Доказать тождество : (с,(a-2c),(2a+b))=(a,b,c) векторы

$(\overline{a}, \overline{b}, \overline{c}) = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} = a_x \begin{vmatrix} b_y & b_z \\ c_y & c_z\end{vmatrix} - a_y \begin{vmatrix} b_x & b_z \\ c_x & c_z\end{vmatrix} + a_z \begin{vmatrix} b_x & b_y \\ c_x & c_y\end{vmatrix} =\\= a_x(b_yc_z - b_zc_y) - a_y(b_xc_z-b_zc_x) + a_z(b_xc_y - b_yc_x)$

$(\overline{c}, \overline{a - 2c}, \overline{2a + b}) = \begin{vmatrix} c_x & c_y & c_z \\ a_x - 2c_x &a_y - 2c_y & a_z - 2c_z \\ 2a_x + b_x & 2a_y + b_y & 2a_z + b_z \end{vmatrix} =\\= \begin{vmatrix} c_x & c_y & c_z \\ a_x&a_y&a_z\\ 2a_x + b_x & 2a_y + b_y & 2a_z + b_z \end{vmatrix} = \begin{vmatrix} c_x & c_y & c_z \\ a_x&a_y&a_z\\ b_x & b_y & b_z \end{vmatrix} = -\begin{vmatrix}a_x&a_y& a_z \\ c_x & c_y & c_z\\ b_x & b_y & b_z \end{vmatrix} =$

$-(-\begin{vmatrix}a_x&a_y&a_z \\ b_x & b_y & b_z\\c_x & c_y & c_z \end{vmatrix}) = \begin{vmatrix}a_x&a_y& a_z \\ b_x & b_y & b_z\\c_x & c_y & c_z \end{vmatrix} =\\= a_x(b_yc_z - b_zc_y) - a_y(b_xc_z-b_zc_x) + a_z(b_xc_y - b_yc_x)$

Очевидно, тождество верно