n 次元極座標のラプラシアンを直接求める

直交座標$(x_1,x_2,\dots ,x_n)$と極座標$(r,{\theta }_1,\dots ,{\theta }_{n-1})$の関係を以下のように定義します．

直交座標におけるラプラシアン${\displaystyle \mathrm{\Delta }=\sum _{i=1}^n\frac{{\partial }^2}{\partial x_i^2}=\frac{{\partial }^2}{\partial x_1^2}+\frac{{\partial }^2}{\partial x_2^2}+\cdots +\frac{{\partial }^2}{\partial x_n^2}}$を極座標に変換した形を求めるとき，$2$次元の場合の変換を繰り返す導出が一般的ですが，ここでは$1$回の変換でまとめて計算します．

$1$階微分

連鎖律より，${\displaystyle \frac{\partial }{\partial r},\frac{\partial }{\partial {\theta }_1},\dots ,\frac{\partial }{\partial {\theta }_{n-1}}}$と${\displaystyle \frac{\partial }{\partial x_1},\frac{\partial }{\partial x_2},\dots ,\frac{\partial }{\partial x_n}}$の間には以下のような関係が成り立ちます．

行列$\left(\begin{array}{cccc}\frac{\partial x_1}{\partial r}& \frac{\partial x_2}{\partial r}& \cdots & \frac{\partial x_n}{\partial r}\\ \frac{\partial x_1}{\partial {\theta }_1}& \frac{\partial x_2}{\partial {\theta }_1}& \dots & \frac{\partial x_n}{\partial {\theta }_1}\\ ⋮& ⋮& & ⋮\\ \frac{\partial x_1}{\partial {\theta }_{n-1}}& \frac{\partial x_2}{\partial {\theta }_{n-1}}& \cdots & \frac{\partial x_n}{\partial {\theta }_{n-1}}\end{array}\right)$を$A$とおきます．

${\displaystyle \frac{\partial }{\partial x_1},\frac{\partial }{\partial x_2},\dots ,\frac{\partial }{\partial x_n}}$を${\displaystyle \frac{\partial }{\partial r},\frac{\partial }{\partial {\theta }_1},\dots ,\frac{\partial }{\partial {\theta }_{n-1}}}$で表したいので，左から$A$の逆行列をかけることを考えます．そのため，$A$の成分を具体的に表し，より簡単な行列の積として書くことを考えます．

$$\begin{array}{rl}A& =\left(\begin{array}{cccc}\frac{\partial x_1}{\partial r}& \frac{\partial x_2}{\partial r}& \cdots & \frac{\partial x_n}{\partial r}\\ \frac{\partial x_1}{\partial {\theta }_1}& \frac{\partial x_2}{\partial {\theta }_1}& \dots & \frac{\partial x_n}{\partial {\theta }_1}\\ ⋮& ⋮& & ⋮\\ \frac{\partial x_1}{\partial {\theta }_{n-1}}& \frac{\partial x_2}{\partial {\theta }_{n-1}}& \cdots & \frac{\partial x_n}{\partial {\theta }_{n-1}}\end{array}\right)\\ & =\left(\begin{array}{ccccccc}\cos {\theta }_1& \sin {\theta }_1\cos {\theta }_2& \sin {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cdots & \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-3}\cos {\theta }_{n-2}& \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}\\ -r\sin {\theta }_1& r\cos {\theta }_1\cos {\theta }_2& r\cos {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cdots & r\cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-3}\cos {\theta }_{n-2}& r\cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& r\cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}\\ 0& -r\sin {\theta }_1\sin {\theta }_2& r\sin {\theta }_1\cos {\theta }_2\cos {\theta }_3& \cdots & r\sin {\theta }_1\cos {\theta }_2\cdots \sin {\theta }_{n-3}\cos {\theta }_{n-2}& r\sin {\theta }_1\cos {\theta }_2\cdots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& r\sin {\theta }_1\cos {\theta }_2\cdots \sin {\theta }_{n-2}\sin {\theta }_{n-1}\\ ⋮& 0& -r\sin {\theta }_1\sin {\theta }_2\sin {\theta }_3& & ⋮& ⋮& ⋮\\ ⋮& ⋮& 0& \ddots & ⋮& ⋮& ⋮\\ ⋮& ⋮& ⋮& \ddots & -r\sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-3}\sin {\theta }_{n-2}& r\sin {\theta }_1\sin {\theta }_2\dots \cos {\theta }_{n-2}\cos {\theta }_{n-1}& r\sin {\theta }_1\sin {\theta }_2\dots \cos {\theta }_{n-2}\sin {\theta }_{n-1}\\ 0& 0& 0& \cdots & 0& -r\sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}& r\sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}\end{array}\right)\end{array}$$
第$1,2,3,\dots ,n$行から，それぞれ$1,r,r\sin {\theta }_1,\dots ,r\sin {\theta }_1\dots \sin {\theta }_{n-2}$をくくり出します．行列の第$1,2,\dots ,n$行の各成分をそれぞれ$a_1,a_2,\dots ,a_n$倍する操作は，左から$\left(\begin{array}{ccccc}{a}_1& 0& \dots & \dots & 0\\ 0& a_2& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & a_{n-1}& 0\\ 0& \cdots & \cdots & 0& a_n\end{array}\right)$を掛ける操作と言い換えられます．
$$\begin{array}{rl}A& =\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& r& \ddots & & & ⋮\\ ⋮& \ddots & r\sin {\theta }_1& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & r\sin {\theta }_1\dots \sin {\theta }_{n-3}& 0\\ 0& \cdots & \cdots & \cdots & 0& r\sin {\theta }_1\dots \sin {\theta }_{n-2}\end{array}\right)\left(\begin{array}{ccccccc}\cos {\theta }_1& \sin {\theta }_1\cos {\theta }_2& \sin {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cdots & \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-3}\cos {\theta }_{n-2}& \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}\\ -\sin {\theta }_1& \cos {\theta }_1\cos {\theta }_2& \cos {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cdots & \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-3}\cos {\theta }_{n-2}& \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}\\ 0& -\sin {\theta }_2& \cos {\theta }_2\cos {\theta }_3& \cdots & \cos {\theta }_2\cdots \sin {\theta }_{n-3}\cos {\theta }_{n-2}& \cos {\theta }_2\cdots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_2\cdots \sin {\theta }_{n-2}\sin {\theta }_{n-1}\\ ⋮& \ddots & -\sin {\theta }_3& \ddots & ⋮& ⋮& ⋮\\ ⋮& & \ddots & \ddots & \cos {\theta }_{n-3}\cos {\theta }_{n-2}& ⋮& ⋮\\ ⋮& & & \ddots & -\sin {\theta }_{n-2}& \cos {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_{n-2}\sin {\theta }_{n-1}\\ 0& \cdots & \cdots & \cdots & 0& -\sin {\theta }_{n-1}& \cos {\theta }_{n-1}\end{array}\right)\end{array}$$
実は右側の行列は直交行列になっているのですが，ここではそれは用いずに，さらに分解を進めます．以下，簡単のため$\sin {\theta }_i,\cos {\theta }_i\ne 0$を仮定します．
第$n-1$列の$-\tan {\theta }_{n-1}$倍を第$n$列に加え，$i=n-2,n-3,\dots ,2,1$の順に第$i$列の$-\tan {\theta }_i\cos {\theta }_{i-1}$倍を第$i+1$列に加えます．列基本変形は，右から基本行列をかけることで表されます．$(i,i)(2\le i\le n-1)$成分は${\displaystyle \cos {\theta }_{i-1}\cos {\theta }_i-\tan {\theta }_{i-1}\cos {\theta }_i(-\sin {\theta }_{i-1})=(\frac{{\cos }^2{\theta }_{i-1}}{\cos {\theta }_{i-1}}+\frac{{\sin }^2{\theta }_{i-1}}{\cos {\theta }_{i-1}})\cos {\theta }_i=\frac{\cos {\theta }_i}{\cos {\theta }_{i-1}}}$，$(n,n)$成分は${\displaystyle \cos {\theta }_{n-1}-\tan {\theta }_{n-1}(-\sin {\theta }_{n-1})=\frac{{\cos }^2{\theta }_{n-1}}{\cos {\theta }_{n-1}}+\frac{{\sin }^2{\theta }_{n-1}}{\cos {\theta }_{n-1}}=\frac{1}{\cos {\theta }_{n-1}}}$となります．

$$\begin{array}{rl}A& =\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& r& \ddots & & & ⋮\\ ⋮& \ddots & r\sin {\theta }_1& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & r\sin {\theta }_1\dots \sin {\theta }_{n-3}& 0\\ 0& \cdots & \cdots & \cdots & 0& r\sin {\theta }_1\dots \sin {\theta }_{n-2}\end{array}\right)\left(\begin{array}{cccccc}\cos {\theta }_1& 0& \cdots & \cdots & \cdots & 0\\ -\sin {\theta }_1& \frac{\cos {\theta }_2}{\cos {\theta }_1}& \ddots & & & ⋮\\ 0& -\sin {\theta }_2& \frac{\cos {\theta }_3}{\cos {\theta }_2}& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & -\sin {\theta }_{n-2}& \frac{\cos {\theta }_{n-1}}{\cos {\theta }_{n-2}}& 0\\ 0& \cdots & \cdots & 0& -\sin {\theta }_{n-1}& \frac{1}{\cos {\theta }_{n-1}}\end{array}\right)\left(\begin{array}{ccccc}1& \tan {\theta }_1\cos {\theta }_2& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & 1& 0\\ 0& \cdots & \cdots & 0& 1\end{array}\right)\dots \left(\begin{array}{cccccc}1& 0& \cdots & 0& 0& 0\\ 0& 1& \ddots & ⋮& ⋮& ⋮\\ ⋮& \ddots & \ddots & 0& 0& 0\\ ⋮& & \ddots & 1& \tan {\theta }_{n-2}\cos {\theta }_{n-1}& 0\\ ⋮& & & \ddots & 1& 0\\ 0& \cdots & \cdots & \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& 1& \ddots & ⋮& ⋮\\ ⋮& \ddots & \ddots & 0& 0\\ ⋮& & \ddots & 1& \tan {\theta }_{n-1}\\ 0& \cdots & \cdots & 0& 1\end{array}\right)\end{array}$$

第$1$行の$\tan {\theta }_1$倍を第$2$行に加え，$i=2,3,\dots ,n-1$の順に第$i$行の$\cos {\theta }_{i-1}\tan {\theta }_i$倍を第$i+1$行に加えます．行基本変形は，左から基本行列をかけることで表されます．

$$\begin{array}{rl}A& =\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& r& \ddots & & & ⋮\\ ⋮& \ddots & r\sin {\theta }_1& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & r\sin {\theta }_1\dots \sin {\theta }_{n-3}& 0\\ 0& \cdots & \cdots & \cdots & 0& r\sin {\theta }_1\dots \sin {\theta }_{n-2}\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ -\tan {\theta }_1& 1& \ddots & & ⋮\\ 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & ⋮\\ 0& -\cos {\theta }_1\tan {\theta }_2& 1& \ddots & & ⋮\\ 0& 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right)\dots \left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ 0& 1& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& 1& 0\\ 0& \cdots & 0& -\cos {\theta }_{n-2}\tan {\theta }_{n-1}& 1\end{array}\right)\left(\begin{array}{ccccc}\cos {\theta }_1& 0& \cdots & \cdots & 0\\ 0& \frac{\cos {\theta }_2}{\cos {\theta }_1}& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & \frac{\cos {\theta }_{n-1}}{\cos {\theta }_{n-2}}& 0\\ 0& \cdots & \cdots & 0& \frac{1}{\cos {\theta }_{n-1}}\end{array}\right)\left(\begin{array}{ccccc}1& \tan {\theta }_1\cos {\theta }_2& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & 1& 0\\ 0& \cdots & \cdots & 0& 1\end{array}\right)\dots \left(\begin{array}{cccccc}1& 0& \cdots & 0& 0& 0\\ 0& 1& \ddots & ⋮& ⋮& ⋮\\ ⋮& \ddots & \ddots & 0& 0& 0\\ ⋮& & \ddots & 1& \tan {\theta }_{n-2}\cos {\theta }_{n-1}& 0\\ ⋮& & & \ddots & 1& 0\\ 0& \cdots & \cdots & \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& 1& \ddots & ⋮& ⋮\\ ⋮& \ddots & \ddots & 0& 0\\ ⋮& & \ddots & 1& \tan {\theta }_{n-1}\\ 0& \cdots & \cdots & 0& 1\end{array}\right)\end{array}$$

$r,\sin {\theta }_i,\cos {\theta }_i\ne 0$のとき，上の積に現れるそれぞれの行列は正則なので，その積である$A$も正則となります．逆行列は以下のようになります．
$$\begin{array}{rl}{A}^{-1}& =\left(\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& 1& \ddots & ⋮& ⋮\\ ⋮& \ddots & \ddots & 0& 0\\ ⋮& & \ddots & 1& -\tan {\theta }_{n-1}\\ 0& \cdots & \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & 0& 0& 0\\ 0& 1& \ddots & ⋮& ⋮& ⋮\\ ⋮& \ddots & \ddots & 0& 0& 0\\ ⋮& & \ddots & 1& -\tan {\theta }_{n-2}\cos {\theta }_{n-1}& 0\\ ⋮& & & \ddots & 1& 0\\ 0& \cdots & \cdots & \cdots & 0& 1\end{array}\right)\dots \left(\begin{array}{ccccc}1& -\tan {\theta }_1\cos {\theta }_2& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & 1& 0\\ 0& \cdots & \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}\frac{1}{\cos {\theta }_1}& 0& \cdots & \cdots & 0\\ 0& \frac{\cos {\theta }_1}{\cos {\theta }_2}& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & \frac{\cos {\theta }_{n-2}}{\cos {\theta }_{n-1}}& 0\\ 0& \cdots & \cdots & 0& \cos {\theta }_{n-1}\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ 0& 1& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& 1& 0\\ 0& \cdots & 0& \cos {\theta }_{n-2}\tan {\theta }_{n-1}& 1\end{array}\right)\dots \left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & ⋮\\ 0& \cos {\theta }_1\tan {\theta }_2& 1& \ddots & & ⋮\\ 0& 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ \tan {\theta }_1& 1& \ddots & & ⋮\\ 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{1}{r}& \ddots & & & ⋮\\ ⋮& \ddots & \frac{1}{r\sin {\theta }_1}& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-3}}& 0\\ 0& \cdots & \cdots & \cdots & 0& \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-2}}\end{array}\right)\end{array}$$

$\left(\begin{array}{ccccc}\frac{1}{\cos {\theta }_1}& 0& \cdots & \cdots & 0\\ 0& \frac{\cos {\theta }_1}{\cos {\theta }_2}& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & \frac{\cos {\theta }_{n-2}}{\cos {\theta }_{n-1}}& 0\\ 0& \cdots & \cdots & 0& \cos {\theta }_{n-1}\end{array}\right)$に対し，$i=2,3,\dots ,n-1$の順に第$i$行の$-\tan {\theta }_{i-1}\cos {\theta }_i$倍を第$i-1$行に加え，第$n$行の$-\tan {\theta }_{n-1}$倍を第$n-1$行に加えます．

$$\begin{array}{rl}{A}^{-1}& =\left(\begin{array}{cccccc}\frac{1}{\cos {\theta }_1}& -\sin {\theta }_1& 0& \cdots & \cdots & 0\\ 0& \frac{\cos {\theta }_1}{\cos {\theta }_2}& -\sin {\theta }_2& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& & \ddots & \ddots & -\sin {\theta }_{n-2}& 0\\ ⋮& & & \ddots & \frac{\cos {\theta }_{n-2}}{\cos {\theta }_{n-1}}& -\sin {\theta }_{n-1}\\ 0& \cdots & \cdots & \cdots & 0& \cos {\theta }_{n-1}\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ 0& 1& \ddots & & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& 1& 0\\ 0& \cdots & 0& \cos {\theta }_{n-2}\tan {\theta }_{n-1}& 1\end{array}\right)\dots \left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & ⋮\\ 0& \cos {\theta }_1\tan {\theta }_2& 1& \ddots & & ⋮\\ 0& 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ \tan {\theta }_1& 1& \ddots & & ⋮\\ 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{1}{r}& \ddots & & & ⋮\\ ⋮& \ddots & \frac{1}{r\sin {\theta }_1}& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-3}}& 0\\ 0& \cdots & \cdots & \cdots & 0& \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-2}}\end{array}\right)\end{array}$$

第$n-1$列に第$n$列の$\cos {\theta }_{n-2}\tan {\theta }_{n-1}$倍を加えます．$(n-1,n-1)$成分は${\displaystyle \frac{\cos {\theta }_{n-2}}{\cos {\theta }_{n-1}}+\cos {\theta }_{n-2}\tan {\theta }_{n-1}(-\sin {\theta }_{n-1})=\cos {\theta }_{n-2}(\frac{1}{\cos {\theta }_{n-1}}-\frac{{\sin }^2{\theta }_{n-1}}{\cos {\theta }_{n-1}})=\cos {\theta }_{n-2}\cos {\theta }_{n-1}}$となります．

$$\begin{array}{rl}{A}^{-1}& =\left(\begin{array}{ccccccc}\frac{1}{\cos {\theta }_1}& -\sin {\theta }_1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{\cos {\theta }_1}{\cos {\theta }_2}& -\sin {\theta }_2& \ddots & & & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & & ⋮\\ ⋮& & \ddots & \ddots & -\sin {\theta }_{n-3}& \ddots & ⋮\\ ⋮& & & \ddots & \frac{\cos {\theta }_{n-3}}{\cos {\theta }_{n-2}}& -\sin {\theta }_{n-2}& 0\\ 0& \cdots & \cdots & \cdots & 0& \cos {\theta }_{n-2}\cos {\theta }_{n-1}& -\sin {\theta }_{n-1}\\ 0& \cdots & \cdots & \cdots & 0& \cos {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_{n-1}\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & ⋮\\ ⋮& \ddots & \ddots & \ddots & & ⋮\\ 0& \cdots & 0& 1& \ddots & ⋮\\ 0& \cdots & 0& \cos {\theta }_{n-3}\tan {\theta }_{n-2}& 1& 0\\ 0& \cdots & 0& 0& 0& 1\end{array}\right)\dots \left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & ⋮\\ 0& \cos {\theta }_1\tan {\theta }_2& 1& \ddots & & ⋮\\ 0& 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ \tan {\theta }_1& 1& \ddots & & ⋮\\ 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{1}{r}& \ddots & & & ⋮\\ ⋮& \ddots & \frac{1}{r\sin {\theta }_1}& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-3}}& 0\\ 0& \cdots & \cdots & \cdots & 0& \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-2}}\end{array}\right)\end{array}$$

第$n-2$列に第$n-1$列の$\cos {\theta }_{n-3}\tan {\theta }_{n-2}$倍を加えます．$(n-2,n-2)$成分は${\displaystyle \frac{\cos {\theta }_{n-3}}{\cos {\theta }_{n-2}}+\cos {\theta }_{n-3}\tan {\theta }_{n-2}(-\sin {\theta }_{n-2})=\cos {\theta }_{n-3}(\frac{1}{\cos {\theta }_{n-2}}-\frac{{\sin }^2{\theta }_{n-2}}{\cos {\theta }_{n-2}})=\cos {\theta }_{n-3}\cos {\theta }_{n-2}}$となります．

$$\begin{array}{rl}{A}^{-1}& =\left(\begin{array}{cccccccc}\frac{1}{\cos {\theta }_1}& -\sin {\theta }_1& 0& \cdots & \cdots & \cdots & \cdots & 0\\ 0& \frac{\cos {\theta }_1}{\cos {\theta }_2}& -\sin {\theta }_2& \ddots & & & & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & & & ⋮\\ ⋮& & \ddots & \ddots & -\sin {\theta }_{n-4}& \ddots & & ⋮\\ ⋮& & & \ddots & \frac{\cos {\theta }_{n-4}}{\cos {\theta }_{n-3}}& -\sin {\theta }_{n-3}& \ddots & ⋮\\ 0& \cdots & \cdots & \cdots & 0& \cos {\theta }_{n-3}\cos {\theta }_{n-2}& -\sin {\theta }_{n-2}& 0\\ 0& \cdots & \cdots & \cdots & 0& \cos {\theta }_{n-3}\sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_{n-2}\cos {\theta }_{n-1}& -\sin {\theta }_{n-1}\\ 0& \cdots & \cdots & \cdots & 0& \cos {\theta }_{n-3}\sin {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_{n-1}\end{array}\right)\left(\begin{array}{ccccccc}1& 0& \cdots & \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & & ⋮\\ ⋮& \ddots & \ddots & \ddots & & & ⋮\\ 0& \cdots & 0& 1& \ddots & & ⋮\\ 0& \cdots & 0& \cos {\theta }_{n-4}\tan {\theta }_{n-3}& 1& \ddots & ⋮\\ 0& \cdots & 0& 0& 0& 1& 0\\ 0& \cdots & 0& 0& 0& 0& 1\end{array}\right)\dots \left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& 1& \ddots & & & ⋮\\ 0& \cos {\theta }_1\tan {\theta }_2& 1& \ddots & & ⋮\\ 0& 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ \tan {\theta }_1& 1& \ddots & & ⋮\\ 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{1}{r}& \ddots & & & ⋮\\ ⋮& \ddots & \frac{1}{r\sin {\theta }_1}& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-3}}& 0\\ 0& \cdots & \cdots & \cdots & 0& \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-2}}\end{array}\right)\end{array}$$

これを繰り返し，第$2$列に第$3$列の$\cos {\theta }_1\tan {\theta }_2$倍を加えるまで進めます．

$$\begin{array}{rl}{A}^{-1}& =\left(\begin{array}{cccccc}\frac{1}{\cos {\theta }_1}& -\sin {\theta }_1& 0& \cdots & \cdots & 0\\ 0& \cos {\theta }_1\cos {\theta }_2& -\sin {\theta }_2& \ddots & & ⋮\\ ⋮& \cos {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cos {\theta }_2\cos {\theta }_3& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & -\sin {\theta }_{n-2}& 0\\ 0& \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_2\sin {\theta }_3\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cdots & \cos {\theta }_{n-2}\cos {\theta }_{n-1}& -\sin {\theta }_{n-1}\\ 0& \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_2\sin {\theta }_3\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}& \cdots & \cos {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_{n-1}\end{array}\right)\left(\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ \tan {\theta }_1& 1& \ddots & & ⋮\\ 0& 0& \ddots & \ddots & ⋮\\ ⋮& ⋮& \ddots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{1}{r}& \ddots & & & ⋮\\ ⋮& \ddots & \frac{1}{r\sin {\theta }_1}& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-3}}& 0\\ 0& \cdots & \cdots & \cdots & 0& \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-2}}\end{array}\right)\end{array}$$

第$1$列に第$2$列の$\tan {\theta }_1$倍を加えます．$(1,1)$成分は${\displaystyle \frac{1}{\cos {\theta }_1}+\tan {\theta }_1(-\sin {\theta }_1)=\frac{1}{\cos {\theta }_1}-\frac{{\sin }^2{\theta }_1}{\cos {\theta }_1}=\cos {\theta }_1}$となります．

$$\begin{array}{rl}{A}^{-1}& =\left(\begin{array}{cccccc}\cos {\theta }_1& -\sin {\theta }_1& 0& \cdots & \cdots & 0\\ \sin {\theta }_1\cos {\theta }_2& \cos {\theta }_1\cos {\theta }_2& -\sin {\theta }_2& \ddots & & ⋮\\ \sin {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cos {\theta }_1\sin {\theta }_2\cos {\theta }_3& \cos {\theta }_2\cos {\theta }_3& \ddots & \ddots & ⋮\\ ⋮& ⋮& ⋮& \ddots & -\sin {\theta }_{n-2}& 0\\ \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cos {\theta }_2\sin {\theta }_3\dots \sin {\theta }_{n-2}\cos {\theta }_{n-1}& \cdots & \cos {\theta }_{n-2}\cos {\theta }_{n-1}& -\sin {\theta }_{n-1}\\ \sin {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_1\sin {\theta }_2\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_2\sin {\theta }_3\dots \sin {\theta }_{n-2}\sin {\theta }_{n-1}& \cdots & \cos {\theta }_{n-2}\sin {\theta }_{n-1}& \cos {\theta }_{n-1}\end{array}\right)\left(\begin{array}{cccccc}1& 0& \cdots & \cdots & \cdots & 0\\ 0& \frac{1}{r}& \ddots & & & ⋮\\ ⋮& \ddots & \frac{1}{r\sin {\theta }_1}& \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & \ddots & \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-3}}& 0\\ 0& \cdots & \cdots & \cdots & 0& \frac{1}{r\sin {\theta }_1\dots \sin {\theta }_{n-2}}\end{array}\right)\end{array}$$