Properties of Determinant

Chapter 4 Class 12 Determinants
Serial order wise

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### Transcript

Question 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