Show that the gaussian curvature $K$ can be written as, $$ \begin{align} K=\frac{1}{\left(E G-F^2\right)^2} \left\{\left|\begin{array}{ccc} \frac12E_{vv}+F_{uv}-\frac12G_{uu} & \frac{1}{2} E_u & F_u-\frac{1}{2} E_v \\ F_v-\frac{1}{2} G_u & E & F \\ \frac{1}{2} G_v & F & G \end{array} \right|\right. \\ & \left.-\left|\begin{array}{ccc} 0 & \frac{1}{2} E_v & \frac{1}{2} G_u \\ \frac{1}{2} E_v & E & F \\ \frac{1}{2} G_u & F & G \end{array}\right|\right\} \\ & \end{align}$$ Solution: We know that, \begin{align} x_v\cdot x_{uu}&=F_u-\frac12E_v\tag A\\ x_v\cdot x_{uv}&=\frac12G_u\tag B \end{align} differentiating (A) w.r.t. $v$ $$x_{vv} \cdot x_{uu}+x_v \cdot x_{uuv}=F_{uv}-\frac{1}{2} E_{vv}$$ differentiating (B) w.r.t $u$ $$x_{u v} \cdot x_{u v}+x_v \cdot x_{uvu}=\frac{1}{2} G_{u u}$$
subtracting we get, $$ {x}_{vv} \cdot x_{u u}-x_{u v} \cdot x_{u v}=-\frac{1}{2} E_{v v}+F_{u v}-\frac{1}{2} G_{u u} $$
Now, $$ \begin{align} K &=\frac{e g-f^2}{E G-F^2}\\ & \frac{1}{\left(E G-F^2\right)^2} \left\{ [x_{uu}\:x_u\:x_v][x_{vv}\:x_u\:x_v]-[x_{uv}\:x_u\:x_v]^2 \right\}\tag1\\ & \frac{1}{\left(E G-F^2\right)^2} \left\{\left|\begin{array}{lll} {x}_{u u} \cdot {x}_{v v} & {x}_{u u} \cdot {x}_u & {x}_{u u}\cdot {x}_v \\ {x}_u \cdot {x}_{v v} & {x}_u \cdot {x}_u & {x}_u \cdot {x}_v \\ {x}_v \cdot {x}_{v v} & {x}_v \cdot {x}_u & {x}_v \cdot {x}_v \end{array}\right|-\left|\begin{array}{lll} {x}_{u v} \cdot {x}_{u v} & {x}_{u v} \cdot {x}_u & {x}_{u v}: {x}_v \\ {x}_u \cdot {x}_{u v} & {x}_u \cdot {x}_u & {x}_u \cdot {x}_v \\ {x}_v \cdot {x}_{u v} & {x}_v \cdot {x}_u & {x}_v \cdot {x}_v \end{array}\right|\right\} \tag2\\ & =\frac{1}{\left(E G-F^2\right)^2} \left\{\left|\begin{array}{ccc} x_{u v} \cdot x_{v v} & \frac{1}{2} E_u & F_u-\frac{1}{2} E_v \\ F_v-\frac{1}{2} G_u & E & F \\ \frac{1}{2} G_v & F & G \end{array} \right|\right. \\ & \left.-\left|\begin{array}{ccc} \mathrm{x}_{u v} \cdot {x}_{u v} & \frac{1}{2} \mathrm{E}_v & \frac{1}{2} G_u \\ \frac{1}{2} E_v & E & F \\ \frac{1}{2} G_u & F & G \end{array}\right|\right\} \\ & =\frac{1}{\left(E G-F^2\right)^2} \left\{\left|\begin{array}{ccc} {x}_{u u} \cdot {x}_{v v}-{x}_{u v} \cdot {x}_{u v} & \frac{1}{2} E_u & F_u-\frac{1}{2} E_v \\ F_v-\frac{1}{2} G_u & E & F \\ \frac{1}{2} G_v & F & G \end{array} \right|\right. \\ & \left.-\left|\begin{array}{ccc} 0 & \frac{1}{2} E_v & \frac{1}{2} G_u \\ \frac{1}{2} E_v & E & F \\ \frac{1}{2} G_u & F & G \end{array}\right|\right\} \\ & = \frac{1}{\left(E G-F^2\right)^2} \left\{\left|\begin{array}{ccc} \frac12E_{vv}+F_{uv}-\frac12G_{uu} & \frac{1}{2} E_u & F_u-\frac{1}{2} E_v \\ F_v-\frac{1}{2} G_u & E & F \\ \frac{1}{2} G_v & F & G \end{array} \right|\right. \\ & \left.-\left|\begin{array}{ccc} 0 & \frac{1}{2} E_v & \frac{1}{2} G_u \\ \frac{1}{2} E_v & E & F \\ \frac{1}{2} G_u & F & G \end{array}\right|\right\} \end{align} $$
Question: I didn't understand how they get from $(1)$ to $(2)$? It will be a great help if anyone explain it.
notation explanation $[a\:b\:c]=a\cdot(b\times c)$
Note that $[a,b,c]$ is the determinant of the matrix whose columns are the vectors $a,b,c$. They then seem to be using $\det(B^\top A) = \det B \det A$ and, similarly, $\det(A^\top A) = \det(A^\top)\det A = (\det A)^2$.