Check sibling questions

Ex 4.2, 7 - Show |-a2 ab ac ba -b2 bc ca cb -c2| = 4a2b2b2

Ex 4.2, 7 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 7 - Chapter 4 Class 12 Determinants - Part 3


Transcript

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.