Question

Evaluate

Answer 1

The determinant is given as,

Δ  =  | x  y x+y y x+y x x+y x y |

Apply the row operation R 1 →  R 1 + R 2 + R 3 in the above determinant,

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

Apply the column operation C 2 → C 2 − C 1 in the above equation,

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

Apply the column operation C 3 → C 3 − C 1 in the above equation,

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

Further simplify the determinant,

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

Δ=2( x+y )( x 2 + y 2 −xy ) =−2( x 3 + y 3 )

Thus, the determinant is −2( x 3 + y 3 ).