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: |■8(𝑎&𝑏−𝑐&𝑐+𝑏@𝑎+𝑐&𝑏&𝑐−𝑎@𝑎−𝑏&𝑏+𝑎&𝑐)| = (𝑎+𝑏+𝑐) (𝑎^2+𝑏^2+𝑐^2) Solving LHS |■8(𝑎&𝑏−𝑐&𝑐+𝑏@𝑎+𝑐&𝑏&𝑐−𝑎@𝑎−𝑏&𝑏+𝑎&𝑐)| Since we cannot solve by adding or subtracting rows We multiply a with C1, b with C2 and c with C3 = 𝑎𝑏𝑐/𝑎𝑏𝑐 |■8(𝑎&𝑏−𝑐&𝑐+𝑏@𝑎+𝑐&𝑏&𝑐−𝑎@𝑎−𝑏&𝑏+𝑎&𝑐)| = 1/𝑎𝑏𝑐 |■8(𝑎(𝑎)&𝑏(𝑏−𝑐)&𝑐(𝑐+𝑏)@𝑎(𝑎+𝑐)&𝑏(𝑏)&𝑐(𝑐−𝑎)@𝑎(𝑎−𝑏)&𝑏(𝑏+𝑎)&𝑐(𝑐))| = 1/𝑎𝑏𝑐 |■8(𝑎^2&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@𝑎^2+𝑎𝑐&𝑏^2&𝑐^2−𝑎𝑐@𝑎^2−𝑎𝑏&𝑏^2+𝑎𝑏&𝑐^2 )| C1 → C1 + C2 + C3 = 1/𝑎𝑏𝑐 |■8(𝑎^2+𝑏^2−𝑏𝑐+𝑐^2+𝑏𝑐&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@𝑎^2+𝑎𝑐+𝑏^2+𝑐^2−𝑎𝑐&𝑏^2&𝑐^2−𝑎𝑐@𝑎^2−𝑎𝑏+𝑏^2+𝑎𝑏+𝑐^2&𝑏^2+𝑎𝑏&𝑐^2 )| = 1/𝑎𝑏𝑐 |■8(𝑎^2+𝑏^2+𝑐^2&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@𝑎^2+𝑏^2+𝑐^2&𝑏^2&𝑐^2−𝑎𝑐@𝑎^2+𝑏^2+𝑐^2&𝑏^2+𝑎𝑏&𝑐^2 )| Taking 𝑎^2+𝑏^2+𝑐^2 common from C1 = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐 |■8(1&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@1&𝑏^2&𝑐^2−𝑎𝑐@1&𝑏^2+𝑎𝑏&𝑐^2 )| R2 → R2 – R1 = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐 |■8(1&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@1−1&𝑏^2−(𝑏^2−𝑏𝑐)&𝑐^2−𝑎𝑐−(𝑐^2+𝑏𝑐)@1&𝑏^2+𝑎𝑏&𝑐^2 )| = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐 |■8(1&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@0&𝑏𝑐&−𝑎𝑐−𝑏𝑐@1&𝑏^2+𝑎𝑏&𝑐^2 )| R3 → R3 – R1 = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐 |■8(1&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@0&𝑏𝑐&−𝑎𝑐−𝑏𝑐@1−1&𝑏^2+𝑎𝑏−(𝑏^2−𝑏𝑐)&𝑐^2−(𝑐^2+𝑏𝑐))| = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐 |■8(1&𝑏^2−𝑏𝑐&𝑐^2+𝑏𝑐@0&𝑏𝑐&−𝑎𝑐−𝑏𝑐@0&𝑎𝑏+𝑏𝑐&−𝑏𝑐)| = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐 |■8(1&𝑏(𝑏−𝑐)&𝑐(𝑐+𝑏)@0&𝑏𝑐&−𝑐(𝑎+𝑏)@0&𝑏(𝑎+𝑐)&−𝑏𝑐)| Taking b common from C2, and c common from C3 = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎𝑏𝑐×𝑏𝑐|■8(1&(𝑏−𝑐)&(𝑐+𝑏)@0&𝑐&−(𝑎+𝑏)@0&(𝑎+𝑐)&−𝑏)| = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 |■8(1&(𝑏−𝑐)&(𝑐+𝑏)@0&𝑐&−(𝑎+𝑏)@0&(𝑎+𝑐)&−𝑏)| = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 |■8(1&(𝑏−𝑐)&(𝑐+𝑏)@0&𝑐&−(𝑎+𝑏)@0&(𝑎+𝑐)&−𝑏)| Finding determinant along C1 = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 × [1 (c × (-b) – (–(a + b))(a + c) + 0 + 0] = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 × [–bc + (a + b)(a + c)] = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 × [–bc + a(a + c) + b(a + c)] = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 × [–bc + a2 + ac + ab + bc] = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 × [a2 + ac + ab] = ((𝑎^2 + 𝑏^2+ 〖 𝑐〗^2))/𝑎 × a[a + c + b] = (𝑎+𝑏+𝑐) (𝑎^2+𝑏^2+𝑐^2)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.