Ex 4.1, 4 - Show that  |3A|  = 27|A|, if A =  [1 0 1 - Ex 4.1

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.1, 4 If A = 1﷮0﷮1﷮0﷮1﷮2﷮0﷮0﷮4﷯﷯ , then show that 3A﷯ = 27 A﷯ We have to prove 3A﷯ = 27 A﷯ Taking L.H.S 3A﷯ First Calculating 3A 3A = 3 1﷮0﷮1﷮0﷮1﷮2﷮0﷮0﷮4﷯﷯ = 3×1﷮3×0﷮3×1﷮3×0﷮3×1﷮3×2﷮3×0﷮3×0﷮3×4﷯﷯ = 3﷮0﷮1﷮0﷮3﷮6﷮0﷮0﷮12﷯﷯ Hence, |3A| = 3﷮0﷮3﷮0﷮3﷮6﷮0﷮0﷮12﷯﷯ = 1 3﷮6﷮0﷮12﷯﷯ – 0 0﷮6﷮0﷮12﷯﷯ + 3 0﷮3﷮0﷮20﷯﷯ = 3(3(12)– 0(6)) – 0 (0(12) – 0(6)) +3 (0(0) – 0(3) = 3(36 – 0) – 0(0) + 3(0) = 3(36) + 0 + 0 = 108 Taking R.H.S 27|A| |A| = 1﷮0﷮1﷮0﷮1﷮2﷮0﷮0﷮4﷯﷯ = 1 1﷮2﷮0﷮4﷯﷯ – 0 0﷮2﷮0﷮4﷯﷯ + 1 0﷮1﷮0﷮0﷯﷯ = 1(1(4) – 0(2) – 0 (0(4) – 0(2) + 1(0 – 0(1)) = 1 (4 – 0) – 0 (0) + 1(0) = 4 Now 27|A| = 27 (4) = 108 = |3A| Hence L.H.S = R.H.S Hence proved

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