Question

Using properties of determinants, prove that:

Answer 1

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 )( x−y )( y−z )( z−x )

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

Further we can take out ( x−y ) from the row R 1 ,

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

Further we can take out ( x−y ) from the row R 1 ,

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

We have to expand the determinant along C 3 ,

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

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 )( x−y )( y−z )( z−x ).