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

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Question 7 By using properties of determinants, show that: |■8(−a2&ab&ac@ba&−b2&bc@ca&cb&−c2)| = 4a2b2c2 Solving L.H.S |■8(−a2&ab&ac@ba&−b2&bc@ca&cb&−c2)| Taking a common from R1, b common from R2 , c common from R3 = abc |■8(−a&b&c@a&−b&c@a&b&−c)| Taking a common from C1, b common from C2 , c common from C3 = abc (abc) |■8(−1&1&1@1&−1&1@1&1&−1)| Applying C2 → C2 + C1 = (abc)2 |■8(−1&1−1&1@1&−1+1&1@1&1+1&−1)| = (abc)2 |■8(−1&0&1@1&0&1@1&2&−1)| Applying C3 → C3 + C1 = (abc)2 |■8(−1&0&1−1@1&0&1+1@1&2&−1+1)| = (abc)2 |■8(−1&0&0@1&0&2@1&2&0)| = (abc)2 (−1|■8(0&2@2&0)|−0|■8(1&2@1&0)|+0|■8(1&0@1&2)|) = (abc)2 (−1|■8(0&2@2&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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.