prove-that-determinant-a-1-a-2-b-c-b-1-b-2-c-a-c-1-c-2-a-b-0-

Question Number 122555 by ZiYangLee last updated on 18/Nov/20

p r o v e t h a t | \begin{matrix} a & 1 & a^{2} (b + c) \\ b & 1 & b^{2} (c + a) \\ c & 1 & c^{2} (a + b) \end{matrix} | = 0

Commented by bramlexs22 last updated on 18/Nov/20

\Rightarrow a | \begin{matrix} 1 b^{2} c + {a b}^{2} \\ 1 {a c}^{2} + {b c}^{2} \end{matrix} | - 1 . | \begin{matrix} b b^{2} c + {a b}^{2} \\ c {a c}^{2} + {b c}^{2} \end{matrix} | + (a^{2} b + a^{2} c) | \begin{matrix} b 1 \\ c 1 \end{matrix} |

Answered by $@y@m last updated on 18/Nov/20

= | \begin{matrix} a & 1 & a^{2} (b + c) \\ b & 1 & b^{2} (c + a) \\ c & 1 & c^{2} (a + b) \end{matrix} |

= | \begin{matrix} a - b & 0 & a^{2} (b + c) - b^{2} (c + a) \\ b - c & 0 & b^{2} (c + a) - c^{2} (a + b) \\ c & 1 & c^{2} (a + b) \end{matrix} |, {b y R}_{1} \to R_{1} - R_{2} & R_{2} \to R_{2} - R_{3}

= | \begin{matrix} a - b & 0 & a b (a - b) + c (a^{2} - b^{2}) \\ b - c & 0 & b c (b - c) + a (b^{2} - c^{2}) \\ 1 & 1 & c^{2} (a + b) \end{matrix} |

= (a - b) (b - c) | \begin{matrix} 1 & 0 & a b + c (a + b) \\ 1 & 0 & b c + a (b + c) \\ 1 & - c & c^{2} (a + b) \end{matrix} |

= 0 (∵ R_{1} = R_{2})

Commented by ZiYangLee last updated on 18/Nov/20

Wow now i understand! thank sir \dots .

Answered by MJS_new last updated on 18/Nov/20

{a c}^{2} (a + b) + a^{2} b (b + c) + b^{2} c (c + a) -

- ({a b}^{2} (c + a) + {b c}^{2} (a + b) + a^{2} c (b + c)) =

= a^{2} c^{2} + {a b c}^{2} + a^{2} b^{2} + a^{2} b c + b^{2} c^{2} + {a b}^{2} c -

- ({a b}^{2} c + a^{2} b^{2} + {a b c}^{2} + b^{2} c^{2} + a^{2} b c + a^{2} c^{2}) = 0

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