Example 30 - Class 12
Last updated at May 29, 2018 by Teachoo
Transcript
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
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11
Example 12
Example 13
Example 14 Important
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24 Important
Example 25
Example 26 Important
Example 27
Example 28
Example 29
Example 30 You are here
Example 31
Example 32 Important
Example 33
Example 34 Important
Chapter 4 Class 12 Determinants
Serial order wise
About the Author
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.