# Misc. 16

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc. 16 Solve the system of the following equations 2x + 3y + 10z = 4 4x + 6y + 5z = 1 6x + 9y + 20z = 2 The system of equations are 2x + 3y + 10z = 4 4x + 6y + 5z = 1 6x + 9y + 20z = 2 Now let 𝟏𝒙 = u , 𝟏𝒚 = v , & 𝟏𝒛 = w The system of equations become 2u + 3v + 10w = 4 4u – 6v + 5w = 1 6u + 9v – 20w = 2 Step 1 Write equation as AX = B 23104−6569−20 𝑢𝑣𝑤 = 412 Hence A = 23104−6569−20 , X = 𝑢𝑣𝑤 & B = 412 Step 2 Calculate |A| |A| = 23104−6569−20 = 2 −659−20 – 3 456−20 + 10 4−669 = 2 ( 120 – 45) – 3 ( – 80 – 30) + 10 ( 36 + 36) = 2 (75) – 3 (– 110) + 10 (72) = 150 + 330 + 720 = 1200 ∴ |A|≠ 0 So, the system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculating X = A-1 B Now, A-1 = 1|A| adj (A) adj (A) = A11A12A13A21A22A23A31A32A33′ = A11A21A31A12A22A32A13A23A33 A = 23104−6569−20 M11 = −659−20 = 120 – 45 = 75 M12 = 456−20 = ( – 80 – 30) = – 110 M13 = 4−669 = 36 – 36 = 72 M21 = 3109−20 = − 60 – 90 = – 150 M22 = 2106−20 = – 40 – 60 = – 100 M23 = 2369 = 18 – 18 = 0 M31 = 310−65 = 15 + 60 = 75 M32 = 21045 = 10 – 40 = – 30 M33 = 234−6 = – 12 – 12 = – 24 A11 = ( –1)1+1 M11 = ( –1)2 . 75 = 75 A12 = ( –1)1+2 M12 = ( –1)3 . ( – 110) = 110 A13 = ( −1)1+3 M13 = ( −1)4 . (72) = 72 A21 = ( −1)2+1 M21 = ( −1)3 . ( – 150) = 150 A22 = ( −1)2+2 M22 = ( –1)4 . ( – 100) = – 100 A23 = (−1)2+3. M23 = (−1)5. 0 = 0 A31 = (−1)3+1. M31 = (−1)4 . 75 = 75 A32 = (−1)3+2 . M32 = (−1)5. ( – 30) = 30 A33 = (−1)3+3 . M33 = ( –1)6 . – 24 = – 24 Thus adj A = 7515075110−11030720−24 Now, A-1 = 1|A| adj A A-1 = 11200 7515075110−11030720−24 X = A-1 B Putting Values 𝑢𝑣𝑤= 11200 7515075110−11030720−24 412 𝑢𝑣𝑤= 11200 75(4)+150(1)+75(4)110(4)+(−110)(1)+30(1)72(4)+0(1)+ −242 𝑢𝑣𝑤 = 11200 300+150+150440−100+60288+0−48 = 11200 600400140 ∴ 𝑢𝑣𝑤 = 12 13 15 Hence u = 12 , v = 13 , & w = 15 Hence x = 2, y = 3 & z = 5

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Ex 4.6, 13 Important

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Example 34 Important

Misc. 2 Important

Misc 11 Important

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Misc. 16 Important You are here

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Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.