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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Question 15 Using properties of determinants, prove that: |โ– 8(๐‘Ž&๐‘โˆ’๐‘&๐‘+๐‘@๐‘Ž+๐‘&๐‘&๐‘โˆ’๐‘Ž@๐‘Žโˆ’๐‘&๐‘+๐‘Ž&๐‘)| = (๐‘Ž+๐‘+๐‘) (๐‘Ž^2+๐‘^2+๐‘^2) Solving LHS |โ– 8(๐‘Ž&๐‘โˆ’๐‘&๐‘+๐‘@๐‘Ž+๐‘&๐‘&๐‘โˆ’๐‘Ž@๐‘Žโˆ’๐‘&๐‘+๐‘Ž&๐‘)| Since we cannot solve by adding or subtracting rows We multiply a with C1, b with C2 and c with C3 = ๐‘Ž๐‘๐‘/๐‘Ž๐‘๐‘ |โ– 8(๐‘Ž&๐‘โˆ’๐‘&๐‘+๐‘@๐‘Ž+๐‘&๐‘&๐‘โˆ’๐‘Ž@๐‘Žโˆ’๐‘&๐‘+๐‘Ž&๐‘)| = 1/๐‘Ž๐‘๐‘ |โ– 8(๐‘Ž(๐‘Ž)&๐‘(๐‘โˆ’๐‘)&๐‘(๐‘+๐‘)@๐‘Ž(๐‘Ž+๐‘)&๐‘(๐‘)&๐‘(๐‘โˆ’๐‘Ž)@๐‘Ž(๐‘Žโˆ’๐‘)&๐‘(๐‘+๐‘Ž)&๐‘(๐‘))| = 1/๐‘Ž๐‘๐‘ |โ– 8(๐‘Ž^2&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@๐‘Ž^2+๐‘Ž๐‘&๐‘^2&๐‘^2โˆ’๐‘Ž๐‘@๐‘Ž^2โˆ’๐‘Ž๐‘&๐‘^2+๐‘Ž๐‘&๐‘^2 )| C1 โ†’ C1 + C2 + C3 = 1/๐‘Ž๐‘๐‘ |โ– 8(๐‘Ž^2+๐‘^2โˆ’๐‘๐‘+๐‘^2+๐‘๐‘&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@๐‘Ž^2+๐‘Ž๐‘+๐‘^2+๐‘^2โˆ’๐‘Ž๐‘&๐‘^2&๐‘^2โˆ’๐‘Ž๐‘@๐‘Ž^2โˆ’๐‘Ž๐‘+๐‘^2+๐‘Ž๐‘+๐‘^2&๐‘^2+๐‘Ž๐‘&๐‘^2 )| = 1/๐‘Ž๐‘๐‘ |โ– 8(๐‘Ž^2+๐‘^2+๐‘^2&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@๐‘Ž^2+๐‘^2+๐‘^2&๐‘^2&๐‘^2โˆ’๐‘Ž๐‘@๐‘Ž^2+๐‘^2+๐‘^2&๐‘^2+๐‘Ž๐‘&๐‘^2 )| Taking ๐‘Ž^2+๐‘^2+๐‘^2 common from C1 = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ |โ– 8(1&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@1&๐‘^2&๐‘^2โˆ’๐‘Ž๐‘@1&๐‘^2+๐‘Ž๐‘&๐‘^2 )| R2 โ†’ R2 โ€“ R1 = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ |โ– 8(1&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@1โˆ’1&๐‘^2โˆ’(๐‘^2โˆ’๐‘๐‘)&๐‘^2โˆ’๐‘Ž๐‘โˆ’(๐‘^2+๐‘๐‘)@1&๐‘^2+๐‘Ž๐‘&๐‘^2 )| = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ |โ– 8(1&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@0&๐‘๐‘&โˆ’๐‘Ž๐‘โˆ’๐‘๐‘@1&๐‘^2+๐‘Ž๐‘&๐‘^2 )| R2 โ†’ R2 โ€“ R1 = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ |โ– 8(1&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@0&๐‘๐‘&โˆ’๐‘Ž๐‘โˆ’๐‘๐‘@1โˆ’1&๐‘^2+๐‘Ž๐‘โˆ’(๐‘^2โˆ’๐‘๐‘)&๐‘^2โˆ’(๐‘^2+๐‘๐‘))| = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ |โ– 8(1&๐‘^2โˆ’๐‘๐‘&๐‘^2+๐‘๐‘@0&๐‘๐‘&โˆ’๐‘Ž๐‘โˆ’๐‘๐‘@0&๐‘Ž๐‘+๐‘๐‘&โˆ’๐‘๐‘)| = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ |โ– 8(1&๐‘(๐‘โˆ’๐‘)&๐‘(๐‘+๐‘)@0&๐‘๐‘&โˆ’๐‘(๐‘Ž+๐‘)@0&๐‘(๐‘Ž+๐‘)&โˆ’๐‘๐‘)| Taking b common from C2, and c common from C3 = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž๐‘๐‘ร—๐‘๐‘|โ– 8(1&(๐‘โˆ’๐‘)&(๐‘+๐‘)@0&๐‘&โˆ’(๐‘Ž+๐‘)@0&(๐‘Ž+๐‘)&โˆ’๐‘)| = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž |โ– 8(1&(๐‘โˆ’๐‘)&(๐‘+๐‘)@0&๐‘&โˆ’(๐‘Ž+๐‘)@0&(๐‘Ž+๐‘)&โˆ’๐‘)| = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž |โ– 8(1&(๐‘โˆ’๐‘)&(๐‘+๐‘)@0&๐‘&โˆ’(๐‘Ž+๐‘)@0&(๐‘Ž+๐‘)&โˆ’๐‘)| Finding determinant along C1 = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž ร— [1 (c ร— (-b) โ€“ (โ€“(a + b))(a + c) + 0 + 0] = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž ร— [โ€“bc + (a + b)(a + c)] = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž ร— [โ€“bc + a(a + c) + b(a + c)] = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž ร— [โ€“bc + a2 + ac + ab + bc] = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž ร— [a2 + ac + ab] = ((๐‘Ž^2 + ๐‘^2+ ใ€– ๐‘ใ€—^2))/๐‘Ž ร— a[a + c + b] = (๐‘Ž+๐‘+๐‘) (๐‘Ž^2+๐‘^2+๐‘^2)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.