Let's calculate \( \mathbf{S} \) matrices as:
\[ \mathbf{S} = \begin{bmatrix}a^2 & ab\cos\gamma & ac\cos\beta \\ ab\cos\gamma & b^2 & bc\cos\alpha \\ ac\cos\beta & bc\cos\alpha & c^2 \\ \end{bmatrix} \]As they are symmetric matrices, we have:
\[ \mathbf{S}_1 = \mathbf{S}^T_1 = \begin{bmatrix}89.1759 & 40.1321 & 33.2226 \\ 40.1321 & 144.8846 & 2.4674 \\ 33.2226 & 2.4674 & 160.187 \\ \end{bmatrix} \]\[ \mathbf{S}_2 = \mathbf{S}^T_2 = \begin{bmatrix}88.8325 & 38.8581 & 34.5862 \\ 38.8581 & 145.3423 & -0.0676 \\ 34.5862 & -0.0676 & 157.4473 \\ \end{bmatrix} \]Hence:
\[ \mathbf{S}_1 \mathbf{S}^T_1 = \mathbf{S}^2_1 = \begin{bmatrix}10666.66774198 & 9475.31283529 & 8383.50582508 \\ 9475.31283529 & 22608.22083033 & 2086.0263713 \\ 8383.50582508 & 2086.0263713 & 26769.70418252 \\ \end{bmatrix} \]\[ \mathbf{S}_2 \mathbf{S}^T_2 = \mathbf{S}^2_2 = \begin{bmatrix}10597.3702223 & 9097.24976876 & 8515.2556112 \\ 9097.24976876 & 22634.34067466 & 1323.48544126 \\ 8515.2556112 & 1323.48544126 & 25985.86207749 \\ \end{bmatrix} \]To find eigenvectors and eigenvalues we have to solve \( \mathbf{S}^2 \mathbf{v} = \lambda \mathbf{v} \) equations.
For that we have to solve \( \det \mathbf{M} = 0 \) equations for \( \lambda \)'s, where:
\[ \mathbf{M}_1 = \mathbf{S}^2_1 - \lambda_1 \mathbf{I} = \begin{bmatrix}10666.66774198-\lambda_1 & 9475.31283529 & 8383.50582508 \\ 9475.31283529 & 22608.22083033-\lambda_1 & 2086.0263713 \\ 8383.50582508 & 2086.0263713 & 26769.70418252-\lambda_1 \\ \end{bmatrix} \]\[ \mathbf{M}_2 = \mathbf{S}^2_2 - \lambda_2 \mathbf{I} = \begin{bmatrix}10597.3702223-\lambda_2 & 9097.24976876 & 8515.2556112 \\ 9097.24976876 & 22634.34067466-\lambda_2 & 1323.48544126 \\ 8515.2556112 & 1323.48544126 & 25985.86207749-\lambda_2 \\ \end{bmatrix} \]By expanding \( \det \mathbf{M} \) the above equations can be expressed as \( -\lambda^3 + A \lambda^2 + B \lambda + C = 0 \), where:
\[ A_1 = 60044.59275483 \]\[ B_1 = -967497074.35397 \]\[ C_1 = 2748224893403.1 \]\[ A_2 = 59217.57297445 \]\[ B_2 = -946397998.64196 \]\[ C_2 = 2627776450233 \]These cubic equations can be expressed in canonical form as: \( \Lambda^3 + p \Lambda + q = 0 \), where:
\[ \Lambda_1 = \lambda_1 - \frac{A_1}{3} = \lambda_1 - 20014.86425161 \]\[ p_1 = -234287298.67715 \]\[ q_1 = 580396989672.07 \]\[ \Lambda_2 = \lambda_2 - \frac{A_2}{3} = \lambda_2 - 19739.190991483 \]\[ p_2 = -222508984.35281 \]\[ q_2 = 671168939843.27 \]By solving these cubic equations with trigonometric method we get:
\[ \Delta_1 = -3.9208690567877E+23 \]\[ \varphi_1 = 2.0047797754921 {\ \rm rad }\]\[ \Delta_2 = -2.954006570189E+23 \]\[ \varphi_2 = 2.1239421619943 {\ \rm rad }\]Thus the roots (eigenvalues) are:
\[ \lambda_{1(1)} = 33887.489220463 \]\[ \lambda_{1(2)} = 3594.3540661482 \]\[ \lambda_{1(3)} = 22562.749468219 \]\[ \lambda_{2(1)} = 32824.135112237 \]\[ \lambda_{2(2)} = 3496.3497437825 \]\[ \lambda_{2(3)} = 22897.088118431 \]To find eigenvectors let's put \( \lambda \)'s to \( \mathbf{M} \begin{bmatrix}x \\ y \\ z \\ \end{bmatrix} = 0 \) and substitute \( x=1 \), \( y=1 \) and \( z=1 \) respectively.
By solving equation systems we get:
\[ \mathbf{v}_{1(1)} = \left[1 , 1.1185213617771 , 1.5056328373899\right] \]\[ \mathbf{v}_{1(2)} = \left[-2.158781232565 , 1 , 0.69091205037158\right] \]\[ \mathbf{v}_{1(3)} = \left[-0.21461481222048 , -1.1542184792236 , 1\right] \]\[ \mathbf{v}_{2(1)} = \left[1 , 1.0817075652382 , 1.4545894518911\right] \]\[ \mathbf{v}_{2(2)} = \left[-2.2172876703584 , 1 , 0.78068770789171\right] \]\[ \mathbf{v}_{2(3)} = \left[-0.17952667330599 , -1.1787499871135 , 1\right] \]Now we can normalize vectors:
\[ \hat{\mathbf{v}}_{1(1)} = \frac{\mathbf{v}_{1(1)}}{|\mathbf{v}_{1(1)}|} = \left[0.47046347501073 , 0.52622344673538 , 0.70834525676873\right] \]\[ \hat{\mathbf{v}}_{1(2)} = \frac{\mathbf{v}_{1(2)}}{|\mathbf{v}_{1(2)}|} = \left[-0.87137671678353 , 0.40364289981723 , 0.27888174353065\right] \]\[ \hat{\mathbf{v}}_{1(3)} = \frac{\mathbf{v}_{1(3)}}{|\mathbf{v}_{1(3)}|} = \left[-0.13916442120164 , -0.7484392383708 , 0.64843810062221\right] \]\[ \hat{\mathbf{v}}_{2(1)} = \frac{\mathbf{v}_{2(1)}}{|\mathbf{v}_{2(1)}|} = \left[0.483034201354 , 0.52250174987339 , 0.70261645419215\right] \]\[ \hat{\mathbf{v}}_{2(2)} = \frac{\mathbf{v}_{2(2)}}{|\mathbf{v}_{2(2)}|} = \left[-0.86796837322008 , 0.39145501272722 , 0.30560411662873\right] \]\[ \hat{\mathbf{v}}_{2(3)} = \frac{\mathbf{v}_{2(3)}}{|\mathbf{v}_{2(3)}|} = \left[-0.11536404731112 , -0.75746610114907 , 0.64260115328096\right] \]to get \( \mathbf{U} \) matrices:
\[ \mathbf{U}_1 = \begin{bmatrix}\hat{\mathbf{v}}_{1(1)} \\ \hat{\mathbf{v}}_{1(2)} \\ \hat{\mathbf{v}}_{1(3)} \\ \end{bmatrix} = \begin{bmatrix}0.47046347501073 & 0.52622344673538 & 0.70834525676873 \\ -0.87137671678353 & 0.40364289981723 & 0.27888174353065 \\ -0.13916442120164 & -0.7484392383708 & 0.64843810062221 \\ \end{bmatrix} \]\[ \mathbf{U}_2 = \begin{bmatrix}\hat{\mathbf{v}}_{2(1)} \\ \hat{\mathbf{v}}_{2(2)} \\ \hat{\mathbf{v}}_{2(3)} \\ \end{bmatrix} = \begin{bmatrix}0.483034201354 & 0.52250174987339 & 0.70261645419215 \\ -0.86796837322008 & 0.39145501272722 & 0.30560411662873 \\ -0.11536404731112 & -0.75746610114907 & 0.64260115328096 \\ \end{bmatrix} \]We calculate \( \mathbf{D} \) matrices as:
\[ \mathbf{D}_1 = \begin{bmatrix}\sqrt{\lambda_{1(1)}} & 0 & 0 \\ 0 & \sqrt{\lambda_{1(2)}} & 0 \\ 0 & 0 & \sqrt{\lambda_{1(3)}} \\ \end{bmatrix} = \begin{bmatrix}184.08554864645 & 0 & 0 \\ 0 & 59.952932089667 & 0 \\ 0 & 0 & 150.20901926389 \\ \end{bmatrix} \]\[ \mathbf{D}_2 = \begin{bmatrix}\sqrt{\lambda_{2(1)}} & 0 & 0 \\ 0 & \sqrt{\lambda_{2(2)}} & 0 \\ 0 & 0 & \sqrt{\lambda_{2(3)}} \\ \end{bmatrix} = \begin{bmatrix}181.17432244178 & 0 & 0 \\ 0 & 59.129939487391 & 0 \\ 0 & 0 & 151.31783807083 \\ \end{bmatrix} \]Now we transpose \( \mathbf{U} \) matrices to get:
\[ \mathbf{U}^T_1 = \begin{bmatrix}0.47046347501073 & -0.87137671678353 & -0.13916442120164 \\ 0.52622344673538 & 0.40364289981723 & -0.7484392383708 \\ 0.70834525676873 & 0.27888174353065 & 0.64843810062221 \\ \end{bmatrix} \]\[ \mathbf{U}^T_2 = \begin{bmatrix}0.483034201354 & -0.86796837322008 & -0.11536404731112 \\ 0.52250174987339 & 0.39145501272722 & -0.75746610114907 \\ 0.70261645419215 & 0.30560411662873 & 0.64260115328096 \\ \end{bmatrix} \]We calculate inverted square root of \( \mathbf{D} \)'s:
\[ \mathbf{D}^{-\frac12}_1 = \begin{bmatrix}0.073703846198208 & 0 & 0 \\ 0 & 0.12915011169443 & 0 \\ 0 & 0 & 0.081592829638985 \\ \end{bmatrix} \]\[ \mathbf{D}^{-\frac12}_2 = \begin{bmatrix}0.074293646853679 & 0 & 0 \\ 0 & 0.13004578565012 & 0 \\ 0 & 0 & 0.081293334174428 \\ \end{bmatrix} \]And then:
\[ \mathbf{N}_1 = \mathbf{U}^T_1 \mathbf{D}^{-\frac12}_1 \mathbf{U}_1 = \begin{bmatrix}0.11595683432241 & -0.018680153252521 & -0.014185953666511 \\ -0.018680153252521 & 0.087156675810616 & 0.0024128564843152 \\ -0.014185953666511 & 0.0024128564843152 & 0.081333277398589 \\ \end{bmatrix} \]\[ \mathbf{N}_2 = \mathbf{U}^T_2 \mathbf{D}^{-\frac12}_2 \mathbf{U}_2 = \begin{bmatrix}0.11638874319205 & -0.018331272847608 & -0.015307425603646 \\ -0.018331272847608 & 0.086853043710618 & 0.0032625551123438 \\ -0.015307425603646 & 0.0032625551123438 & 0.082390979775559 \\ \end{bmatrix} \]We can now calculate orthogonalized values of unit cell's parameters as:
\[ a^\circ_1 = \mathbf{S}_{1(1,1)} \mathbf{N}_{1(1,1)} + \mathbf{S}_{1(2,1)} \mathbf{N}_{1(2,1)} + \mathbf{S}_{1(3,1)} \mathbf{N}_{1(3,1)} = 9.1196 \ \unicode{x212B} \]\[ b^\circ_1 = \mathbf{S}_{1(1,2)} \mathbf{N}_{1(1,2)} + \mathbf{S}_{1(2,2)} \mathbf{N}_{1(2,2)} + \mathbf{S}_{1(3,2)} \mathbf{N}_{1(3,2)} = 11.8839 \ \unicode{x212B} \]\[ c^\circ_1 = \mathbf{S}_{1(1,3)} \mathbf{N}_{1(1,3)} + \mathbf{S}_{1(2,3)} \mathbf{N}_{1(2,3)} + \mathbf{S}_{1(3,3)} \mathbf{N}_{1(3,3)} = 12.5632 \ \unicode{x212B} \]\[ a^\circ_2 = \mathbf{S}_{2(1,1)} \mathbf{N}_{2(1,1)} + \mathbf{S}_{2(2,1)} \mathbf{N}_{2(2,1)} + \mathbf{S}_{2(3,1)} \mathbf{N}_{2(3,1)} = 9.0974 \ \unicode{x212B} \]\[ b^\circ_2 = \mathbf{S}_{2(1,2)} \mathbf{N}_{2(1,2)} + \mathbf{S}_{2(2,2)} \mathbf{N}_{2(2,2)} + \mathbf{S}_{2(3,2)} \mathbf{N}_{2(3,2)} = 11.9109 \ \unicode{x212B} \]\[ c^\circ_2 = \mathbf{S}_{2(1,3)} \mathbf{N}_{2(1,3)} + \mathbf{S}_{2(2,3)} \mathbf{N}_{2(2,3)} + \mathbf{S}_{2(3,3)} \mathbf{N}_{2(3,3)} = 12.4426 \ \unicode{x212B} \]As \( a^\circ_1 + b^\circ_1 + c^\circ_1 > a^\circ_2 + b^\circ_2 + c^\circ_2 \) we calculate unit cell identity parameter as:
\[ \Pi = \frac{a^\circ_1 + b^\circ_1 + c^\circ_1}{a^\circ_2 + b^\circ_2 + c^\circ_2} - 1 = 0.0035 \]