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Example 13 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at March 16, 2023 by Teachoo
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Chapter 4 Class 12 Determinants
Serial order wise
Transcript
Example 13 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)
