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

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  1. Class 12
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Ex 4.2, 7 By using properties of determinants, show that: −a2﷮ab﷮ac﷮ba﷮−b2﷮bc﷮ca﷮cb﷮−c2﷯﷯ = 4a2b2c2 Taking L.H.S −a2﷮ab﷮ac﷮ba﷮−b2﷮bc﷮ca﷮cb﷮−c2﷯﷯ Taking a common from R1, b common from R2 , c common from R3 = abc −a﷮b﷮c﷮a﷮−b﷮c﷮a﷮b﷮−c﷯﷯ Taking a common from C1, b common from C2 , c common from C3 = abc (abc) −1﷮1﷮1﷮1﷮−1﷮1﷮1﷮1﷮−1﷯﷯ Applying C2 → C2 + C1 = (abc)2 −1﷮1−1﷮1﷮1﷮−1+1﷮1﷮1﷮1+1﷮−1﷯﷯ = (abc)2 −1﷮0﷮1﷮1﷮0﷮1﷮1﷮2﷮−1﷯﷯ Applying C3 → C3 + C1 = (abc)2 −1﷮0﷮1−1﷮1﷮0﷮1+1﷮1﷮2﷮−1+1﷯﷯ = (abc)2 −1﷮0﷮0﷮1﷮0﷮2﷮1﷮2﷮0﷯﷯ = (abc)2 −1 0﷮2﷮2﷮0﷯﷯−0 1﷮2﷮1﷮0﷯﷯+0 1﷮0﷮1﷮2﷯﷯﷯ = (abc)2 −1 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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