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Example 7 - Verify Property 2 for |2 -3 5 6 0 4 1 5 -7| - Examples

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Example 7, Verify Property 2 for Δ = 2﷮−3﷮5﷮6﷮0﷮4﷮1﷮5﷮−7﷯﷯ ∆ = 2﷮−3﷮5﷮6﷮0﷮4﷮1﷮5﷮−7﷯﷯ ∆ = 2 0﷮4﷮5﷮−7﷯﷯ - (-3) 6﷮4﷮1﷮−7﷯﷯ + 5 6﷮0﷮1﷮5﷯﷯ = 2(0(– 7) – 5(4) + 3 (6( – 7) – (1) (4)) +5 (6(5) – 1 (0)) = 2(0 – 20) + 3 (-42 – 4) + 5 (30 – 0) = 2 (– 20) +3 (– 46) + 5 (30) = – 40 – 138 + 150 = – 28 Interchanging R2 and R3 i.e. R2 ⟷ R3, We get ∆1 = 2﷮−3﷮5﷮1﷮5﷮−7﷮6﷮0﷮4﷯﷯ = 2 5﷮−7﷮0﷮4﷯﷯ – (– 3) 1﷮−7﷮6﷮4﷯﷯ + 5 1﷮5﷮6﷮0﷯﷯ = 2(5(4) – 0( – 7) + 3 (1(4) – 6( – 7)) + 5(1 (0) – 6(5)) = 2 (20 – 0) + 3 ( 4 + 42) + 5 ( 0 – 30) = 2 (20) + 3 (46) + 5 ( – 30) = 40 + 138 – 150 = 28 ∆1 = 28 = –(-28) = – ∆ Therefore ∆1= –∆ Hence property 2nd is verified i.e. If any two rows ( or columns) of a determinant are interchanged , then sign of determinant changes.

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