LinAlg4.2.24

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If {\begin{vmatrix}a&a^{2}&1+a^{3}\\b&b^{2}&1+b^{3}\\c&c^{2}&1+c^{3}\end{vmatrix}}=0\, and the vectors {\bar  {A}}=(1,a,a^{2}),{\bar  {B}}=(1,b,b^{2}),{\bar  {C}}=(1,c,c^{2})\, are non-coplanar,then prove that abc=-1\,

Given {\bar  {A}}=(1,a,a^{2}),{\bar  {B}}=(1,b,b^{2}),{\bar  {C}}=(1,c,c^{2})\, are non-coplanar

[ABC]\not \equiv 0\, implies {\begin{vmatrix}1&a&a^{2}\\1&b&b^{2}\\1&c&c\end{vmatrix}}\not \equiv 0\,

Also given that {\begin{vmatrix}a&a^{2}&1+a^{3}\\b&b^{2}&1+b^{3}\\c&c^{2}&1+c^{3}\end{vmatrix}}=0\, implies {\begin{vmatrix}a&a^{2}&1\\b&b^{2}&1\\c&c^{2}&1\end{vmatrix}}+{\begin{vmatrix}a&a^{2}&a^{3}\\b&b^{2}&b^{3}\\c&c^{2}&c^{3}\end{vmatrix}}=0\,

{\begin{vmatrix}1&a&a^{2}\\1&b&b^{2}\\1&c&c^{2}\end{vmatrix}}+abc{\begin{vmatrix}1&a&a^{2}\\1&b&b^{2}\\1&c&c^{2}\end{vmatrix}}=0\,

By applying C_{1}\leftrightarrow \ C_{2},C_{2}\leftrightarrow \ C_{3}\, in the first,also taking out a,b,c common from rows one,two and three respectively in the second determinant,we get

(1+abc){\begin{vmatrix}1&a&a^{2}\\1&b&b^{2}\\1&c&c^{2}\end{vmatrix}}=0\, which is equal to 1+abc=0\,(since from the given,the determinant in this is equal to zero).

Hence abc=-1\,


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