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Question

Using properties of determinants, prove that:

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Solution

The determinant is given as,

Δ=| x x 2 1+p x 3 y y 2 1+p y 3 z z 2 1+p z 3 |

To prove,

| x x 2 1+p x 3 y y 2 1+p y 3 z z 2 1+p z 3 |=( 1+pxyz )( xy )( yz )( zx )

Simplify the left hand side of the above equation,

| x x 2 1+p x 3 y y 2 1+p y 3 z z 2 1+p z 3 |

Further we can write from the property of sum of two elements in the

determinant,

Δ=| x x 2 1 y y 2 1 z z 2 1 |+| x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 |

Further we can take out p from the column C 3 ,

Δ=| x x 2 1 y y 2 1 z z 2 1 |+p| x x 2 x 3 y y 2 y 3 z z 2 z 3 |

Further we can take out x, y and z from the rows R 1 , R 2 and R 3 respectively,

Δ=| x x 2 1 y y 2 1 z z 2 1 |+pxyz| 1 x x 2 1 y y 2 1 z z 2 |

Further apply the column transformation C 1 C 2 in the above determinant,

Δ=| x x 2 1 y y 2 1 z z 2 1 |pxyz| x 1 x 2 y 1 y 2 z 1 z 2 |

Further apply the column transformation C 2 C 3 in the above determinant,

Δ=| x x 2 1 y y 2 1 z z 2 1 |( 1 )pxyz| x x 2 1 y y 2 1 z z 2 1 | =| x x 2 1 y y 2 1 z z 2 1 |+pxyz| x x 2 1 y y 2 1 z z 2 1 |

Further we can take out | x x 2 1 y y 2 1 z z 2 1 | from the above determinant,

( 1+pxyz )| x x 2 1 y y 2 1 z z 2 1 |

Further apply the row transformation R 1 R 1 R 2 in the above determinant

Δ=( 1+pxyz )| xy x 2 y 2 11 y y 2 1 z z 2 1 | =( 1+pxyz )| xy ( xy )( x+y ) 0 y y 2 1 z z 2 1 |

Further we can take out ( xy ) from the row R 1 ,

Δ=( 1+pxyz )( xy )| 1 ( x+y ) 0 y y 2 1 z z 2 1 |

Further apply the row transformation R 2 R 2 R 3 in the above determinant,

Δ=( 1+pxyz )( xy )| 1 ( x+y ) 0 yz ( yz )( y+z ) 11 z z 2 1 | =( 1+pxyz )( xy )| 1 ( x+y ) 0 yz ( yz )( y+z ) 0 z z 2 1 |

Further we can take out ( xy ) from the row R 1 ,

Δ=( 1+pxyz )( xy )( yz )| 1 ( x+y ) 0 1 ( y+z ) 0 z z 2 1 |

Further apply the row transformation R 1 R 1 R 2 in the above determinant,

Δ=( 1+pxyz )( xy )( yz )| 11 ( x+yyz ) 0 1 ( y+z ) 0 z z 2 1 | =( 1+pxyz )( xy )( yz )| 0 ( xz ) 0 1 ( y+z ) 0 z z 2 1 |

We have to expand the determinant along C 3 ,

Δ=( 1+pxyz )( xy )( yz )( 0| 1 y+z z z 2 |0| 0 xz z z 2 |+1| 0 xy 1 y+z | ) =( 1+pxyz )( xy )( yz )( zx ) =( 1+pxyz )( xy )( yz )( zx )

Hence, it is proved that | x x 2 1+p x 3 y y 2 1+p y 3 z z 2 1+p z 3 |=( 1+pxyz )( xy )( yz )( zx ).


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