Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6 Deleted for CBSE Board 2023 Exams

Example 7 Deleted for CBSE Board 2023 Exams

Example 8 Deleted for CBSE Board 2023 Exams

Example 9 Important Deleted for CBSE Board 2023 Exams

Example 10 Important Deleted for CBSE Board 2023 Exams

Example 11 Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Important Deleted for CBSE Board 2023 Exams

Example 16 Important Deleted for CBSE Board 2023 Exams

Example 17

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important

Example 23

Example 24 Important

Example 25

Example 26 Important

Example 27

Example 28 Important

Example 29

Example 30 You are here

Example 31 Important Deleted for CBSE Board 2023 Exams

Example 32 Important Deleted for CBSE Board 2023 Exams

Example 33 Important

Example 34 Important Deleted for CBSE Board 2023 Exams

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at March 16, 2023 by Teachoo

Example 30 If a, b, c are positive and unequal, show that value of the determinant Δ = abcbcacab is negative Δ = abcbcacab Applying C1→ C1 + C2 + C3 = 𝐚+𝐛+𝐜bc𝐚+𝐛+𝐜ca𝐚+𝐛+𝐜ab Taking common a + b + c from C1 = (𝐚+𝐛+𝐜) 1bc1ca1ab Applying R2 → R2 – R1 = (a+b+c) 1bc𝟏−𝟏c−ba−c1ab = (a+b+c) 1bc𝟎c−ba−c1ab Applying R3 → R2 – R1 = (a+b+c) 1bc0c−ba−c𝟏−𝟏a−bb−c = (a+b+c) 1bc0c−ba−c𝟎a−bb−c Expanding determinant along C1 = (a + b + c ) 1 c−ba−cb−cb−c−0 bc𝑎−𝑏b−c+0 bcc−ba−c = (a + b + c ) 1 𝑐−𝑎 𝑏−𝑐− 𝑎−𝑏 𝑎−𝑐−0+0 = (a + b + c ) 𝑏−𝑐 − 𝑏−𝑐−(𝑎−𝑏)(𝑎−𝑐) = (a + b + c ) − 𝑏2+𝑐2−2𝑏𝑐−(𝑎2−𝑎𝑐−𝑏𝑎+𝑏𝑐) = (a + b + c ) −a2−b2 −𝑐2+𝑎𝑏+𝑏𝑐+𝑐𝑎 = – (a + b + c ) a2+b2+𝑐2−𝑎𝑏−𝑏𝑐−𝑐𝑎 Multiplying & Dividing by 2 = – 1 × 22 (a + b + c ) a2+b2+𝑐2−𝑎𝑏−𝑏𝑐−𝑐𝑎 = −12 (a + b + c ) 2a2+2b2+2𝑐2−2𝑎𝑏−2𝑏𝑐−2𝑐𝑎 = −12 (a + b + c ) 𝑎2+𝑎2+𝑏2+𝑏2+𝑐2+𝑐2−2𝑎𝑏−2𝑏𝑐−2𝑐𝑎 = −12 (a + b + c ) 𝒂𝟐+𝒄𝟐−𝟐𝒄𝒂+𝑎2+𝑏2−2𝑎𝑏+𝑏2+𝑐2−2𝑏𝑐 = −12 (a + b + c ) 𝒂−𝒄𝟐+ 𝑎−𝑏2+ 𝑐−𝑎2 Now 𝑎−𝑐2+ 𝑎−𝑏2+ 𝑐−𝑎2 > 0 & a + b + c > 0 ∴ ∆ = −12(a + b + c ) 𝑎−𝑐2+ 𝑎−𝑏2+ 𝑐−𝑎2 < 0 Hence ∆ is negative Hence Shown