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

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Example 25 If A = 2﷮3﷮1﷮−4﷯﷯ and B = 1﷮−2﷮−1﷮3﷯﷯ , then verify that (AB)-1 = B-1 A-1 Taking L .H.S (AB) –1 First calculating AB AB = 2﷮3﷮1﷮−4﷯﷯ 1﷮−2﷮−1﷮3﷯﷯ = 2 1﷯+3(−1)﷮2 −2﷯+3(3)﷮1 1﷯+( −4)(−1)﷮1 −2﷯+ −4﷯3﷯﷯ = 2−3﷮−4+9﷮1+4﷮−2−12﷯﷯ = −1﷮5﷮5﷮−14﷯﷯ Now, (AB)-1 = 1﷮|AB|﷯ adj (AB) exists if |AB| ≠ 0 |AB| = −1﷮5﷮5﷮−14﷯﷯ = (-1)(-14) – 5(5) = 14 – 25 = – 11 Since |AB| ≠ 0 (AB)-1 exists AB = −1﷮5﷮5﷮−14﷯﷯ adj (AB) = −1﷮5﷮5﷮−14﷯﷯ = −14﷮−5﷮−5﷮−1﷯﷯ Now, (AB)–1 = 1﷮|AB|﷯ adj (AB) Putting values = 1﷮−11﷯ −14﷮−5﷮−5﷮−1﷯﷯ = 1﷮11﷯ 14﷮5﷮5﷮1﷯﷯ Taking R.H.S B-1A-1 First Calculating B–1 B = 1﷮−2﷮−1﷮3﷯﷯ B = 1﷮|B|﷯ adj (B) exists if |B| ≠ 0 |B| = 1﷮−2﷮−1﷮3﷯﷯ = 3 – 2 = 1 Since |B| ≠ 0 , B-1 exist B = 1﷮−2﷮−1﷮3﷯﷯ adj (B) = 1﷮−2﷮−1﷮3﷯﷯ = 3﷮2﷮1﷮1﷯﷯ Now, (B)–1 = 1﷮|B|﷯ adj (B) Putting values = 1﷮1﷯ 3﷮2﷮1﷮1﷯﷯ = 3﷮2﷮1﷮1﷯﷯ Finding A-1 A-1 = 1﷮|A|﷯ adj (A) exists if |A| ≠ 0 |A| = 2﷮3﷮1﷮−4﷯﷯ = 2 ( – 4) – 1( 3) = – 8 – 3 = – 11 Since |A| ≠ 0 , A–1 exists A = 2﷮3﷮1﷮−4﷯﷯ adj (A) = 2﷮3﷮1﷮−4﷯﷯ = −4﷮−3﷮−1﷮2﷯﷯ Now, A-1 = 1﷮|A|﷯ adj (A) = 1﷮−11﷯ −4﷮−3﷮−1﷮2﷯﷯ = 1﷮11﷯ 4﷮3﷮1﷮−2﷯﷯ Thus, B-1A-1 = 3﷮2﷮1﷮1﷯﷯ × 1﷮11﷯ 4﷮3﷮1﷮−2﷯﷯ = 1﷮11﷯ 3﷮2﷮1﷮1﷯﷯ 4﷮3﷮1﷮−2﷯﷯ = 1﷮11﷯ 3 4﷯+2(1)﷮3 3﷯+2(−2)﷮ 1 4﷯+1(1)﷮1 3﷯+1(−2)﷯﷯ = 1﷮11﷯ 12+2﷮9−4﷮4+1﷮3−2﷯﷯ = 1﷮11﷯ 14﷮5﷮5﷮1﷯﷯ = L.H.S ∴ L.H.S = R.H.S Hence proved