Ex 4.2, 4 - Using property of determinants |1 bc a(b + c)|

Ex 4.2, 4 - Chapter 4 Class 12 Determinants - Part 2

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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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