Check sibling questions

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

Get live Maths 1-on-1 Classs - Class 6 to 12


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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.