Chapter 4 Class 12 Determinants
Chapter 4 Class 12 Determinants
Last updated at December 16, 2024 by Teachoo
Transcript
Misc 7 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 š/š = u , š/š = v , & š/š = w The system of equations become 2u + 3v + 10w = 4 4u ā 6v + 5w = 1 6u + 9v ā 20w = 2 Writing equation as AX = B [ā 8(2&3&10@4&ā6&5@6&9&ā20)] [ā 8(š¢@š£@š¤)] = [ā 8(4@1@2)] Hence A = [ā 8(2&3&10@4&ā6&5@6&9&ā20)] , X = [ā 8(š¢@š£@š¤)] & B = [ā 8(4@1@2)] Calculating |A| |A| = |ā 8(2&3&10@4&ā6&5@6&9&ā20)| = 2 |ā 8(ā6&5@9&ā20)| ā 3 |ā 8(4&5@6&ā20)| + 10 |ā 8(4&ā6@6&9)| = 2 (120 ā 45) ā3 (ā80 ā 30) + 10 ( 36 + 36) = 2 (75) ā3 (ā110) + 10 (72) = 150 + 330 + 720 = 1200 ā“ |A|ā 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) = [ā 8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^ā² = [ā 8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [ā 8(2&3&10@4&ā6&5@6&9&ā20)] M11 = |ā 8(ā6&5@9&ā20)| = 120 ā 45 = 75 M12 = |ā 8(4&5@6&ā20)| = (ā80 ā 30) = ā110 M13 = |ā 8(4&ā6@6&9)| = 36 ā36 = 72 M21 = |ā 8(3&10@9&ā20)| = ā60 ā 90 = ā150 M22 = |ā 8(2&10@6&ā20)| = ā40 ā 60 = ā100 M23 = |ā 8(2&3@6&9)| = 18 ā 18 = 0 M31 = |ā 8(3&10@ā6&5)| = 15 + 60 = 75 M32 = |ā 8(2&10@4&5)| = 10 ā 40 = ā30 M33 = |ā 8(2&3@4&ā6)| = ā12 ā 12 = ā24 Now, A11 = ć"(ā1)" ć^(1+1) M11 = (ā1)2 . 75 = 75 A12 = ć"(ā1)" ć^"1+2" M12 = ć"(ā1)" ć^3 . (ā110) = 110 A13 = ć(ā1)ć^(1+3) M13 = ć(ā1)ć^4 . (72) = 72 A21 = ć(ā1)ć^(2+1) M21 = ć(ā1)ć^3 . (ā150) = 150 A22 = ć(ā1)ć^(2+2) M22 = (ā1)4 . (ā100) = ā100 A23 = ć(ā1)ć^(2+3). M23 = ć(ā1)ć^5. 0 = 0 A31 = ć(ā1)ć^(3+1). M31 = ć(ā1)ć^4 . 75 = 75 A32 = ć(ā1)ć^(3+2) . M32 = ć(ā1)ć^5. (ā30) = 30 A33 = ć(ā1)ć^(3+3) . M33 = (ā1)6 . ā24 = ā24 Thus, adj A = [ā 8(75&150&75@110&ā110&30@72&0&ā24)] Now, A-1 = 1/(|A|) adj A A-1 = š/šššš [ā 8(šš&ššš&šš@ššš&āššš&šš@šš&š&āšš)] Also, X = Aā1 B Putting Values [ā 8(š¢@š£@š¤)]= 1/1200 [ā 8(75&150&75@110&ā110&30@72&0&ā24)] [ā 8(4@1@2)] [ā 8(š¢@š£@š¤)]= 1/1200 [ā 8(75(4)+150(1)+75(4)@110(4)+(ā110)(1)+30(1)@72(4)+0(1)+(ā24)2)] [ā 8(š¢@š£@š¤)] = 1/1200 [ā 8(300+150+150@440ā100+60@288+0ā48)] = 1/1200 [ā 8(600@400@140)] [ā 8(š@š@š)] = [ā 8(š/š@š/š@š/š)] Hence u = 1/2 , v = 1/3 , & w = 1/5 Thus, x = 2, y = 3 & z = 5 Putting u = š/š 1/2 = 1/š„ x = 2 Putting v = š/š 1/3 = 1/š¦ y = 3 Putting w = š/š 1/5 = 1/š§ z = 5