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Matrices and Determinants - Formula Sheet and Summary Important

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at March 16, 2023 by Teachoo

Misc 8 Let A = 1−21−231115 verify that (i) [adj A]-1 = adj (A-1) First we will calculate adj (A) & A-1 adj A = A11 A12 A13 A21 A22 A23 A31 A32 A33′= A11 A21 A31 A12 A22 A32 A13 A23 A33 A = 1−21−231115 M11 = 3115 = 15 – 1 = 14 M12 = −2115 = – 10 – 1 = – 11 M13 = −2311 = – 2 – 3 = – 5 M21 = −2115 = – 10 – 4 = – 11 M22 = 1115 = 5 – 1 = 4 M23 = 1−211 = 1 + 2 = 3 M31 = −2131 = – 2 – 3 = – 5 M32 = 11−21 = 1 + 2 = 3 M33 = 1−2−23 = 3 – 4 = – 5 A11 = ( – 1)1 + 1 M11 = ( – 1)2 – 14 = 14 A12 = ( – 1)1+2 M12 = ( – 1)3 ( – 11) = 11 A13 = ( – 1)1+3 M13 = ( – 1)4 ( – 5) = –5 A21 = ( – 1)2+1 M21 = ( – 1)3 . ( – 11) = 11 A22 = ( – 1)2+2 M22= ( – 1)4 . 4 = 4 A23 = ( – 1)2+3 M23 = ( – 1)5 (3) = – 3 A31 = ( – 1)3+1 M31 = ( – 1)4 . ( – 5) = – 5 A32 = ( – 1)3+2 M32 = ( – 1)5 . (3) = – 3 A33 = ( – 1)3+3 M33= ( – 1)6 . ( – 5) = – 1 Thus, adj (A) = A11 A21 A31 A12 A22 A32 A13 A23 A33 = 𝟏𝟒𝟏𝟏−𝟓𝟏𝟏𝟒−𝟑−𝟓−𝟑−𝟏 Now, A-1 = 1|A| adj (B) Finding |A| |A| = 1−21−231115 = 1 (15 – 1) + 2 ( – 10 – 1) + 1 ( – 2 – 3) = 14 – 22 – 5 = – 13 Therefore A-1 = 1|A| adj (A) = 𝟏𝟏𝟑 𝟏𝟒𝟏𝟏−𝟓𝟏𝟏𝟒−𝟑−𝟓−𝟑−𝟏 We need to verify [adj A] −1 = adj (A-1) Taking L.H.S (adj A)-1 Let B = adj (A) B = 1411−5114−3−5−3−1 Now, B-1 = 1|B| adj (B) exists if |B| ≠ 0 |B| = 1411−5114−3−5−3−1 = 14 ( – 4 – 9) +1 ( – 11 – 15) – 5 ( – 33 + 20) = 14( – 13) – 11 ( – 26) – 5( – 13) = – 182 + 286 + 65 = 169 Thus |B| = 169 ≠ 0 ∴ B-1 exist Now, calculating adj B adj B = A11 A12 A13 A21 A22 A23 A31 A32 A33′= A11 A21 A31 A12 A22 A32 A13 A23 A33 Here Aij are the cofactors of matrix B B = 1411−5114−3−5−3−1 M11 = 4−3−3−1 = – 4 – 9 = – 13 M12 = 11−3−5−1 = – 11 – 15 = – 26 M13 = 114−5−3 = – 33 + 20 = – 13 M21 = 11−5−3−1 = – 11 – 15 = – 26 M22 = 14−5−5−1 = – 14 – 25 = – 39 M23 = 1411−5−3 = ( – 42 + 55) = + 13 M31 = 11−54−3 = – 33 – 20 = – 13 M32 = 145−113 = – 42 + 55 = 13 M33 = 1411−114 = 56 – 121 = – 65 A11 = ( – 1)1 + 1 M11 = ( – 1)2 (– 13) = – 13 A12 = ( – 1)1+2 M12 = ( – 1)3 ( – 26) = 26 A13 = ( – 1)1+3 M13 = ( – 1)4 . ( – 13) = – 13 A21 = ( – 1)2+1 M21 = ( – 1)3 . ( – 26) = 26 A22 = ( – 1)2+2 M22= ( – 1)4 . ( – 39) = – 39 A23 = ( – 1)2+3 M23 = ( – 1)5 . ( – 13) = – 13 A31 = ( – 1)3+1 M31 = ( – 1)4 . ( – 13) = – 13 A32 = ( – 1)3+2 M32 = ( – 1)5 . (13) = – 13 A33 = ( – 1)3+3 M33 = ( – 1)6 . ( – 65) = – 65 Thus, adj B = A11 A21 A31 A12 A22 A32 A13 A23 A33 = −13261326−39−13−13−13−65 Now, B-1 = 1|B| (adj B) = 1169 −13261326−39−13−13−13−65 Taking 13 common from all elements of the matrix = 𝟏𝟑169 −12−12−3−1−1−1−5 = 113 −12−12−3−1−1−1−5 Thus, [adj A] -1 = B-1 = 𝟏𝟏𝟑 −𝟏𝟐−𝟏𝟐−𝟑−𝟏−𝟏−𝟏−𝟓 Taking R.H.S adj (A-1) A-1 = 113 −14−115−11−43531 Let C = A-1 C = 113 −14−115−11−43531 = −1413 −1113 513 −1113 −413 313 513 313 113 Now, adj C = adj (A-1) adj C = A11 A12 A13 A21 A22 A23 A31 A32 A33′ = A11 A21 A31 A12 A22 A32 A13 A23 A33 Here Aij are the cofactors of matrix C C = −1413 −1113 513 −1113 −413 313 513 313 113 M11 = −413 −313 313 113 = −4169 – 9169 = −13169 = −113 M12 = −1113 313 513 113 = −11169 – 15169 = −26169 = −213 M13 = −1113 −413 513 313 = −33169 + 20169 = −13169 = −113 M21 = −1113 513 313 113 = −11169 – 55169 = −26169 = −213 M22 = −1413 513 513 113 = −14169 – 25169 = −39169 = −313 M23 = −1413 −1113 513 313 = ( −42169 + 55169) = 13169 = 113 M31 = −1113 513 −413 313 = −33169 + 20167 = −13169 = −1169 M32 = −1413 513 −1113 313 = −42169 + 55169 = 13169 = 113 M33 = −1413 −1113 −1113 −413 = 56169 – −121169 = −65169 = −513 A11 = ( – 1)1 + 1 M11 = ( – 1)2 −113 = −113 A12 = ( – 1)1+2 M12 = ( – 1)3 ( −213) = 213 A13 = ( – 1)1+3 M13 = ( – 1)4 . −113 = −113 A21 = ( – 1)2+1 M21 = ( – 1)3 . ( −213) = 213 A22 = ( – 1)2+2 M22 = ( – 1)4 . ( −313) = −313 A23 = ( – 1)2+3 M23 = ( – 1)5 . ( 113) = −113 A31 = ( – 1)3+1 M31 = ( – 1)4 . ( −1169) = −113 A32 = ( – 1)3+2 M32 = ( – 1)5 . ( 113 ) = −113 A33 = (– 1)3+3 M33 = ( – 1)6 . ( −513) = −513 Thus, adj C = −113 213 −113 213 −313 −113 −113 −113 −513 = 113 −12−12−3−1−1−1−5 ∴ adj (A-1) = adj C = 𝟏𝟏𝟑 −𝟏𝟐−𝟏𝟐−𝟑−𝟏−𝟏−𝟏−𝟓 = R.H.S Hence L.H.S = R.H.S ∴ (adj A)−1 = adj (A-1) Misc. 8 Let A = 1−21−231115 verify that (ii) (A-1)-1 = A We have to find (A-1)-1 So, (A-1)-1 = 1| A−1| adj (A-1) From First part, A-1 = 𝟏𝟏𝟑 −𝟏𝟒−𝟏𝟏𝟓−𝟏𝟏−𝟒𝟑𝟓𝟑𝟏 Calculating |A-1| |A-1| = 113 −14−115−11−43531 Using |KA| = Kn |A| Where n is order of A = 1133 −14 −4331− −11 −11351+5 −11−453 = 1133( –14 ( – 4 – 9) + 11 ( – 11 – 15) + 5 ( – 33 + 20)) = 1133( –14 ( – 13) + 11 ( – 26) + 5 ( – 13)) = 1133(182 – 286 – 65) = 1133( – 169) = −16913 ×13 ×13 = 1−13 Now, (A-1)-1 = 1| A−1| (adj A-1) Putting values = 1 −113 × 𝟏𝟏𝟑 −𝟏𝟐−𝟏𝟐−𝟑−𝟏−𝟏−𝟏−𝟓 = –13 × 113 −12−12−3−1−1−1−5 = – −12−12−3−1−1−1−5 = −12−12−3−1−1−1−5 = A Thus, (A-1)-1 = A Hence Proved Now, (A-1)-1 = 1| A−1| (adj A-1) Putting values = 1 −113 × 𝟏𝟏𝟑 −𝟏𝟐−𝟏𝟐−𝟑−𝟏−𝟏−𝟏−𝟓 = –13 × 113 −12−12−3−1−1−1−5 = – −12−12−3−1−1−1−5 = −12−12−3−1−1−1−5 = A Thus, (A-1)-1 = A Hence Proved