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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 )

Using properties of determinant, prove that (a + b + c) (a2 + b2 + c2)

Question 15 - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 15 - CBSE Class 12 Sample Paper for 2019 Boards - Part 3
Question 15 - CBSE Class 12 Sample Paper for 2019 Boards - Part 4
Question 15 - CBSE Class 12 Sample Paper for 2019 Boards - Part 5

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Transcript

Question 15 Using properties of determinants, prove that: |β– 8(π‘Ž&π‘βˆ’π‘&𝑐+𝑏@π‘Ž+𝑐&𝑏&π‘βˆ’π‘Ž@π‘Žβˆ’π‘&𝑏+π‘Ž&𝑐)| = (π‘Ž+𝑏+𝑐) (π‘Ž^2+𝑏^2+𝑐^2) Solving LHS |β– 8(π‘Ž&π‘βˆ’π‘&𝑐+𝑏@π‘Ž+𝑐&𝑏&π‘βˆ’π‘Ž@π‘Žβˆ’π‘&𝑏+π‘Ž&𝑐)| Since we cannot solve by adding or subtracting rows We multiply a with C1, b with C2 and c with C3 = π‘Žπ‘π‘/π‘Žπ‘π‘ |β– 8(π‘Ž&π‘βˆ’π‘&𝑐+𝑏@π‘Ž+𝑐&𝑏&π‘βˆ’π‘Ž@π‘Žβˆ’π‘&𝑏+π‘Ž&𝑐)| = 1/π‘Žπ‘π‘ |β– 8(π‘Ž(π‘Ž)&𝑏(π‘βˆ’π‘)&𝑐(𝑐+𝑏)@π‘Ž(π‘Ž+𝑐)&𝑏(𝑏)&𝑐(π‘βˆ’π‘Ž)@π‘Ž(π‘Žβˆ’π‘)&𝑏(𝑏+π‘Ž)&𝑐(𝑐))| = 1/π‘Žπ‘π‘ |β– 8(π‘Ž^2&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@π‘Ž^2+π‘Žπ‘&𝑏^2&𝑐^2βˆ’π‘Žπ‘@π‘Ž^2βˆ’π‘Žπ‘&𝑏^2+π‘Žπ‘&𝑐^2 )| C1 β†’ C1 + C2 + C3 = 1/π‘Žπ‘π‘ |β– 8(π‘Ž^2+𝑏^2βˆ’π‘π‘+𝑐^2+𝑏𝑐&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@π‘Ž^2+π‘Žπ‘+𝑏^2+𝑐^2βˆ’π‘Žπ‘&𝑏^2&𝑐^2βˆ’π‘Žπ‘@π‘Ž^2βˆ’π‘Žπ‘+𝑏^2+π‘Žπ‘+𝑐^2&𝑏^2+π‘Žπ‘&𝑐^2 )| = 1/π‘Žπ‘π‘ |β– 8(π‘Ž^2+𝑏^2+𝑐^2&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@π‘Ž^2+𝑏^2+𝑐^2&𝑏^2&𝑐^2βˆ’π‘Žπ‘@π‘Ž^2+𝑏^2+𝑐^2&𝑏^2+π‘Žπ‘&𝑐^2 )| Taking π‘Ž^2+𝑏^2+𝑐^2 common from C1 = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘ |β– 8(1&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@1&𝑏^2&𝑐^2βˆ’π‘Žπ‘@1&𝑏^2+π‘Žπ‘&𝑐^2 )| R2 β†’ R2 – R1 = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘ |β– 8(1&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@1βˆ’1&𝑏^2βˆ’(𝑏^2βˆ’π‘π‘)&𝑐^2βˆ’π‘Žπ‘βˆ’(𝑐^2+𝑏𝑐)@1&𝑏^2+π‘Žπ‘&𝑐^2 )| = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘ |β– 8(1&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@0&𝑏𝑐&βˆ’π‘Žπ‘βˆ’π‘π‘@1&𝑏^2+π‘Žπ‘&𝑐^2 )| R3 β†’ R3 – R1 = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘ |β– 8(1&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@0&𝑏𝑐&βˆ’π‘Žπ‘βˆ’π‘π‘@1βˆ’1&𝑏^2+π‘Žπ‘βˆ’(𝑏^2βˆ’π‘π‘)&𝑐^2βˆ’(𝑐^2+𝑏𝑐))| = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘ |β– 8(1&𝑏^2βˆ’π‘π‘&𝑐^2+𝑏𝑐@0&𝑏𝑐&βˆ’π‘Žπ‘βˆ’π‘π‘@0&π‘Žπ‘+𝑏𝑐&βˆ’π‘π‘)| = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘ |β– 8(1&𝑏(π‘βˆ’π‘)&𝑐(𝑐+𝑏)@0&𝑏𝑐&βˆ’π‘(π‘Ž+𝑏)@0&𝑏(π‘Ž+𝑐)&βˆ’π‘π‘)| Taking b common from C2, and c common from C3 = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Žπ‘π‘Γ—π‘π‘|β– 8(1&(π‘βˆ’π‘)&(𝑐+𝑏)@0&𝑐&βˆ’(π‘Ž+𝑏)@0&(π‘Ž+𝑐)&βˆ’π‘)| = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž |β– 8(1&(π‘βˆ’π‘)&(𝑐+𝑏)@0&𝑐&βˆ’(π‘Ž+𝑏)@0&(π‘Ž+𝑐)&βˆ’π‘)| = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž |β– 8(1&(π‘βˆ’π‘)&(𝑐+𝑏)@0&𝑐&βˆ’(π‘Ž+𝑏)@0&(π‘Ž+𝑐)&βˆ’π‘)| Finding determinant along C1 = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž Γ— [1 (c Γ— (-b) – (–(a + b))(a + c) + 0 + 0] = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž Γ— [–bc + (a + b)(a + c)] = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž Γ— [–bc + a(a + c) + b(a + c)] = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž Γ— [–bc + a2 + ac + ab + bc] = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž Γ— [a2 + ac + ab] = ((π‘Ž^2 + 𝑏^2+ γ€– 𝑐〗^2))/π‘Ž Γ— a[a + c + b] = (π‘Ž+𝑏+𝑐) (π‘Ž^2+𝑏^2+𝑐^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 12 years. He provides courses for Maths and Science at Teachoo.