射影幾何/經典/射影變換/射影平面的變換

二維射影變換是一種自同構，將射影平面對映到自身。

平面變換可以用以下方式合成地定義：點X在“主觀”平面上必須變換到點T，該點也在主觀平面上。變換使用這些工具：一對“觀察點”P和Q，以及一個“客觀”平面。主觀平面和客觀平面以及這兩個點都位於三維空間中，這兩個平面可以在某條線上相交。

畫出經過點P和X的直線l₁。直線l₁與客觀平面在點R處相交。畫出經過點Q和R的直線l₂。直線l₂與射影平面在點T處相交。然後T是X的射影變換。

分析

設xy平面為“主觀”平面，平面m為“客觀”平面。設平面m由以下式子描述

$z=f(x,y)=mx+ny+b$

其中常數m和n是偏斜率，b是z截距。

設有一對“觀察”點P和Q，

$P:(P_{x},P_{y},P_{z}),$

$Q:(Q_{x},Q_{y},Q_{z}).$

設點X位於“主觀”平面上

$X:(x,y,0).$

點X必須變換到點T，

$T:(T_{x},T_{y},0)$

也在“主觀”平面上。

分析結果是一對方程，一個用於橫座標 T_x，另一個用於縱座標 T_y

$T_{x}={x(-mQ_{x}P_{z}-nQ_{z}P_{y}+Q_{z}(P_{z}-b))+(ny+b)(Q_{z}P_{x}-Q_{x}P_{z}) \over (mx+ny)(Q_{z}-P_{z})-(mP_{x}+nP_{y})Q_{z}+(Q_{z}-b)P_{z}},\qquad \qquad (12)$

$T_{y}={y(-nQ_{y}P_{z}-mQ_{z}P_{x}+Q_{z}(P_{z}-b))+(mx+b)(Q_{z}P_{y}-Q_{y}P_{z}) \over (ny+mx)(Q_{z}-P_{z})-(nP_{y}+mP_{x})Q_{z}+(Q_{z}-b)P_{z}}.\qquad \qquad (13)$

二維變換最多有九個自由度：P_x、P_y、P_z、Q_x、Q_y、Q_z、m、n、b。注意公式 (12) 和 (13) 有相同的分子，並且可以透過交換m 和 n，以及交換x 和 y（包括P 和 Q 的下標）從T_x 獲得T_y。

三線性分數變換

設

$\alpha =-mQ_{x}P_{z}-nQ_{z}P_{y}+Q_{z}(P_{z}-b),$

$\beta =n(Q_{z}P_{x}-Q_{x}P_{z}),$

$\gamma =b(Q_{z}P_{x}-Q_{x}P_{z}),$

$\delta =m(Q_{z}-P_{z}),$

$\epsilon =n(Q_{z}-P_{z}),$

$\zeta =-(mP_{x}+nP_{y})Q_{z}+(Q_{z}-b)P_{z},$

使得

$T_{x}={\alpha x+\beta y+\gamma \over \delta x+\epsilon y+\zeta }.\qquad \qquad (14)$

同時設

$\eta =m(Q_{z}P_{y}-Q_{y}P_{z}),$

$\theta =-mQ_{z}P_{x}-nQ_{y}P_{z}+Q_{z}(P_{z}-b),$

$\kappa =b(Q_{z}P_{y}-Q_{y}P_{z}),$

使得

$T_{y}={\eta x+\theta y+\kappa \over \delta x+\epsilon y+\zeta }.\qquad \qquad (15)$

公式 (14) 和 (15) 共同描述了三線性分數變換。

三線性變換的複合

如果一個變換由公式 (14) 和 (15) 給出，那麼這樣的變換可以用九個係數來表徵，這些係數可以排列成一個係數矩陣

$M_{T}={\begin{bmatrix}\alpha &\beta &\gamma \\\eta &\theta &\kappa \\\delta &\epsilon &\zeta \end{bmatrix}}.$

如果有一對平面變換 T₁ 和 T₂，它們的係數矩陣分別為 $M_{T_{1}}$ 和 $M_{T_{2}}$，那麼這兩個變換的複合是另一個平面變換 T₃，

$T_{3}=T_{2}\circ T_{1},$

使得

$T_{3}(x,y)=T_{2}(T_{1}(x,y)).$

T₃ 的係數矩陣可以透過將 T₂ 和 T₁ 的係數矩陣相乘得到

$M_{T_{3}}=M_{T_{2}}\,M_{T_{1}}.$

證明

給定由以下公式定義的 T₁

$T_{1x}={\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}},$

$T_{1y}={\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}},$

並且給定由以下定義的 T₂

$T_{2x}={\alpha _{2}x+\beta _{2}y+\gamma _{2} \over \delta _{2}x+\epsilon _{2}y+\zeta _{2}},$

$T_{2y}={\eta _{2}x+\theta _{2}y+\kappa _{2} \over \delta _{2}x+\epsilon _{2}y+\zeta _{2}},$

然後可以透過將 T₁ 代入 T₂ 來計算 T₃

$T_{3x}=T_{2x}(T_{1x},T_{1y})={\alpha _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\beta _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\gamma _{2} \over \delta _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\epsilon _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\zeta _{2}}.$

用相同的 trinomial 乘以分子和分母

$T_{3x}={\alpha _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\beta _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\gamma _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1}) \over \delta _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\epsilon _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\zeta _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1})}.$

將 x、y 和 1 的係數分組

$T_{3x}={x(\alpha _{2}\alpha _{1}+\beta _{2}\eta _{1}+\gamma _{2}\delta _{1})+y(\alpha _{2}\beta _{1}+\beta _{2}\theta _{1}+\gamma _{2}\epsilon _{1})+(\alpha _{2}\gamma _{1}+\beta _{2}\kappa _{1}+\gamma _{2}\zeta _{1}) \over x(\delta _{2}\alpha _{1}+\epsilon _{2}\eta _{1}+\zeta _{2}\delta _{1})+y(\delta _{2}\beta _{1}+\epsilon _{2}\theta _{1}+\zeta _{2}\epsilon _{1})+(\delta _{2}\gamma _{1}+\epsilon _{2}\kappa _{1}+\zeta _{2}\zeta _{1})}={\alpha _{3}x+\beta _{3}y+\gamma _{3} \over \delta _{3}x+\epsilon _{3}y+\zeta _{3}}.$

T₃ 的六個係數與透過以下乘積獲得的係數相同

${\begin{bmatrix}\alpha _{2}&\beta _{2}&\gamma _{2}\\\eta _{2}&\theta _{2}&\kappa _{2}\\\delta _{2}&\epsilon _{2}&\zeta _{2}\end{bmatrix}}{\begin{bmatrix}\alpha _{1}&\beta _{1}&\gamma _{1}\\\eta _{1}&\theta _{1}&\kappa _{1}\\\delta _{1}&\epsilon _{1}&\zeta _{1}\end{bmatrix}}={\begin{bmatrix}\alpha _{3}&\beta _{3}&\gamma _{3}\\\eta _{3}&\theta _{3}&\kappa _{3}\\\delta _{3}&\epsilon _{3}&\zeta _{3}\end{bmatrix}}.\qquad \qquad (16)$

其餘三個係數可以透過以下方式驗證

$T_{3y}=T_{2y}(T_{1x},T_{1y})={\eta _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\theta _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\kappa _{2} \over \delta _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\epsilon _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\zeta _{2}}.$

用相同的 trinomial 乘以分子和分母

$T_{3y}={\eta _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\theta _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\kappa _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1}) \over \delta _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\epsilon _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\zeta _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1})}.$

將x、y和1的係數分組

$T_{3x}={x(\eta _{2}\alpha _{1}+\theta _{2}\eta _{1}+\kappa _{2}\delta _{1})+y(\eta _{2}\beta _{1}+\theta _{2}\theta _{1}+\kappa _{2}\epsilon _{1})+(\eta _{2}\gamma _{1}+\theta _{2}\kappa _{1}+\kappa _{2}\zeta _{1}) \over x(\delta _{2}\alpha _{1}+\epsilon _{2}\eta _{1}+\zeta _{2}\delta _{1})+y(\delta _{2}\beta _{1}+\epsilon _{2}\theta _{1}+\zeta _{2}\epsilon _{1})+(\delta _{2}\gamma _{1}+\epsilon _{2}\kappa _{1}+\zeta _{2}\zeta _{1})}={\eta _{3}x+\theta _{3}y+\kappa _{3} \over \delta _{3}x+\epsilon _{3}y+\zeta _{3}}.$

所得到的三個剩餘係數與透過公式 (16) 獲得的係數相同。證畢。

直線的平面變換

由公式 (14) 和 (15) 給出的三線性變換將一條直線

$y=mx+b$

變換成另一條直線

$T_{y}=nT_{x}+c$

其中 n 和 c 是常數，等於

$n={m(\epsilon \kappa -\zeta \theta )+b(\delta \theta -\epsilon \eta )+(\delta \kappa -\zeta \eta ) \over m(\epsilon \gamma -\zeta \beta )+b(\delta \beta -\epsilon \alpha )+(\delta \gamma -\zeta \alpha )}$

和

$c={m(\beta \kappa -\gamma \theta )+b(\alpha \theta -\beta \eta )+(\alpha \kappa -\gamma \eta ) \over m(\beta \zeta -\gamma \epsilon )+b(\alpha \epsilon -\beta \delta )+(\alpha \zeta -\gamma \delta )}.$

證明

給定 y = m x + b，然後將其代入公式 (14) 和 (15) 得出

$T_{x}={\alpha x+\beta (mx+b)+\gamma \over \delta x+\epsilon (mx+b)+\zeta }={(\alpha +\beta m)x+(\beta b+\gamma ) \over (\delta +\epsilon m)x+(\epsilon b+\zeta )},$

和

$T_{y}={(\eta +\theta m)x+(\theta b+\kappa ) \over (\delta +\epsilon m)x+(\epsilon b+\zeta )}.$

如果T_y = n T_x + c 且n 和c 為常數，則

${\partial T_{y} \over \partial x}=n{\partial T_{x} \over \partial x}$

使得

$n={\partial T_{y}/\partial x \over \partial T_{x}/\partial y}.$

計算表明

${\partial T_{x} \over \partial x}={(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m) \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )]^{2}}$

和

${\partial T_{y} \over \partial x}={(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m) \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )]^{2}}$

因此

$n={\partial T_{y}/\partial x \over \partial T_{x}/\partial y}={(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m) \over (\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)}.$

現在我們應該得到c 為

$c=T_{y}-nT_{x}$

$={(\eta +\theta m)x+(\theta b+\kappa )-\left[{(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m) \over (\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)}\right]\cdot [(\alpha +\beta m)x+(\beta b+\gamma )] \over (\delta +\epsilon m)x+(\epsilon b+\zeta )}.$

將分子中的兩個分數相加。

$c={\left\{[(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)][(\eta +\theta m)x+(\theta b+\kappa )]-[(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m)][(\alpha +\beta m)x+(\beta b+\gamma )]\right\} \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )][(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)]}.$

將分子括號中的二項式展開，然後抵消相等且相反的項。

$c={-(\beta b+\gamma )(\delta +\epsilon m)(\eta +\theta m)x+(\epsilon b+\zeta )(\alpha +\beta m)(\theta b+\kappa )+(\theta b+\kappa )(\delta +\epsilon m)(\alpha +\beta m)x-(\epsilon b+\zeta )(\eta +\theta m)(\beta b+\gamma ) \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )][(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)]}.$

將分子分解成兩項，其中只有一項包含“數目”(x)。在分母中還有一個“數目”。現在目標是使這兩項都抵消。

$c={\left\{[(\theta b+\kappa )(\alpha +\beta m)-(\beta b+\gamma )(\eta +\theta m)](\delta +\epsilon m)x+[(\alpha +\beta m)(\theta b+\kappa )-(\eta +\theta m)(\beta b+\gamma )](\epsilon b+\zeta )\right\} \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )][(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)]}.$

對分子進行因式分解

$c={[(\theta b+\kappa )(\alpha +\beta m)-(\beta b+\gamma )(\eta +\theta m)][(\delta +\epsilon m)x+(\epsilon b+\zeta )] \over [(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)][(\delta +\epsilon m)x+(\epsilon b+\zeta )]}.$

含有未知數項的因式相互抵消，因此

$c={(\alpha +\beta m)(\theta b+\kappa )-(\beta b+\gamma )(\eta +\theta m) \over (\alpha +\beta m)(\epsilon b+\zeta )-(\beta b+\gamma )(\delta +\epsilon m)}$

是一個常數。 Q.E.D.

比較c和n，注意到它們的denominator相同。 此外，n可以透過交換以下係數從c得到

$\alpha \leftrightarrow \delta ,\ \beta \leftrightarrow \epsilon ,\ \gamma \leftrightarrow \zeta .$

在n的分子和分母之間也存在以下對稱性

$\alpha \leftrightarrow \eta ,\ \beta \leftrightarrow \theta ,\ \gamma \leftrightarrow \kappa .$

c的分子和分母也具有交換對稱性：$\{\eta \leftrightarrow \delta ,\ \theta \leftrightarrow \epsilon ,\ \kappa \leftrightarrow \zeta \}.$

n和c之間的交換對稱性可以分解成二項式

$n\leftrightarrow c\equiv \{(\alpha +m\beta )\leftrightarrow (\delta +m\epsilon ),\ (\gamma +b\beta )\leftrightarrow (\zeta +b\epsilon )\}.$

所有這些交換對稱性相當於在係數矩陣中交換成對的行。

圓錐曲線的平面變換

像方程（14）和（15）給出的三線性變換T 將把一個圓錐曲線

$Ax^{2}+By^{2}+Cx+Dy+Exy+F=0\qquad \qquad (17)$

轉換成另一個圓錐曲線

$A'T_{x}^{2}+B'T_{y}^{2}+C'T_{x}+D'T_{y}+E'T_{x}T_{y}+F'=0.\qquad \qquad (18)$

證明

設有一個由方程（17）描述的圓錐曲線和一個由方程（15）和（16）描述的平面變換T，它將點(x,y) 轉換成點(T_x,T_y)。

可以找到一個逆變換T′，它將點(T_x,T_y) 轉換回點(x,y)。這個逆變換有一個係數矩陣

$M_{T'}={\begin{bmatrix}\alpha '&\beta '&\gamma '\\\eta '&\theta '&\kappa '\\\delta '&\epsilon '&\zeta '\end{bmatrix}}.$

方程（17）可以用逆變換表示

$A\left({\alpha 'T_{x}+\beta 'T_{y}+\gamma ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)^{2}+B\left({\eta 'T_{x}+\theta 'T_{y}+\kappa ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)^{2}+C\left({\alpha 'T_{x}+\beta 'T_{y}+\gamma ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)+D\left({\eta 'T_{x}+\theta 'T_{y}+\kappa ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)+E\left({\alpha 'T_{x}+\beta 'T_{y}+\gamma ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)\left({\eta 'T_{x}+\theta 'T_{y}+\kappa ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)+F=0.$

方程兩邊乘以三項式的平方可以“消去”分母

$A(\alpha 'T_{x}+\beta 'T_{y}+\gamma ')^{2}+B(\eta 'T_{x}+\theta 'T_{y}+\kappa ')^{2}+C(\alpha 'T_{x}+\beta 'T_{y}+\gamma ')(\delta 'T_{x}+\epsilon 'T_{y}+\zeta ')+D(\eta 'T_{x}+\theta 'T_{y}+\kappa ')(\delta 'T_{x}+\epsilon 'T_{y}+\zeta ')+E(\alpha 'T_{x}+\beta 'T_{y}+\gamma ')(\eta 'T_{x}+\theta 'T_{y}+\kappa ')+F(\delta 'T_{x}+\epsilon 'T_{y}+\zeta ')^{2}=0.$

展開三項式乘積並收集T_x 和 T_y 的公共冪

${\begin{matrix}(A\alpha '^{2}+B\eta '^{2}+C\alpha '\delta '+D\eta '\delta '+E\alpha '\eta '+F\delta '^{2})T_{x}^{2}\\+(A\beta '^{2}+B\theta '^{2}+C\beta '\epsilon '+D\theta '\epsilon '+E\beta '\theta '+F\epsilon '^{2})T_{y}^{2}\\+(2A\alpha '\gamma '+2B\eta '\kappa '+C(\alpha '\zeta '+\gamma '\delta ')+D(\eta '\zeta '+\kappa '\delta ')+E(\alpha '\kappa '+\gamma '\eta ')+2F\delta '\zeta ')T_{x}\\+(2A\beta '\gamma '+2B\theta '\kappa '+C(\beta '\zeta '+\gamma '\epsilon ')+D(\theta '\zeta '+\kappa '\epsilon ')+E(\beta '\kappa '+\gamma '\theta ')+2F\epsilon '\zeta ')T_{y}\\+(2A\alpha '\beta '+2B\eta '\theta '+C(\alpha '\epsilon '+\beta '\delta ')+D(\eta '\epsilon '+\theta '\delta ')+E(\alpha '\theta '+\beta '\eta ')+2F\delta '\epsilon ')T_{x}T_{y}\\+(A\gamma '^{2}+B\kappa '^{2}+C\gamma '\zeta '+D\kappa '\zeta '+E\gamma '\kappa '+F\zeta '^{2})=0.\end{matrix}}\qquad \qquad (19)$

公式 (19) 與公式 (18) 形式相同。

剩下要做的就是用非帶撇係數表示帶撇係數。為此，將克萊姆法則應用於係數矩陣 M_T，以獲得逆變換的帶撇矩陣。

$M_{T'}={1 \over \Delta }{\begin{bmatrix}\left|{\begin{matrix}\theta &\kappa \\\epsilon &\zeta \end{matrix}}\right|&\left|{\begin{matrix}\epsilon &\zeta \\\beta &\gamma \end{matrix}}\right|&\left|{\begin{matrix}\beta &\gamma \\\theta &\kappa \end{matrix}}\right|\\\quad &\quad &\quad \\\left|{\begin{matrix}\kappa &\eta \\\zeta &\delta \end{matrix}}\right|&\left|{\begin{matrix}\zeta &\delta \\\gamma &\alpha \end{matrix}}\right|&\left|{\begin{matrix}\gamma &\alpha \\\kappa &\eta \end{matrix}}\right|\\\quad &\quad &\quad \\\left|{\begin{matrix}\eta &\theta \\\delta &\epsilon \end{matrix}}\right|&\left|{\begin{matrix}\delta &\epsilon \\\alpha &\beta \end{matrix}}\right|&\left|{\begin{matrix}\alpha &\beta \\\eta &\theta \end{matrix}}\right|\end{bmatrix}}\qquad \qquad (20)$

其中 *Δ* 是未加撇的係數矩陣的行列式。

等式 (20) 允許用未加撇的係數表示加撇的係數。但將這些替換應用於等式 (19) 中的加撇係數後，可以注意到行列式 *Δ* 自己抵消了，因此可以完全忽略它。所以

$A'=A(\theta \zeta -\kappa \epsilon )^{2}+B(\kappa \delta -\eta \zeta )^{2}+C(\theta \zeta -\kappa \epsilon )(\eta \epsilon -\theta \delta )+D(\kappa \delta -\eta \zeta )(\eta \epsilon -\theta \delta )+E(\theta \zeta -\kappa \epsilon )(\kappa \delta -\eta \zeta )+F(\eta \epsilon -\theta \delta )^{2}$

$B'=A(\epsilon \gamma -\zeta \beta )^{2}+B(\zeta \alpha -\delta \gamma )^{2}+C(\epsilon \gamma -\zeta \beta )(\delta \beta -\epsilon \alpha )+D(\zeta \alpha -\delta \gamma )(\delta \beta -\epsilon \alpha )+E(\epsilon \gamma -\zeta \beta )(\zeta \alpha -\delta \gamma )+F(\delta \beta -\epsilon \alpha )^{2}$

$C'=2A(\theta \zeta -\kappa \epsilon )(\beta \kappa -\gamma \theta )+2B(\kappa \delta -\eta \zeta )(\gamma \eta -\alpha \kappa )+C[(\theta \zeta -\kappa \epsilon )(\alpha \theta -\beta \eta )+(\beta \kappa -\gamma \theta )(\eta \epsilon -\theta \delta )]+D[(\kappa \delta -\eta \zeta )(\alpha \theta -\beta \eta )+(\gamma \eta -\alpha \kappa )(\eta \epsilon -\theta \delta )]+E[(\theta \zeta -\kappa \epsilon )(\gamma \eta -\alpha \kappa )+(\beta \kappa -\gamma \theta )(\kappa \delta -\eta \zeta )]+2F(\eta \epsilon -\theta \delta )(\alpha \theta -\beta \eta )$

$E'=2A(\theta \zeta -\kappa \epsilon )(\epsilon \gamma -\zeta \beta )+2B(\kappa \delta -\eta \zeta )(\zeta \alpha -\delta \gamma )+C[(\theta \zeta -\kappa \epsilon )(\delta \beta -\epsilon \alpha )+(\epsilon \gamma -\zeta \beta )(\eta \epsilon -\theta \delta )]+D[(\kappa \delta -\eta \zeta )(\delta \beta -\epsilon \alpha )+(\zeta \alpha -\delta \gamma )(\eta \epsilon -\theta \delta )]+E[(\theta \zeta -\kappa \epsilon )(\zeta \alpha -\delta \gamma )+(\epsilon \gamma -\zeta \beta )(\kappa \delta -\eta \zeta )]+2F(\eta \epsilon -\theta \delta )(\delta \beta -\epsilon \alpha )$