Question 7 - Chapter 4 Class 12 Determinants (Important Question)

Last updated at April 16, 2024 by

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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