Question 8 - Examples - Chapter 4 Class 12 Determinants
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 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)
Examples
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Example 7 Important
Example 8
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Example 11 Important
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Example 17 Important
Example 18
Example 19 Important
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Question 8 You are here
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Question 14 Important
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo