Question 4 Using the property of determinants and without expanding, prove that: |■8(1&𝑏𝑐&𝑎(𝑏+𝑐)@1&𝑐𝑎&𝑏(𝑐+𝑎)@1&𝑎𝑏&𝑐(𝑎+𝑏))| = 0 |■8(1&𝑏𝑐&𝑎(𝑏+𝑐)@1&𝑐𝑎&𝑏(𝑐+𝑎)@1&𝑎𝑏&𝑐(𝑎+𝑏))| = |■8(1&𝑏𝑐&𝑎𝑏+𝑎𝑐@1&𝑐𝑎&𝑏𝑐+𝑏𝑎@1&𝑎𝑏&𝑐𝑎+𝑐𝑏)| C3 → C3 + C2 = |■8(1&𝑏𝑐&𝑎𝑏+𝑎𝑐+𝑏𝑐@1&𝑐𝑎&𝑏𝑐+𝑏𝑎+𝑏𝑐@1&𝑎𝑏&𝑐𝑎+𝑐𝑏+𝑏𝑐)| Taking (𝑎𝑏+𝑎𝑐+𝑏𝑐) common from C3 = (𝑎𝑏+𝑎𝑐+𝑏𝑐) |■8(𝟏&𝑏𝑐&𝟏@𝟏&𝑐𝑎&𝟏@𝟏&𝑎𝑏&𝟏)| C1 and C3 is same = 0 By Property: if any two row or columns of a determinant are identical then value of determinant is zero

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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.

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