Question 11 - Making whole row/column one and simplifying - Chapter 4 Class 12 Determinants

Last updated at Dec. 16, 2024 by

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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Question 11 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 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