Question

without expandind the determinant show that [1,bc,b+c],[1,ca,c+a],[1,ab,a+b]|=|[1,a,a^2],[1,b,b^2],[1,c,c^2]|`

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

|. 1. bc. b+c. |

|. 1. ca. c+a. |

|. 1 ab. a+b.|

multiplying the first row by a, second by b and third by c

1/abc*. |. a. abc. ab+ac. |

|. b. abc. bc+ab. |

|. c. abc. ac+ab. |

taking abc out from column 2

|. a. 1. ab+ac. |

|. b. 1. bc+ab. |

|. c. 1. ac+ab. |

|. a. 1. ab+bc+ac - bc. |

|. b. 1. ab+bc+ac - ac. |

|. c. 1. ab+bc+ac - ab. |

|. a. 1. ab+bc+ac |. |. a. 1. bc. |

|. b. 1 ab+bc+ac | - |. b. 1. ac |

|. c. 1. ab+bc+ac.|. |. c. 1. ab.|

after taking (ab+bc+ac) out from third column of first determinant, second and third columns will become same and the value of first determinant will be zero. thus, second determinant will remain. now multiplying first row of left determinant by a, second by b and third by c, we have

|. a^2. a abc. |

-1/(abc) |. b^2. b. abc. |

|. c^2. c. abc. |

taking abc out from third column and cancelling we have,

|. a^2. a. 1. |

-1 |. b^2. b. 1. |

|. c^2. c. 1. |

switching first and third column ( this will create a negative sign effect which will cancel the already existing negative sign . thus

|. 1. a. a^2. |

|. 1. b. b^2. |

|. 1. c. c^2. |

proved