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 )




CBSE Class 12 Sample Paper for 2019 Boards
CBSE Class 12 Sample Paper for 2019 Boards
Last updated at Dec. 16, 2024 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 )
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)