Question 8 - Examples - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Examples
Example 2
Example 3
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8
Example 9
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18
Example 19 Important
Question 1
Question 2
Question 3
Question 4 Important
Question 5 Important
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Question 8 You are here
Question 9 Important
Question 10 Important
Question 11 Important
Question 12
Question 13 Important
Question 14 Important
Question 15 Important
Last updated at April 16, 2024 by Teachoo
Question 8 Evaluate ∆ = 1𝑎𝑏𝑐1𝑏𝑐𝑎1𝑐𝑎𝑏 ∆ = 1𝑎𝑏𝑐1𝑏𝑐𝑎1𝑐𝑎𝑏 We need to obtain 0 in 2nd row as well as 3rd row Applying R2 → R2 – R1 ∆ = 1𝑎𝑏𝑐𝟏 –𝟏𝑏−𝑎𝑐𝑎−𝑏𝑐1𝑐𝑎𝑏 = 1𝑎𝑏𝑐𝟎𝑏−𝑎𝑐(𝑎−𝑏)1𝑐𝑎𝑏 Applying R3 → R3 – R1 = 1𝑎𝑏𝑐0𝑏−𝑎𝑐(𝑎−𝑏)𝟏−𝟏𝑐−𝑎𝑎𝑏−𝑏𝑐 = 1𝑎𝑏𝑐0(𝑏−𝑎)𝑐(𝑎−𝑏)𝟎(𝑐−𝑎)𝑏(𝑎−𝑐) Expanding it along C1 = 1 𝑏−𝑎𝑐 𝑎−𝑏 𝑐−𝑎𝑏 𝑎−𝑐–0 𝑎𝑏𝑐 𝑐−𝑎𝑏 𝑎−𝑐 +0 𝑎𝑏𝑐 𝑏−𝑎𝑐 𝑎−𝑏 = 1 b−𝑎c 𝑎−𝑏 c−a𝑏 𝑎−𝑐 – 0 + 0 = 1 (b – a) b(a – c) – (c – a) (c) (a – b) = –(a – b) b ( – (c – a)) – (c – a) c (a – b) = (a – b) b (c – a) – (c – a) c (a – b) = (a – b)(c – a) (b – c)) = (a – b) (b – c) (c – a) Thus ∆ = (a – b) (b – c) (c – a)