Slide75.JPG

Slide76.JPG
Slide77.JPG Slide78.JPG Slide79.JPG Slide80.JPG Slide81.JPG Slide82.JPG

  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

Transcript

Misc 16 Solve the system of the following equations 2/x + 3/y + 10/z = 4 4/x + 6/y + 5/z = 1 6/x + 9/y + 20/z = 2 The system of equations are 2/x + 3/y + 10/z = 4 4/x + 6/y + 5/z = 1 6/x + 9/y + 20/z = 2 Now let ๐Ÿ/๐’™ = u , ๐Ÿ/๐’š = v , & ๐Ÿ/๐’› = w The system of equations become 2u + 3v + 10w = 4 4u โ€“ 6v + 5w = 1 6u + 9v โ€“ 20w = 2 Writing equation as AX = B [โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] = [โ– 8(4@1@2)] Hence A = [โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)] , X = [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] & B = [โ– 8(4@1@2)] Calculating |A| |A| = |โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)| = 2 |โ– 8(โˆ’6&5@9&โˆ’20)| โ€“ 3 |โ– 8(4&5@6&โˆ’20)| + 10 |โ– 8(4&โˆ’6@6&9)| = 2 (120 โ€“ 45) โ€“3 (โ€“80 โ€“ 30) + 10 ( 36 + 36) = 2 (75) โ€“3 (โ€“110) + 10 (72) = 150 + 330 + 720 = 1200 โˆด |A|โ‰  0 So, the system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj (A) = [โ– 8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^โ€ฒ = [โ– 8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)] M11 = |โ– 8(โˆ’6&5@9&โˆ’20)| = 120 โ€“ 45 = 75 M12 = |โ– 8(4&5@6&โˆ’20)| = (โ€“80 โ€“ 30) = โ€“110 M13 = |โ– 8(4&โˆ’6@6&9)| = 36 โ€“36 = 72 M21 = |โ– 8(3&10@9&โˆ’20)| = โˆ’60 โ€“ 90 = โ€“150 M22 = |โ– 8(2&10@6&โˆ’20)| = โ€“40 โ€“ 60 = โ€“100 M23 = |โ– 8(2&3@6&9)| = 18 โ€“ 18 = 0 M31 = |โ– 8(3&10@โˆ’6&5)| = 15 + 60 = 75 M32 = |โ– 8(2&10@4&5)| = 10 โ€“ 40 = โ€“30 M33 = |โ– 8(2&3@4&โˆ’6)| = โ€“12 โ€“ 12 = โ€“24 Now, A11 = ใ€–"(โ€“1)" ใ€—^(1+1) M11 = (โ€“1)2 . 75 = 75 A12 = ใ€–"(โ€“1)" ใ€—^"1+2" M12 = ใ€–"(โ€“1)" ใ€—^3 . (โ€“110) = 110 A13 = ใ€–(โˆ’1)ใ€—^(1+3) M13 = ใ€–(โˆ’1)ใ€—^4 . (72) = 72 A21 = ใ€–(โˆ’1)ใ€—^(2+1) M21 = ใ€–(โˆ’1)ใ€—^3 . (โ€“150) = 150 A22 = ใ€–(โˆ’1)ใ€—^(2+2) M22 = (โ€“1)4 . (โ€“100) = โ€“100 A23 = ใ€–(โˆ’1)ใ€—^(2+3). M23 = ใ€–(โˆ’1)ใ€—^5. 0 = 0 A31 = ใ€–(โˆ’1)ใ€—^(3+1). M31 = ใ€–(โˆ’1)ใ€—^4 . 75 = 75 A32 = ใ€–(โˆ’1)ใ€—^(3+2) . M32 = ใ€–(โˆ’1)ใ€—^5. (โ€“30) = 30 A33 = ใ€–(โˆ’1)ใ€—^(3+3) . M33 = (โ€“1)6 . โ€“24 = โ€“24 Thus, adj A = [โ– 8(75&150&75@110&โˆ’110&30@72&0&โˆ’24)] Now, A-1 = 1/(|A|) adj A A-1 = 1/1200 [โ– 8(75&150&75@110&โˆ’110&30@72&0&โˆ’24)] Also, X = Aโˆ’1 B Putting Values [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)]= 1/1200 [โ– 8(75&150&75@110&โˆ’110&30@72&0&โˆ’24)] [โ– 8(4@1@2)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)]= 1/1200 [โ– 8(75(4)+150(1)+75(4)@110(4)+(โˆ’110)(1)+30(1)@72(4)+0(1)+(โˆ’24)2)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] = 1/1200 [โ– 8(300+150+150@440โˆ’100+60@288+0โˆ’48)] = 1/1200 [โ– 8(600@400@140)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] = [โ– 8(1/2@1/3@1/5)] Hence u = 1/2 , v = 1/3 , & w = 1/5 Thus, x = 2, y = 3 & z = 5 Putting u = ๐Ÿ/๐’™ 1/2 = 1/๐‘ฅ x = 2 Putting v = ๐Ÿ/๐’š 1/3 = 1/๐‘ฆ y = 3 Putting w = ๐Ÿ/๐’› 1/5 = 1/๐‘ง z = 5

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.