# Ex 4.2, 7 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 22, 2020 by Teachoo

Ex 4.2

Ex 4.2, 1
Deleted for CBSE Board 2023 Exams

Ex 4.2, 2 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 3 Deleted for CBSE Board 2023 Exams

Ex 4.2, 4 Deleted for CBSE Board 2023 Exams

Ex 4.2, 5 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 7 Important Deleted for CBSE Board 2023 Exams You are here

Ex 4.2, 8 (i) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 8 (ii) Deleted for CBSE Board 2023 Exams

Ex 4.2, 9 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 10 (i) Deleted for CBSE Board 2023 Exams

Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 11 (i) Deleted for CBSE Board 2023 Exams

Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 12 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 13 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2023 Exams

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at Jan. 22, 2020 by Teachoo

Ex 4.2, 7 By using properties of determinants, show that: |■8(−a2&ab&[email protected]&−b2&[email protected]&cb&−c2)| = 4a2b2c2 Solving L.H.S |■8(−a2&ab&[email protected]&−b2&[email protected]&cb&−c2)| Taking a common from R1, b common from R2 , c common from R3 = abc |■8(−a&b&[email protected]&−b&[email protected]&b&−c)| Taking a common from C1, b common from C2 , c common from C3 = abc (abc) |■8(−1&1&[email protected]&−1&[email protected]&1&−1)| Applying C2 → C2 + C1 = (abc)2 |■8(−1&1−1&[email protected]&−1+1&[email protected]&1+1&−1)| = (abc)2 |■8(−1&0&[email protected]&0&[email protected]&2&−1)| Applying C3 → C3 + C1 = (abc)2 |■8(−1&0&1−[email protected]&0&[email protected]&2&−1+1)| = (abc)2 |■8(−1&0&[email protected]&0&[email protected]&2&0)| = (abc)2 (−1|■8(0&[email protected]&0)|−0|■8(1&[email protected]&0)|+0|■8(1&[email protected]&2)|) = (abc)2 (−1|■8(0&[email protected]&0)|−0+0) = (abc)2 ( – 1(0(0) – 2(2))) = (abc)2 (4) = 4 (abc)2 = 4a2b2c2 = R.H.S Hence proved