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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

Transcript

Example 16 Show that |■8(1+a&1&1@1&1+b&1@1&1&1+c)| = abc (1+ 1/a + 1/b + 1/c) = abc + bc + ca + ab Solving L.H.S |■8(1+a&1&1@1&1+b&1@1&1&1+c)| Taking out a, b, c, common from R1, R2, & R3 respectively = abc |■8(1/a+1&1/a&1/a@1/b&1/b+1&1/b@1/c&1/c&1/a+1)| Applying R1→ R1 + R2 + R3 = abc |■8(1+1/a+1/b+1/c&1/a+1/b+1+1/c&1/a+1/b+1/c+1@1/b&1/b+1&1/b@1/c&1/c&1/c+1)| = abc |■8(𝟏+𝟏/𝐚+𝟏/𝐛+𝟏/𝐜&𝟏+𝟏/𝐚+𝟏/𝐛+𝟏/𝐜&𝟏+𝟏/𝐚+𝟏/𝐛+𝟏/𝐜@1/b&1/b+1&1/b@1/c&1/c&1/c+1)| Taking (1+1/𝑎+1/𝑏+1/𝑐) common from R1 = abc (𝟏+𝟏/𝐚+𝟏/𝐛+𝟏/𝐜) |■8(1&1&1@1/b&1/b+1&1/b@1/c&1/c&1/c+1)| = abc (𝟏+𝟏/𝐚+𝟏/𝐛+𝟏/𝐜) |■8(1&1&1@1/b&1/b+1&1/b@1/c&1/c&1/c+1)| Applying C3 → C3 – C1 = abc (1+1/a+1/b+1/c) |■8(1&0&𝟏−𝟏@1/b&1&1/b−1/b@1/c&0&1/c+1−1/𝑐)| = abc (1+1/a+1/b+1/c) |■8(1&0&𝟎@1/b&1&0@1/c&0&1)| Expanding determinant along R1 = abc (1+1/a+1/b+1/c) ( 1|■8(1&0@0&1)|−0|■8(1/𝑏&0@1/𝑐&1)|+0|■8(1/𝑏&1@1/𝑐&0)|) = abc (1+1/a+1/b+1/c) (1(1 – 0) – 0 + 0) = abc (1+1/a+1/b+1/c) (1(1)) = abc (1+1/a+1/b+1/c) = abc ((𝑎𝑏𝑐 + 𝑏𝑐 + 𝑎𝑐 + 𝑎𝑏)/𝑎𝑏𝑐) = 𝑎𝑏𝑐+𝑏𝑐+𝑎𝑐+𝑎𝑏 = R.H.S Hence Proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.