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&[email protected]&1+b&[email protected]&1&1+c)| = abc (1+ 1/a + 1/b + 1/c) = abc + bc + ca + ab Solving L.H.S |■8(1+a&1&[email protected]&1+b&[email protected]&1&1+c)| Taking out a, b, c, common from R1, R2, & R3 respectively = abc |■8(1/a+1&1/a&1/[email protected]/b&1/b+1&1/[email protected]/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/[email protected]/b&1/b+1&1/[email protected]/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/[email protected]/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&[email protected]/b&1/b+1&1/[email protected]/c&1/c&1/c+1)| = abc (𝟏+𝟏/𝐚+𝟏/𝐛+𝟏/𝐜) |■8(1&1&[email protected]/b&1/b+1&1/[email protected]/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/[email protected]/c&0&1/c+1−1/𝑐)| = abc (1+1/a+1/b+1/c) |■8(1&0&𝟎@1/b&1&[email protected]/c&0&1)| Expanding determinant along R1 = abc (1+1/a+1/b+1/c) ( 1|■8(1&[email protected]&1)|−0|■8(1/𝑏&[email protected]/𝑐&1)|+0|■8(1/𝑏&[email protected]/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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.