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$$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$$ $$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$$

$$k(a_{11}(a_{33}a_{22}-a_{23}a_{32})+a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{22}a_{31}-a_{21}a_{32})) = k \det \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = 1. $$

Does someone know how to obtain the last equation combining the first and second ones?

Bumblebee
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Riobaldo Tatarana
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1 Answers1

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The trick is to express $a_{1i}^2$ as $a_{1i}.k(...)$

$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$

$a_{11}^2=a_{11}k({a_{33}a_{22}-a_{23}a_{32}})$

$a_{12}^2=a_{12}k({a_{21}a_{33}-a_{31}a_{23}})$

$a_{13^2}=a_{13}k({a_{22}a_{31}-a_{21}a_{32}})$

$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$

$a_{11}^2 + a_{12}^2 + a_{13}^2$

$ = k(a_{11}({a_{33}a_{22}-a_{23}a_{32}})$$+a_{12}({a_{21}a_{33}-a_{31}a_{23}})+a_{13}({a_{22}a_{31}-a_{21}a_{32}}))$

=k $\det\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$=1

Star Bright
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