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Misc 8 - Verify [adj A]-1 = adj (A-1) - Chapter 4 NCERT - Inverse of two matrices and verifying properties

Misc 8 - Chapter 4 Class 12 Determinants - Part 2
Misc 8 - Chapter 4 Class 12 Determinants - Part 3
Misc 8 - Chapter 4 Class 12 Determinants - Part 4
Misc 8 - Chapter 4 Class 12 Determinants - Part 5
Misc 8 - Chapter 4 Class 12 Determinants - Part 6
Misc 8 - Chapter 4 Class 12 Determinants - Part 7
Misc 8 - Chapter 4 Class 12 Determinants - Part 8
Misc 8 - Chapter 4 Class 12 Determinants - Part 9
Misc 8 - Chapter 4 Class 12 Determinants - Part 10
Misc 8 - Chapter 4 Class 12 Determinants - Part 11
Misc 8 - Chapter 4 Class 12 Determinants - Part 12
Misc 8 - Chapter 4 Class 12 Determinants - Part 13
Misc 8 - Chapter 4 Class 12 Determinants - Part 14
Misc 8 - Chapter 4 Class 12 Determinants - Part 15
Misc 8 - Chapter 4 Class 12 Determinants - Part 16

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Transcript

Misc 8 Let A = 1﷮−2﷮1﷮−2﷮3﷮1﷮1﷮1﷮5﷯﷯ verify that (i) [adj A]-1 = adj (A-1) First we will calculate adj (A) & A-1 adj A = A﷮11﷯﷮ A﷮12﷯﷮ A﷮13﷯﷮ A﷮21﷯﷮ A﷮22﷯﷮ A﷮23﷯﷮ A﷮31﷯﷮ A﷮32﷯﷮ A﷮33﷯﷯﷯﷮′﷯= A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ A = 1﷮−2﷮1﷮−2﷮3﷮1﷮1﷮1﷮5﷯﷯ M11 = 3﷮1﷮1﷮5﷯﷯ = 15 – 1 = 14 M12 = −2﷮1﷮1﷮5﷯﷯ = – 10 – 1 = – 11 M13 = −2﷮3﷮1﷮1﷯﷯ = – 2 – 3 = – 5 M21 = −2﷮1﷮1﷮5﷯﷯ = – 10 – 4 = – 11 M22 = 1﷮1﷮1﷮5﷯﷯ = 5 – 1 = 4 M23 = 1﷮−2﷮1﷮1﷯﷯ = 1 + 2 = 3 M31 = −2﷮1﷮3﷮1﷯﷯ = – 2 – 3 = – 5 M32 = 1﷮1﷮−2﷮1﷯﷯ = 1 + 2 = 3 M33 = 1﷮−2﷮−2﷮3﷯﷯ = 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) = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ = 𝟏𝟒﷮𝟏𝟏﷮−𝟓﷮𝟏𝟏﷮𝟒﷮−𝟑﷮−𝟓﷮−𝟑﷮−𝟏﷯﷯ Now, A-1 = 1﷮|A|﷯ adj (B) Finding |A| |A| = 1﷮−2﷮1﷮−2﷮3﷮1﷮1﷮1﷮5﷯﷯ = 1 (15 – 1) + 2 ( – 10 – 1) + 1 ( – 2 – 3) = 14 – 22 – 5 = – 13 Therefore A-1 = 1﷮|A|﷯ adj (A) = 𝟏﷮𝟏𝟑﷯ 𝟏𝟒﷮𝟏𝟏﷮−𝟓﷮𝟏𝟏﷮𝟒﷮−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 = 14﷮11﷮−5﷮11﷮4﷮−3﷮−5﷮−3﷮−1﷯﷯ Now, B-1 = 1﷮|B|﷯ adj (B) exists if |B| ≠ 0 |B| = 14﷮11﷮−5﷮11﷮4﷮−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 = A﷮11﷯﷮ A﷮12﷯﷮ A﷮13﷯﷮ A﷮21﷯﷮ A﷮22﷯﷮ A﷮23﷯﷮ A﷮31﷯﷮ A﷮32﷯﷮ A﷮33﷯﷯﷯﷮′﷯= A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ Here Aij are the cofactors of matrix B B = 14﷮11﷮−5﷮11﷮4﷮−3﷮−5﷮−3﷮−1﷯﷯ M11 = 4﷮−3﷮−3﷮−1﷯﷯ = – 4 – 9 = – 13 M12 = 11﷮−3﷮−5﷮−1﷯﷯ = – 11 – 15 = – 26 M13 = 11﷮4﷮−5﷮−3﷯﷯ = – 33 + 20 = – 13 M21 = 11﷮−5﷮−3﷮−1﷯﷯ = – 11 – 15 = – 26 M22 = 14﷮−5﷮−5﷮−1﷯﷯ = – 14 – 25 = – 39 M23 = 14﷮11﷮−5﷮−3﷯﷯ = ( – 42 + 55) = + 13 M31 = 11﷮−5﷮4﷮−3﷯﷯ = – 33 – 20 = – 13 M32 = 14﷮5﷮−11﷮3﷯﷯ = – 42 + 55 = 13 M33 = 14﷮11﷮−11﷮4﷯﷯ = 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 = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ = −13﷮26﷮13﷮26﷮−39﷮−13﷮−13﷮−13﷮−65﷯﷯ Now, B-1 = 1﷮|B|﷯ (adj B) = 1﷮169﷯ −13﷮26﷮13﷮26﷮−39﷮−13﷮−13﷮−13﷮−65﷯﷯ Taking 13 common from all elements of the matrix = 𝟏𝟑﷮169﷯ −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ = 1﷮13﷯ −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ Thus, [adj A] -1 = B-1 = 𝟏﷮𝟏𝟑﷯ −𝟏﷮𝟐﷮−𝟏﷮𝟐﷮−𝟑﷮−𝟏﷮−𝟏﷮−𝟏﷮−𝟓﷯﷯ Taking R.H.S adj (A-1) A-1 = 1﷮13﷯ −14﷮−11﷮5﷮−11﷮−4﷮3﷮5﷮3﷮1﷯﷯ Let C = A-1 C = 1﷮13﷯ −14﷮−11﷮5﷮−11﷮−4﷮3﷮5﷮3﷮1﷯﷯ = −14﷮13﷯﷮ −11﷮13﷯﷮ 5﷮13﷯﷮ −11﷮13﷯﷮ −4﷮13﷯﷮ 3﷮13﷯﷮ 5﷮13﷯﷮ 3﷮13﷯﷮ 1﷮13﷯﷯﷯ Now, adj C = adj (A-1) adj C = A﷮11﷯﷮ A﷮12﷯﷮ A﷮13﷯﷮ A﷮21﷯﷮ A﷮22﷯﷮ A﷮23﷯﷮ A﷮31﷯﷮ A﷮32﷯﷮ A﷮33﷯﷯﷯﷮′﷯ = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ Here Aij are the cofactors of matrix C C = −14﷮13﷯﷮ −11﷮13﷯﷮ 5﷮13﷯﷮ −11﷮13﷯﷮ −4﷮13﷯﷮ 3﷮13﷯﷮ 5﷮13﷯﷮ 3﷮13﷯﷮ 1﷮13﷯﷯﷯ M11 = −4﷮13﷯﷮ −3﷮13﷯﷮ 3﷮13﷯﷮ 1﷮13﷯﷯﷯ = −4﷮169﷯ – 9﷮169﷯ = −13﷮169﷯ = −1﷮13﷯ M12 = −11﷮13﷯﷮ 3﷮13﷯﷮ 5﷮13﷯﷮ 1﷮13﷯﷯﷯ = −11﷮169﷯ – 15﷮169﷯ = −26﷮169﷯ = −2﷮13﷯ M13 = −11﷮13﷯﷮ −4﷮13﷯﷮ 5﷮13﷯﷮ 3﷮13﷯﷯﷯ = −33﷮169﷯ + 20﷮169﷯ = −13﷮169﷯ = −1﷮13﷯ M21 = −11﷮13﷯﷮ 5﷮13﷯﷮ 3﷮13﷯﷮ 1﷮13﷯﷯﷯ = −11﷮169﷯ – 55﷮169﷯ = −26﷮169﷯ = −2﷮13﷯ M22 = −14﷮13﷯﷮ 5﷮13﷯﷮ 5﷮13﷯﷮ 1﷮13﷯﷯﷯ = −14﷮169﷯ – 25﷮169﷯ = −39﷮169﷯ = −3﷮13﷯ M23 = −14﷮13﷯﷮ −11﷮13﷯﷮ 5﷮13﷯﷮ 3﷮13﷯﷯﷯ = ( −42﷮169﷯ + 55﷮169﷯) = 13﷮169﷯ = 1﷮13﷯ M31 = −11﷮13﷯﷮ 5﷮13﷯﷮ −4﷮13﷯﷮ 3﷮13﷯﷯﷯ = −33﷮169﷯ + 20﷮167﷯ = −13﷮169﷯ = −1﷮169﷯ M32 = −14﷮13﷯﷮ 5﷮13﷯﷮ −11﷮13﷯﷮ 3﷮13﷯﷯﷯ = −42﷮169﷯ + 55﷮169﷯ = 13﷮169﷯ = 1﷮13﷯ M33 = −14﷮13﷯﷮ −11﷮13﷯﷮ −11﷮13﷯﷮ −4﷮13﷯﷯﷯ = 56﷮169﷯ – −121﷮169﷯ = −65﷮169﷯ = −5﷮13﷯ A11 = ( – 1)1 + 1 M11 = ( – 1)2 −1﷮13﷯﷯ = −1﷮13﷯ A12 = ( – 1)1+2 M12 = ( – 1)3 ( −2﷮13﷯) = 2﷮13﷯ A13 = ( – 1)1+3 M13 = ( – 1)4 . −1﷮13﷯ = −1﷮13﷯ A21 = ( – 1)2+1 M21 = ( – 1)3 . ( −2﷮13﷯) = 2﷮13﷯ A22 = ( – 1)2+2 M22 = ( – 1)4 . ( −3﷮13﷯) = −3﷮13﷯ A23 = ( – 1)2+3 M23 = ( – 1)5 . ( 1﷮13﷯) = −1﷮13﷯ A31 = ( – 1)3+1 M31 = ( – 1)4 . ( −1﷮169﷯) = −1﷮13﷯ A32 = ( – 1)3+2 M32 = ( – 1)5 . ( 1﷮13﷯ ) = −1﷮13﷯ A33 = (– 1)3+3 M33 = ( – 1)6 . ( −5﷮13﷯) = −5﷮13﷯ Thus, adj C = −1﷮13﷯﷮ 2﷮13﷯﷮ −1﷮13﷯﷮ 2﷮13﷯﷮ −3﷮13﷯﷮ −1﷮13﷯﷮ −1﷮13﷯﷮ −1﷮13﷯﷮ −5﷮13﷯﷯﷯ = 1﷮13﷯ −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ ∴ adj (A-1) = adj C = 𝟏﷮𝟏𝟑﷯ −𝟏﷮𝟐﷮−𝟏﷮𝟐﷮−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﷮−2﷮1﷮−2﷮3﷮1﷮1﷮1﷮5﷯﷯ 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 = 𝟏﷮𝟏𝟑﷯ −𝟏𝟒﷮−𝟏𝟏﷮𝟓﷮−𝟏𝟏﷮−𝟒﷮𝟑﷮𝟓﷮𝟑﷮𝟏﷯﷯ Calculating |A-1| |A-1| = 1﷮13﷯ −14﷮−11﷮5﷮−11﷮−4﷮3﷮5﷮3﷮1﷯﷯﷯ Using |KA| = Kn |A| Where n is order of A = 1﷮13﷯﷯﷮3﷯ −14 −4﷮3﷮3﷮1﷯﷯− −11﷯ −11﷮3﷮5﷮1﷯﷯+5 −11﷮−4﷮5﷮3﷯﷯﷯ = 1﷮13﷯﷯﷮3﷯( –14 ( – 4 – 9) + 11 ( – 11 – 15) + 5 ( – 33 + 20)) = 1﷮13﷯﷯﷮3﷯( –14 ( – 13) + 11 ( – 26) + 5 ( – 13)) = 1﷮13﷯﷯﷮3﷯(182 – 286 – 65) = 1﷮13﷯﷯﷮3﷯( – 169) = −169﷮13 ×13 ×13﷯ = 1﷮−13﷯ Now, (A-1)-1 = 1﷮| A﷮−1﷯|﷯ (adj A-1) Putting values = 1﷮ −1﷮13﷯﷯ × 𝟏﷮𝟏𝟑﷯ −𝟏﷮𝟐﷮−𝟏﷮𝟐﷮−𝟑﷮−𝟏﷮−𝟏﷮−𝟏﷮−𝟓﷯﷯ = –13 × 1﷮13﷯ −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ = – −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ = −1﷮2﷮−1﷮2﷮−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﷮ −1﷮13﷯﷯ × 𝟏﷮𝟏𝟑﷯ −𝟏﷮𝟐﷮−𝟏﷮𝟐﷮−𝟑﷮−𝟏﷮−𝟏﷮−𝟏﷮−𝟓﷯﷯ = –13 × 1﷮13﷯ −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ = – −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ = −1﷮2﷮−1﷮2﷮−3﷮−1﷮−1﷮−1﷮−5﷯﷯ = A Thus, (A-1)-1 = A Hence Proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.