Misc 7 - Chapter 4 Class 12 Determinants

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Misc 7 If A-1 = 3﷮−1﷮1﷮−15﷮6﷮−5﷮5﷮−2﷮2﷯﷯ and B = 1﷮2﷮−2﷮−1﷮3﷮0﷮0﷮−2﷮1﷯﷯ , Find (AB)-1 We know that (AB)-1 = B-1 A-1 Calculating B-1 We know that B-1 = 1﷮|B|﷯ adj (B) exists if |B| ≠ 0 |B| = 1﷮2﷮−2﷮−1﷮3﷮0﷮0﷮−2﷮1﷯﷯ = 1 (3 – 0) – 2( – 1 – 0) – 2 (2 – 0) = 1 (3) – 2 ( – 1) – 2(2) = 3 + 2 – 4 = 1 Since |B| ≠ 0 Thus, B-1 exists Calculating adj B Now, adj B = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ B = 1﷮2﷮−2﷮−1﷮3﷮0﷮0﷮−2﷮1﷯﷯ M11 = 3﷮0﷮−2﷮1﷯﷯ =3(1) – (-2)0 = 3 M12 = −1﷮0﷮0﷮1﷯﷯ = -1(1) – 0(0)= –1 M13 = −1﷮3﷮0﷮−2﷯﷯ =(-1)(-2) – 0(3)= 2 M21 = 2﷮−2﷮−2﷮1﷯﷯ =2(1)–(-2)(-2)= -2 M22 = 1﷮−2﷮0﷮1﷯﷯ = 1(1) – 0(-2) = 1 A11 = ( – 1)1 + 1 M11 = ( – 1)2 . 3 = 3 A12 = ( – 1)1+2 M12 = ( – 1)3 ( – 1) = 1 A13 = ( – 1)1+3 M13 = ( – 1)4 2 = 2 A21 = ( – 1)2+1 M21 = ( – 1)3 ( – 2) = 2 A22 = ( – 1)2+2 M22 = ( – 1)4 . 1 = 1 A23 = ( – 1)2+3 M23 = ( – 1)5 ( – 2) = 2 A31 = ( – 1)3+1 M31 = ( – 1)4 . 6 = 6 A32 = ( – 1)3+2 M32 = ( – 1)5 ( – 2) = 2 A33 = ( – 1)3+3 M33 = ( – 1)6 . 5 = 5 Thus, adj (B) = 3﷮2﷮6﷮1﷮1﷮2﷮2﷮2﷮5﷯﷯ Now, B-1 = 1﷮|B|﷯ adj (B) Putting values = 1﷮1﷯ 3﷮2﷮6﷮1﷮1﷮2﷮2﷮2﷮5﷯﷯ = 3﷮2﷮6﷮1﷮1﷮2﷮2﷮2﷮5﷯﷯ Now, (AB)-1 = B-1 A-1 = 3﷮2﷮6﷮1﷮1﷮2﷮2﷮2﷮5﷯﷯ 3﷮−1﷮1﷮−15﷮6﷮−5﷮5﷮−2﷮2﷯﷯ = 3 3﷯+2 −15﷯+6(−5)﷮3 −1﷯+2 6﷯+6(−2)﷮3 1﷯+2 −5﷯+6(2)﷮1 3﷯+1 −15﷯+2(−5)﷮1 −1﷯+1 6﷯+2(−2)﷮1 1﷯+1 −5﷯+2(2)﷮2 3﷯+2 −15﷯+5(−5)﷮2 −1﷯+2 6﷯+5(−2)﷮2 1﷯+2 −5﷯+5(2)﷯﷯ = 9−30+30﷮−3+12−12﷮3−10+12﷮3−15+10﷮−1+6−4﷮1−5+4﷮6−30+25﷮−2+12−10﷮2−10+10﷯﷯ = 𝟗﷮−𝟑﷮𝟓﷮−𝟐﷮𝟏﷮𝟎﷮𝟏﷮𝟎﷮𝟐﷯﷯