdot v_t & y_u\cdot u_t+y_v\cdot v_t
\end{vmatrix}=\begin{vmatrix}
x_u & y_u\\
x_v & y_v
\end{vmatrix}\cdot\begin{vmatrix}
u_s & u_t\\
v_s & v_t
\end{vmatrix}
\]
这说明
\[
r_s\times r_t=(r_u\times r_v)\frac{\partial(u,v)}{\partial(s,t)}.
\]