Question

Using Cofactors of elements of third column, evaluate

Answer 1

Given determinant is,

| 1 x yz 1 y zx 1 z xy |

Since,

Δ= a 13 A 13 + a 23 A 23 + a 33 A 23

Minor of a 13 is,

M 13 =| 1 y 1 z | =z−y

Minor of a 23 is,

M 23 =| 1 x 1 z | =z−x

Minor of a 33 is,

M 33 =| 1 x 1 y | =y−x

Cofactor of a 13 is,

A 13 = ( −1 ) 1+3 ( z−y ) =z−y

Cofactor of a 23 is,

A 23 = ( −1 ) 2+3 ( z−x ) =x−z

Cofactor of a 33 is,

A 33 = ( −1 ) 3+3 ( y−x ) =y−x

The value of determinant is,

Δ= a 13 A 13 + a 23 A 23 + a 33 A 33 =yz( z−y )+zx( x−z )+xy( y−x ) =y z 2 − y 2 z+z x 2 − z 2 x+x y 2 − x 2 y =z( x 2 − y 2 )+ z 2 ( y−x )+xy( y−x )

Simplify further,

Δ=z( x−y )( x+y )+ z 2 ( y−x )+xy( y−x ) =( x−y )( zx+zy− z 2 −xy ) =( x−y )[ z( x−z )+y( z−x ) ] =( x−y )( y−z )( z−x )

Therefore, the value of the given determinant is ( x−y )( y−z )( z−x ).