Check sibling questions

Example 16 - Show that Determinant = abc (1 + 1/a + 1/b + 1/c) = abc +

Example 16 - Chapter 4 Class 12 Determinants - Part 2
Example 16 - Chapter 4 Class 12 Determinants - Part 3
Example 16 - Chapter 4 Class 12 Determinants - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.