Example 6  - Verify Property 1 for |2 -3 5 6 0 4 1 5 -7| - Verifying properties of a determinant

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Example 6 Verify Property 1 for Δ = 2﷮−3﷮5﷮6﷮0﷮4﷮1﷮5﷮−7﷯﷯ Expanding the determinant along first row, ∆ = 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 By interchanging rows and columns , we get ∆1 = 2﷮6﷮1﷮−3﷮0﷮5﷮5﷮4﷮−7﷯﷯ ∆1 = 2 0﷮5﷮4﷮−7﷯﷯ – 6 −3﷮5﷮5﷮−7﷯﷯ + 1 −3﷮0﷮5﷮4﷯﷯ = 2(0(– 7) – 4(5) – 6( – 3 (– 7) – 5 (5)) +1 (– 3 (4) – 5 (0)) = 2 (0 – 20) – 6 (+ 21 – 25 ) +1 (– 12 + 0) = 2 ( – 20) – 6 ( – 4) + 1 ( – 12) = – 40 + 24 – 12 = – 52 + 24 = – 28 Hence ∆1 = ∆ Hence ∆1 = ∆ ∴ Property 1 is verified. i.e. If we interchange row & Columns, value of determinant remains unchanged

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