1. Chapter 4 Class 12 Determinants
2. Serial order wise

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Example 24 If A = 1﷮3﷮3﷮1﷮4﷮3﷮1﷮3﷮4﷯﷯, then verify that A adj A = |A| I. Also find A–1. Taking L.H.S A (adj A) First Calculating adj A adj A = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ Now, A = 1﷮3﷮3﷮1﷮4﷮3﷮1﷮3﷮4﷯﷯ M11 = 4﷮3﷮3﷮4﷯﷯ = 4(4) – 3(3) = 7 M12 = 1﷮3﷮1﷮4﷯﷯ = 1(4) –1(3) = 1 M13 = 1﷮4﷮1﷮3﷯﷯ = 1(3) – 1(4) = – 1 M21 = 3﷮3﷮3﷮4﷯﷯ = 3(4) – 3(3) = 3 M22 = 1﷮3﷮1﷮4﷯﷯ = 1(4) – 1(3) = 1 Thus, adj (A) = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ = 7﷮−3﷮−3﷮−1﷮1﷮0﷮−1﷮0﷮1﷯﷯ Finding A (adj A) A adj (A) = 1﷮3﷮3﷮1﷮4﷮3﷮1﷮3﷮4﷯﷯ 7﷮3﷮−3﷮−1﷮1﷮0﷮−1﷮0﷮1﷯﷯ = 1 7﷯+3 −1﷯+3(−1)﷮1 −3﷯+3 1﷯+3(0)﷮1 −3﷯+3 0﷯+3(1)﷮1 7﷯+4 −1﷯+3(−1)﷮1 −3﷯+4 1﷯+3(0)﷮1 −3﷯+4 0﷯+3(1)﷮1 7﷯+3 −1﷯+4(−1)﷮1 −3﷯+3 1﷯+4(0)﷮1 −3﷯+3 0﷯+4(1)﷯﷯ = 7−3−3﷮−3+3+0﷮−3+0+3﷮7−4−3﷮−3+4+0﷮−3+0+3﷮7−3−4﷮−3+3+0﷮−3+0+4﷯﷯ = 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯ Taking R.H.S |A| I Calculating |A| |A| = 1﷮3﷮3﷮1﷮4﷮3﷮1﷮3﷮4﷯﷯ = 1 (4(4) – 3(3)) – 3(1(4) – 1(3)) + 3(1(4) – 1(3)) = 1(7) – 3(1) +3( – 1) = 7 – 3 – 3 = 1 Now, |A| I = 1 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯ = 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯ = L.H.S Thus, A(adj A) = |A| I Hence proved Finding A-1 We know that A-1 = 1﷮|A|﷯ (adj A) exists if |A| ≠ 0 Here, |A| = 1 ≠ 0 Thus A-1 exists So, A-1 = 1﷮|A|﷯ (adj A) = 1﷮1﷯ 7﷮−3﷮−3﷮−1﷮1﷮0﷮−1﷮0﷮1﷯﷯ = 𝟕﷮−𝟑﷮−𝟑﷮−𝟏﷮𝟏﷮𝟎﷮−𝟏﷮𝟎﷮𝟏﷯﷯