# Misc 8 - Chapter 4 Class 12 Determinants

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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

Miscellaneous

Misc 1

Misc. 2 Important Deleted for CBSE Board 2021 Exams only

Misc 3

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8 You are here

Misc 9

Misc 10

Misc 11 Important Deleted for CBSE Board 2021 Exams only

Misc 12 Important Deleted for CBSE Board 2021 Exams only

Misc. 13 Deleted for CBSE Board 2021 Exams only

Misc 14 Deleted for CBSE Board 2021 Exams only

Misc. 15 Important Deleted for CBSE Board 2021 Exams only

Misc. 16 Important

Misc 17 Important Deleted for CBSE Board 2021 Exams only

Misc 18

Misc 19 Important

Matrices and Determinants - Formula Sheet and Summary

Chapter 4 Class 12 Determinants

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.