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Question Number 13 by user1 last updated on 25/Jan/15
Expand the determnent △= determinant ((a,h,g),(h,b,f),(g,f,c))
$$\mathrm{Expand}\:\mathrm{the}\:\mathrm{determnent}\: \\ $$$$\:\:\bigtriangleup=\begin{vmatrix}{{a}}&{{h}}&{{g}}\\{{h}}&{{b}}&{{f}}\\{{g}}&{{f}}&{{c}}\end{vmatrix} \\ $$
Answered by user1 last updated on 30/Oct/14
Expanding by 1^(st) row, we have: △=a∙ determinant ((b,f),(f,c))−h∙ determinant ((h,f),(g,c))+g∙ determinant ((h,b),(g,f)) =a(bc−f^2 )−h(ch−fg)+g(fh−bg) =(abc+2fgh−af^2 −bg^2 −ch^2 )
$$\mathrm{Expanding}\:\mathrm{by}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{row},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\bigtriangleup={a}\centerdot\begin{vmatrix}{{b}}&{{f}}\\{{f}}&{{c}}\end{vmatrix}−{h}\centerdot\begin{vmatrix}{{h}}&{{f}}\\{{g}}&{{c}}\end{vmatrix}+{g}\centerdot\begin{vmatrix}{{h}}&{{b}}\\{{g}}&{{f}}\end{vmatrix} \\ $$$$\:\:\:\:\:={a}\left({bc}−{f}^{\mathrm{2}} \right)−{h}\left({ch}−{fg}\right)+{g}\left({fh}−{bg}\right) \\ $$$$\:\:\:\:=\left({abc}+\mathrm{2}{fgh}−{af}^{\mathrm{2}} −{bg}^{\mathrm{2}} −{ch}^{\mathrm{2}} \right) \\ $$

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