KCET2017MathematicsIndefinite Integration
Let ( = | array ccc A x & x² & 1 B y & y² & 1 C z & z² & 1 array | ) and ( ₁= | array cccc A & B & C & x & y & z & z y & z x & x y & array | ) then
Options
- A( ₁=- )
- B( ₁= )
- C( ₁ )
- D( ₁=2 )
Correct answer
B. ( ₁= )
Step-by-step solution
Given that, ( = | array ccc A x & x² & 1 B y & y² & 1 C z & z² & 1 array | (1) ) [ and ₁= | array ccc A & B & C x & y & z z y & z x & x y array | (2) ] Take out common ( x, y ) and ( z ) in Eq. (1) from ( R₁, R₂ ) and ( R₃ ) respectively we get ( =x y z | array ccc A & x & 1 x B & y & 1 y C & z & 1 z array | ) ( C₃ x y z C₃ ), we get ( = | array ccc A & x & y z B & y & z x C & z & x y array | ) ( = | array ccc A & B & C x & y & z y z & z x & x y array |= ₁ ) Therefore, ( = ₁ )