Question 15
Using properties of determinants, prove that:
|a (b - c) (c + b) (a + c) b (c - a) (a - b) (b + a) c| = (a + b + c) (a 2 + b 2 + c 2 )




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Last updated at Oct. 1, 2019 by Teachoo
Question 15
Using properties of determinants, prove that:
|a (b - c) (c + b) (a + c) b (c - a) (a - b) (b + a) c| = (a + b + c) (a 2 + b 2 + c 2 )
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Transcript
Question 15 Using properties of determinants, prove that: |โ 8(๐&๐โ๐&๐+๐@๐+๐&๐&๐โ๐@๐โ๐&๐+๐&๐)| = (๐+๐+๐) (๐^2+๐^2+๐^2) Solving LHS |โ 8(๐&๐โ๐&๐+๐@๐+๐&๐&๐โ๐@๐โ๐&๐+๐&๐)| Since we cannot solve by adding or subtracting rows We multiply a with C1, b with C2 and c with C3 = ๐๐๐/๐๐๐ |โ 8(๐&๐โ๐&๐+๐@๐+๐&๐&๐โ๐@๐โ๐&๐+๐&๐)| = 1/๐๐๐ |โ 8(๐(๐)&๐(๐โ๐)&๐(๐+๐)@๐(๐+๐)&๐(๐)&๐(๐โ๐)@๐(๐โ๐)&๐(๐+๐)&๐(๐))| = 1/๐๐๐ |โ 8(๐^2&๐^2โ๐๐&๐^2+๐๐@๐^2+๐๐&๐^2&๐^2โ๐๐@๐^2โ๐๐&๐^2+๐๐&๐^2 )| C1 โ C1 + C2 + C3 = 1/๐๐๐ |โ 8(๐^2+๐^2โ๐๐+๐^2+๐๐&๐^2โ๐๐&๐^2+๐๐@๐^2+๐๐+๐^2+๐^2โ๐๐&๐^2&๐^2โ๐๐@๐^2โ๐๐+๐^2+๐๐+๐^2&๐^2+๐๐&๐^2 )| = 1/๐๐๐ |โ 8(๐^2+๐^2+๐^2&๐^2โ๐๐&๐^2+๐๐@๐^2+๐^2+๐^2&๐^2&๐^2โ๐๐@๐^2+๐^2+๐^2&๐^2+๐๐&๐^2 )| Taking ๐^2+๐^2+๐^2 common from C1 = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ |โ 8(1&๐^2โ๐๐&๐^2+๐๐@1&๐^2&๐^2โ๐๐@1&๐^2+๐๐&๐^2 )| R2 โ R2 โ R1 = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ |โ 8(1&๐^2โ๐๐&๐^2+๐๐@1โ1&๐^2โ(๐^2โ๐๐)&๐^2โ๐๐โ(๐^2+๐๐)@1&๐^2+๐๐&๐^2 )| = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ |โ 8(1&๐^2โ๐๐&๐^2+๐๐@0&๐๐&โ๐๐โ๐๐@1&๐^2+๐๐&๐^2 )| R3 โ R3 โ R1 = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ |โ 8(1&๐^2โ๐๐&๐^2+๐๐@0&๐๐&โ๐๐โ๐๐@1โ1&๐^2+๐๐โ(๐^2โ๐๐)&๐^2โ(๐^2+๐๐))| = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ |โ 8(1&๐^2โ๐๐&๐^2+๐๐@0&๐๐&โ๐๐โ๐๐@0&๐๐+๐๐&โ๐๐)| = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ |โ 8(1&๐(๐โ๐)&๐(๐+๐)@0&๐๐&โ๐(๐+๐)@0&๐(๐+๐)&โ๐๐)| Taking b common from C2, and c common from C3 = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐๐๐ร๐๐|โ 8(1&(๐โ๐)&(๐+๐)@0&๐&โ(๐+๐)@0&(๐+๐)&โ๐)| = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ |โ 8(1&(๐โ๐)&(๐+๐)@0&๐&โ(๐+๐)@0&(๐+๐)&โ๐)| = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ |โ 8(1&(๐โ๐)&(๐+๐)@0&๐&โ(๐+๐)@0&(๐+๐)&โ๐)| Finding determinant along C1 = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ ร [1 (c ร (-b) โ (โ(a + b))(a + c) + 0 + 0] = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ ร [โbc + (a + b)(a + c)] = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ ร [โbc + a(a + c) + b(a + c)] = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ ร [โbc + a2 + ac + ab + bc] = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ ร [a2 + ac + ab] = ((๐^2 + ๐^2+ ใ ๐ใ^2))/๐ ร a[a + c + b] = (๐+๐+๐) (๐^2+๐^2+๐^2)
CBSE Class 12 Sample Paper for 2019 Boards
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