NCERT Classes
Previous-Year-Papers
Entrance Exams
State Boards
Home > Determinants
Using properties of determinants show the following:
b + c 2 ab caab a + c 2 bcac bc a + b 2 = 2abc ( a + b + c )3
Consider,
∆ = ( b + c )2 ab acab ( a + c )2 bcac bc ( a + b )2By performing R1 → a R1, R2 → b R2, R3 → c R3 and dividing the determinant by abc, we get∆ = 1abc a ( b + c )2 a2b a2cab2 b ( a + c )2 b2cac2 bc2 c ( a + b )2
Now, taking a, b, c common from C1, C2, and C3
∆ = abcabc ( b + c )2 a2 a2b2 ( a + c )2 b2c2 c2 ( a + b )2⇒ ∆ = ( b + c )2 a2 a2b2 ( c + a )2 b2c2 c2 ( a + b )2Applying C1 → C1 - C2, C2 → C2 - C3⇒ ∆ = ( b + c )2 - a2 0 a2 b2 -( c + a )2 ( c + a )2 - b2 b20 c2 - ( a + b )2 ( a + b )2∆ = a + b + c 2 b + c - a 0 a2 b - c - a c + a - b b20 c - a - b ( a + b )2
Applying R3 → R3 - ( R1 + R2 )∆ = a + b + c 2 b + c -a 0 a2b - c -a c + a - b b2 2a - 2b-2a 2ab
Applying C1 → C1 + C2∆ = a + b + c 2 b + c - a 0 a20 c + a - b b2-2b -2a 2ab Applying C3 → C3 + bC2∆ = a + b + c 2 b + c - a 0 a20 c + a - b bc + ab-2b -2a 0
Applying C1 → aC1 and C2 → bC2∆ = a + b + c 2ab ab + ac - a20 a20 bc + ab - b2 bc + ab-2ab -2ab 0Applying C1 → C1 - C2∆ = a + b + c 2ab ab + ac - a20 a2-bc - ab + b2 bc + ab - b2 bc + ab0 -2ab 0
Expanding along R3
= a + b + c 2ab 2ab ab2c + a2b2 + abc2 + a2bc - a2bc - a3b + a2bc + a3b - a2b2 = 2 ( a + b + c )2 ( ab2c + abc2 + a2bc )= 2 ( a + b + c )3 abc= 2 abc ( a + b + c )3 = R.H.S.
Mock Test Series
