Question

Prove that

Answer 1

The matrix of the order 3×3 is,

| a 2 bc ac+ c 2 a 2 +ab b 2 ac ab b 2 +bc c 2 |

It has to be proved that,

| a 2 bc ac+ c 2 a 2 +ab b 2 ac ab b 2 +bc c 2 |=4 a 2 b 2 c 2

Simplify the left hand side of the above equation.

| a×a b×c c( a+c ) a( a+b ) b×b a×c a×b b( b+c ) c×c |

The terms a, b and c can be taken out from the columns C 1 , C 2 and C 3 respectively.

| a c ( a+c ) ( a+b ) b a b ( b+c ) c |

Further apply the column transformation C 3 → C 3 − C 1 in the above determinant.

Δ = 2abc  | a+c c ( a+c )−( a+c ) ( a+b ) b a−( a+b ) ( b+c ) ( b+c ) c−( b+c ) | = 2abc  | a+c c 0 ( a+b ) b −b ( b+c ) ( b+c ) −b |

Further apply the column transformation C 2 → C 2 − C 1 in the above determinant.

Δ =2abc | ( a+c ) −a 0 ( a+b ) −a −b ( b+c ) 0 −b |

Further, −a and −b can be taken out from the columns C 2 and C 3 respectively.

Δ = 2abc( −a )( −b )  | ( a+c ) 1 0 ( a+b ) 1 1 ( b+c ) 0 1 |

Expand the determinant along the row R 1 .

Δ  = 2 a 2 b 2 c( ( a+c )| 1 1 0 1 |−1| a+b 1 b+c 1 |+0| a+b 1 b+c 0 | ) = 2 a 2 b 2 c( ( a+c )| 1 1 0 1 |−1| a+b 1 b+c 1 |+0 ) =2 a 2 b 2 c( ( a+c )−1( a+b−b−c ) ) =2 a 2 b 2 c( 2c )

Δ=4 a 2 b 2 c 2

Hence, it is proved that | a 2 bc ac+ c 2 a 2 +ab b 2 ac ab b 2 +bc c 2 |=4 a 2 b 2 c 2 .