4.6 APPLICATION OF MATRICES AND DETERMINANTS .pptx
- 1. MATHEMATICS VIDEO TUTORIAL SERIES TOPIC: APPLICATION OF MATRICES AND DETERMINANTS CHAPTER – 4 DETERMINANTS
- 2. CONSISTENCY OF SYSTEM OF LINEAR EQUATIONS CONSISTENT SYSTEM A system of equations is said to be consistent if its solution (one or more) exists INCONSISTENT SYSTEM A system of equations is said to be inconsistent if its solution does not exist.
- 3. SOLUTION OF SYSTEM OF LINEAR EQUATIONS USING INVERSE OF A MATRIX A System of linear equations can be expressed as matrix equation and can be solved using the inverse of the coefficient matrix
- 4. SOLUTION OF SYSTEM OF LINEAR EQUATIONS Consider the system of linear equations in two variables Let 𝑿 = (𝒙 𝒚 )𝑩= (𝒄𝟏 𝒄𝟐 ) Then the system of equations can be written as AX = B (𝒂𝟏 𝒃𝟏 𝒂𝟐 𝒃𝟐 ) (𝒙 𝒚 )¿ (𝒄 𝟏 𝒄 𝟐 )
- 5. SOLUTION OF SYSTEM OF LINEAR EQUATIONS Consider the system of linear equations in three variables 𝑳𝒆𝒕 𝑨= ( 𝒂𝟏 𝒃𝟏 𝒄𝟏 𝒂𝟐 𝒃𝟐 𝒄𝟐 𝒂𝟑 𝒃𝟑 𝒄𝟑 )𝑿 = ( 𝒙 𝒚 𝒛 ) 𝑩= ( 𝒅𝟏 𝒅𝟐 𝒅𝟑 ) Then the system of equations can be written as ( 𝒂𝟏 𝒃𝟏 𝒄𝟏 𝒂𝟐 𝒃𝟐 𝒄𝟐 𝒂𝟑 𝒃𝟑 𝒄𝟑 )( 𝒙 𝒚 𝒛 )¿ ( 𝒅𝟏 𝒅𝟐 𝒅𝟑 ) AX = B
- 6. SOLUTION OF SYSTEM OF LINEAR EQUATIONS If A is a non-singular matrix then AX = B 𝑨−𝟏 ( 𝑨𝑿 )=𝑨−𝟏 𝑩 𝑷𝒓𝒆𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒚𝒊𝒏𝒈 𝒃𝒚 𝑨−𝟏 = 𝑩𝒚 𝑨𝒔𝒔𝒐𝒄𝒊𝒂𝒕𝒆 𝑷𝒓𝒐𝒑𝒆𝒓𝒕𝒚 = 𝑨 𝑨−𝟏 = 𝑰 I X = This matrix equation gives unique solution to the system of linear equations
- 7. SOLUTION OF SYSTEM OF LINEAR EQUATIONS If A is singular then We calculate adj(A)B If adj(A)B = O then the system will have infinite solutions If adj(A)B ≠ O then the system will have no solution
- 8. SOLUTION OF SYSTEM OF LINEAR EQUATIONS AX = B |𝑨|≠ 𝟎 The system is consistent and has unique solution |𝑨|=𝟎 adj(A)B = O The system is consistent and has many solutions |𝑨|=𝟎 adj(A)B ≠ O The system is inconsistent and has NO solutions 𝑿= 𝑨−𝟏 𝑩
- 9. SOLUTION OF SYSTEM OF LINEAR EQUATIONS Consider the system of linear equations in two variables 𝑨= (𝟑 −𝟒 𝟒 𝟐 ) 𝑿 = (𝒙 𝒚 )𝑩= (𝟖 𝟓) AX = B = 6 + 16 = 22 ≠ 0 The system is consistent and has unique solution
- 10. SOLUTION OF SYSTEM OF LINEAR EQUATIONS Consider the system of linear equations in two variables 𝑨= (𝟑 − 𝟒 𝟔 −𝟖)𝑿 = (𝒙 𝒚 )𝑩 = ( 𝟖 𝟏𝟔 ) AX = B = -24 + 24 = 0 The system is consistent and has many solutions 𝒂𝒅𝒋 ( 𝑨)= (−𝟖 𝟒 −𝟔 𝟑) ¿ (−𝟔𝟒 +𝟔𝟒 −𝟒𝟖 +𝟒𝟖) = O
- 11. SOLUTION OF SYSTEM OF LINEAR EQUATIONS Consider the system of linear equations in two variables 𝑨= (𝟑 − 𝟒 𝟔 −𝟖)𝑿 = (𝒙 𝒚 )𝑩 = (𝟖 𝟔) AX = B = -24 + 24 = 0 The system is inconsistent and has no solutions 𝒂𝒅𝒋 ( 𝑨)= (−𝟖 𝟒 −𝟔 𝟑) ¿ (−𝟔𝟒 +𝟔𝟒 −𝟒𝟖+𝟖 ) ≠ O
- 12. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 1. Examine the consistency of the system of equations : x + 2y = 2 ; 2x + 3y = 3 2 3 𝑨= (𝟏 𝟐 𝟐 𝟑) 𝑿 = (𝒙 𝒚 )𝑩= (𝟐 𝟑) AX = B = 3 – 4 = - 1 ≠ 0 Then the system of equations can be written as 𝑨𝒔|𝑨|≠ 𝟎 , 𝑨−𝟏 𝒆𝒙𝒊𝒔𝒕𝒔 The system is consistent and has unique solution
- 13. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 2. Examine the consistency of the system of equations : 4 5 𝑨= (𝟐 − 𝟏 𝟏 𝟏 )𝑿 = (𝒙 𝒚 )𝑩= (𝟓 𝟒) AX = B = 2 + 1= 3 ≠ 0 Then the system of equations can be written as 𝑨𝒔|𝑨|≠ 𝟎 , 𝑨−𝟏 𝒆𝒙𝒊𝒔𝒕𝒔 The system is consistent and has unique solution
- 14. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 3. Examine the consistency of the system of equations : 8 5 𝑨= (𝟏 𝟑 𝟐 𝟔) 𝑿 = (𝒙 𝒚 )𝑩= (𝟓 𝟖) AX = B = 6 – 6 = 0 Then the system of equations can be written as The system is inconsistent and has no solution 𝒂𝒅𝒋 ( 𝑨)= ( 𝟔 − 𝟑 −𝟐 𝟏 ) ¿ (𝟑 𝟎 −𝟐𝟒 −𝟏𝟎+ 𝟔 ) ≠ O
- 15. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 4. Examine the consistency of the system of equations : 𝑨= ( 𝟏 𝟏 𝟏 𝟐 𝟑 𝟐 𝒂 𝒂 𝟐 𝒂)𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟏 𝟐 𝟒) AX = B Then the system of equations can be written as |𝑨|= | 𝟏 𝟏 𝟏 𝟐 𝟑 𝟐 𝒂 𝒂 𝟐 𝒂| ¿ 𝟏 ×|𝟑 𝟐 𝒂 𝟐 𝒂|−𝟏 ×|𝟐 𝟐 𝒂 𝟐 𝒂|+𝟏×|𝟐 𝟑 𝒂 𝒂| ¿(𝟔 𝒂−𝟐𝒂)− (𝟒𝒂− 𝟐𝒂)+(𝟐𝒂−𝟑 𝒂) 𝑨𝒔|𝑨|≠𝟎, 𝑨−𝟏 𝒆𝒙𝒊𝒔𝒕𝒔 The system is consistent and has unique solution
- 16. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 5. Examine the consistency of the system of equations : 3 𝑨= ( 𝟑 −𝟏 −𝟐 𝟎 𝟐 −𝟏 𝟑 −𝟓 𝟎 )𝑿= ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟏 𝟐 𝟑) AX = B Then the system of equations can be written as 𝑨|= | 𝟑 −𝟏 −𝟐 𝟎 𝟐 −𝟏 𝟑 −𝟓 𝟎 | ¿ 𝟑 ×| 𝟐 − 𝟏 −𝟓 𝟎 |+𝟑 ×|−𝟏 −𝟐 𝟐 −𝟏| Expanding along ¿𝟑(𝟎– 𝟓)+𝟑(𝟏+𝟒) 15 0 𝑨𝒊𝒔 𝒂𝒔𝒊𝒏𝒈𝒖𝒍𝒂𝒓 𝒎𝒂𝒕𝒓𝒊𝒙 𝒂𝒏𝒅 𝑨−𝟏 𝒅𝒐𝒆𝒔𝒏𝒐𝒕𝒆𝒙𝒊𝒔𝒕
- 17. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒂𝒅𝒋 ( 𝑨 ) 𝑩 = 0 - 5 = -5 = - (0 + 3) = - 3 = 0 - 6 = -6 = -(0 -10) = 10 = 0 + 6 = 6 = -(-15 + 3) = 12 = 1 + 4 = 5 = -(-3 - 0) = 3 = 6 – 0 = 6 𝑨= ( 𝟑 −𝟏 −𝟐 𝟎 𝟐 −𝟏 𝟑 −𝟓 𝟎 )
- 18. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 -5 - 3 -6 = 10 6 12 5 3 6 𝑨𝒊𝒋 = ( −𝟓 −𝟑 − 𝟔 𝟏𝟎 𝟔 𝟏𝟐 𝟓 𝟑 𝟔 ) 𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( −𝟓 𝟏𝟎 𝟓 −𝟑 𝟔 𝟑 −𝟔 𝟏𝟐 𝟔) 𝒂𝒅𝒋 ( 𝑨) × 𝑩= ( − 𝟓 𝟏𝟎 𝟓 − 𝟑 𝟔 𝟑 − 𝟔 𝟏𝟐 𝟔)( 𝟐 − 𝟏 𝟑 ) ¿ ( −𝟏𝟎 − 𝟏𝟎+𝟏𝟓 −𝟔 −𝟔+ 𝟗 −𝟏𝟐 − 𝟏𝟐+𝟏𝟖) ¿ ( − 𝟓 − 𝟑 − 𝟔) ≠ O The system is inconsistent and has no solution
- 19. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 6. Examine the consistency of the system of equations : 5 5-1 𝑨= ( 𝟓 − 𝟏 𝟒 𝟐 𝟑 𝟓 𝟓 − 𝟐 𝟔)𝑿= ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟓 𝟐 −𝟏) AX = B Then the system of equations can be written as |𝑨|= | 𝟓 −𝟏 𝟒 𝟐 𝟑 𝟓 𝟓 −𝟐 𝟔| ¿ 𝟓 ×| 𝟑 𝟓 −𝟐 𝟔|−(−𝟏)|𝟐 𝟓 𝟓 𝟔|+𝟒 ×|𝟐 𝟑 𝟓 − 𝟐| =5(18 + 10) +(12 - 25)+4(-4 - 15) =140 – 13 - 76 =51 ≠0𝑨𝒔|𝑨|≠𝟎, 𝑨−𝟏 𝒆𝒙𝒊𝒔𝒕𝒔 The system is consistent and has unique solution
- 20. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 7. Solve the system of equations using matrix method 4 𝑨= (𝟓 𝟐 𝟕 𝟑) 𝑿 = (𝒙 𝒚 )𝑩= (𝟒 𝟓) = 15 – 14 = 1 Then the system of equations can be written as 𝒂𝒅𝒋 ( 𝑨)= ( 𝟑 − 𝟐 −𝟕 𝟓 ) AX = B 𝑿=𝑨−𝟏 𝑩 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋 ( 𝑨)= 𝟏 𝟏 ( 𝟑 −𝟐 −𝟕 𝟓 ) 𝑿=𝑨−𝟏 𝑩 (𝒙 𝒚 )= ( 𝟑 − 𝟐 −𝟕 𝟓 )(𝟒 𝟓)(𝒙 𝒚 )= ( 𝟏𝟐 −𝟏𝟎 −𝟐𝟖+𝟐𝟓 ) (𝒙 𝒚 )= ( 𝟐 −𝟑) 𝒙=𝟐; 𝒚 =−𝟑
- 21. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 8. Solve the system of equations using matrix method 3 -2 3 𝑨= (𝟐 − 𝟏 𝟑 𝟒 )𝑿= (𝒙 𝒚 )𝑩= (−𝟐 𝟑 ) = 8 + 3 = 11 Then the system of equations can be written as 𝒂𝒅𝒋 ( 𝑨)= ( 𝟒 𝟏 −𝟑 𝟐) AX = B 𝑿=𝑨−𝟏 𝑩 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋 ( 𝑨)= 𝟏 𝟏𝟏 ( 𝟒 𝟏 −𝟑 𝟐) 𝑿=𝑨−𝟏 𝑩(𝒙 𝒚 )= 𝟏 𝟏𝟏 ( 𝟒 𝟏 −𝟑 𝟐)(−𝟐 𝟑 )(𝒙 𝒚 )= 𝟏 𝟏𝟏 (−𝟖+𝟑 𝟔+𝟔 ) (𝒙 𝒚 )= 𝟏 𝟏𝟏 (−𝟓 𝟏𝟐) 𝒙=− 𝟓 𝟏𝟏 ; 𝒚 = 𝟏𝟐 𝟏𝟏
- 22. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 9. Solve the system of equations using matrix method 7 3 7 𝑨= (𝟒 −𝟑 𝟑 −𝟓)𝑿 = (𝒙 𝒚 )𝑩= (𝟑 𝟕) = -20 + 9 = - 11 Then the system of equations can be written as 𝒂𝒅𝒋 ( 𝑨)= (−𝟓 𝟑 −𝟑 𝟒) AX = B 𝑿=𝑨−𝟏 𝑩 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋 ( 𝑨)=− 𝟏 𝟏𝟏 (− 𝟓 𝟑 − 𝟑 𝟒) 𝑿=𝑨−𝟏 𝑩(𝒙 𝒚 )=− 𝟏 𝟏𝟏 (−𝟓 𝟑 −𝟑 𝟒)(𝟑 𝟕)(𝒙 𝒚 )=− 𝟏 𝟏𝟏 (−𝟏𝟓+𝟐𝟏 −𝟗+𝟐𝟖 ) (𝒙 𝒚 )=− 𝟏 𝟏𝟏 ( 𝟔 𝟏𝟗) 𝒙=− 𝟔 𝟏𝟏 ; 𝒚=− 𝟏𝟗 𝟏𝟏
- 23. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 10. Solve the system of equations using matrix method 5 3 5 𝑨= (𝟓 𝟐 𝟑 𝟐) 𝑿 = (𝒙 𝒚 )𝑩= (𝟑 𝟓) = 10 – 6 = 4 Then the system of equations can be written as 𝒂𝒅𝒋 ( 𝑨)= ( 𝟐 − 𝟐 −𝟑 𝟓 ) AX = B 𝑿=𝑨−𝟏 𝑩 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋 ( 𝑨)= 𝟏 𝟒 ( 𝟐 −𝟐 −𝟑 𝟓 ) 𝑿=𝑨−𝟏 𝑩(𝒙 𝒚 )= 𝟏 𝟒 ( 𝟐 −𝟐 −𝟑 𝟓 )(𝟑 𝟓)(𝒙 𝒚 )= 𝟏 𝟒 ( 𝟔− 𝟏𝟎 −𝟗+ 𝟐𝟓) (𝒙 𝒚 )= 𝟏 𝟒 (−𝟒 𝟏𝟔) 4
- 24. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 11. Solve the system of equations using matrix method ; 𝑨= ( 𝟐 𝟏 𝟏 𝟏 −𝟐 −𝟏 𝟎 𝟑 −𝟓)𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟏 𝟑 𝟐 𝟗 ) Then the system of equations can be written as AX = B 𝑿=𝑨−𝟏 𝑩 |𝑨|= | 𝟐 𝟏 𝟏 𝟏 −𝟐 −𝟏 𝟎 𝟑 −𝟓| ¿ 𝟐 ×|−𝟐 − 𝟏 𝟑 − 𝟓|−𝟏 ×|𝟏 𝟏 𝟑 − 𝟓| Expanding along = 2 (10 + 3) -1 (-5 -3) = 2 (13) -1 (-8) = 34 |𝑨|=𝟑𝟒
- 25. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝑨−𝟏 = 10 +3 = 13 = - (-5 - 0) = 5 = 3 – 0 = 3 = -(-5 - 3) = 8 = -10 - 0 = -10 = -(6 - 0) = -6 = -1 + 2 = 1 = -(-2 - 1) = 3 = - 4 -1 = - 5 𝑨= ( 𝟐 𝟏 𝟏 𝟏 −𝟐 −𝟏 𝟎 𝟑 −𝟓)
- 26. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 13 5 3 = 8 -10 -6 1 3 -5 𝑨𝒊𝒋= ( 𝟏 𝟑 𝟓 𝟑 𝟖 −𝟏𝟎 − 𝟔 𝟏 𝟑 − 𝟓) 𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( 𝟏 𝟑 𝟖 𝟏 𝟓 −𝟏𝟎 𝟑 𝟑 −𝟔 − 𝟓) 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋( 𝑨) 𝑨 −𝟏 = 𝟏 𝟑𝟒 ( 𝟏 𝟑 𝟖 𝟏 𝟓 −𝟏𝟎 𝟑 𝟑 − 𝟔 −𝟓)
- 27. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑨 −𝟏 = 𝟏 𝟑𝟒 ( 𝟏 𝟑 𝟖 𝟏 𝟓 −𝟏𝟎 𝟑 𝟑 − 𝟔 −𝟓) 𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟏 𝟑 𝟐 𝟗 ) 𝑿=𝑨−𝟏 𝑩 = = = = 𝒙=𝟏 ; 𝒚 = 𝟏 𝟐 ; 𝒛 =− 𝟑 𝟐
- 28. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 12. Solve the system of equations using matrix method 0 ; 0 𝑨= ( 𝟏 − 𝟏 𝟏 𝟐 𝟏 − 𝟑 𝟏 𝟏 𝟏 )𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟒 𝟎 𝟐) Then the system of equations can be written as AX = B 𝑿=𝑨−𝟏 𝑩 |𝑨|= | 𝟏 −𝟏 𝟏 𝟐 𝟏 −𝟑 𝟏 𝟏 𝟏 | ¿ 𝟏 ×|𝟏 −𝟑 𝟏 𝟏 |−(−𝟏 )×|𝟐 − 𝟑 𝟏 𝟏 |+𝟏 ×|𝟐 𝟏 𝟏 𝟏 = 1 (1 + 3) +1 (2 + 3) + 1 (2 - 1) = 4 + 5 + 1 = 10 |𝑨|=𝟏𝟎
- 29. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝑨−𝟏 = 1 + 3 = 4 = - (2 + 3) = -5 = 2 – 1 = 1 = -(-1 -1) = 2 = 1 – 1 = 0 = -(1 +1) = -2 = 3 – 1 = 2 = -(-3 - 2) = 5 = 1 + 3 = 4 𝑨= ( 𝟏 − 𝟏 𝟏 𝟐 𝟏 − 𝟑 𝟏 𝟏 𝟏 )
- 30. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 4 - 5 1 = 2 0 - 2 2 5 4 𝑨𝒊𝒋 = ( 𝟒 −𝟓 𝟏 𝟐 𝟎 −𝟐 𝟐 𝟓 𝟒 ) 𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( 𝟒 𝟐 𝟐 −𝟓 𝟎 𝟓 𝟏 −𝟐 𝟒) 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋( 𝑨) 𝑨 −𝟏 = 𝟏 𝟏𝟎 ( 𝟒 𝟐 𝟐 − 𝟓 𝟎 𝟓 𝟏 −𝟐 𝟒)
- 31. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑨 −𝟏 = 𝟏 𝟏𝟎 ( 𝟒 𝟐 𝟐 − 𝟓 𝟎 𝟓 𝟏 −𝟐 𝟒) 𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟒 𝟎 𝟐) 𝑿=𝑨−𝟏 𝑩 = = = = 1
- 32. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 13. Solve the system of equations using matrix method -4 ; - 4 3 𝑨= ( 𝟐 𝟑 𝟑 𝟏 − 𝟐 𝟏 𝟑 − 𝟏 − 𝟐)𝑿= ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟓 −𝟒 𝟑 ) Then the system of equations can be written as AX = B 𝑿=𝑨−𝟏 𝑩 |𝑨|= | 𝟐 𝟑 𝟑 𝟏 −𝟐 𝟏 𝟑 −𝟏 −𝟐| ¿𝟐×|−𝟐 𝟏 −𝟏 − 𝟐|−𝟑×|𝟏 𝟏 𝟑 − 𝟐|+𝟑×|𝟏 − 𝟐 𝟑 − 𝟏| = 2 (4 + 1) - 3 (-2 - 3) + 3 (-1 +6) = 10 + 15 + 15 = 40 |𝑨|=𝟒𝟎
- 33. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝑨−𝟏 = 4 + 1 = 5 = - (-2 - 3) = 5 = -1 + 6 = 5 = -(-6 + 3) = 3 = -4 – 9 = - 13 = -(-2 – 9 ) = 11 = 3 + 6 = 9 = -(2 – 3) = 1 = - 4 – 3 = - 7 𝑨= ( 𝟐 𝟑 𝟑 𝟏 − 𝟐 𝟏 𝟑 − 𝟏 − 𝟐)
- 34. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 5 5 5 = 3 - 13 11 9 1 -7 𝑨𝒊𝒋= ( 𝟓 𝟓 𝟓 𝟑 − 𝟏𝟑 𝟏𝟏 𝟗 𝟏 − 𝟕) 𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( 𝟓 𝟑 𝟗 𝟓 − 𝟏𝟑 𝟏 𝟓 𝟏𝟏 − 𝟕) 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋( 𝑨) 𝑨 −𝟏 = 𝟏 𝟒𝟎 ( 𝟓 𝟑 𝟗 𝟓 −𝟏𝟑 𝟏 𝟓 𝟏𝟏 −𝟕)
- 35. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑨 −𝟏 = 𝟏 𝟒𝟎 ( 𝟓 𝟑 𝟗 𝟓 −𝟏𝟑 𝟏 𝟓 𝟏𝟏 −𝟕) 𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟓 − 𝟒 𝟑 ) 𝑿=𝑨−𝟏 𝑩 = = = = -1
- 36. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 14. Solve the system of equations using matrix method -5 ; - 5 𝑨= ( 𝟏 − 𝟏 𝟐 𝟑 𝟒 − 𝟓 𝟐 − 𝟏 𝟑 )𝑿= ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟕 −𝟓 𝟏𝟐) Then the system of equations can be written as AX = B 𝑿=𝑨−𝟏 𝑩 |𝑨|= | 𝟏 −𝟏 𝟐 𝟑 𝟒 −𝟓 𝟐 −𝟏 𝟑 | ¿ 𝟏 ×| 𝟒 − 𝟓 −𝟏 𝟑 |−(−𝟏)×|𝟑 −𝟓 𝟐 𝟑 |+𝟐 ×|𝟑 𝟒 𝟐 −𝟏 = 1 (12 – 5) + 1 (9 + 10) + 2 (- 3 – 8) = 7 + 19 - 22 = 4 4
- 37. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝑨−𝟏 = 12 – 5 = 7 = - (9 + 10) = - 19 = - 3 – 8 = -11 = -(- 3 + 2) = 1 = 3 – 4 = -1 = -(-1 + 2) = - 1 = 5 – 8 = -3 = -(-5 - 6) = 11 = 4 + 3 = 7 𝑨= ( 𝟏 − 𝟏 𝟐 𝟑 𝟒 − 𝟓 𝟐 − 𝟏 𝟑 )
- 38. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 7 -19 -11 = 1 - 1 -1 -3 11 7 𝑨𝒊𝒋= ( 𝟕 −𝟏𝟗 −𝟏𝟏 𝟏 − 𝟏 − 𝟏 −𝟑 𝟏𝟏 𝟕 ) 𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( 𝟕 𝟏 −𝟑 −𝟏𝟗 −𝟏 𝟏𝟏 −𝟏𝟏 −𝟏 𝟕 ) 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋( 𝑨) 𝑨 −𝟏 = 𝟏 𝟒 ( 𝟕 𝟏 − 𝟑 − 𝟏𝟗 − 𝟏 𝟏𝟏 − 𝟏𝟏 − 𝟏 𝟕 )
- 39. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑨 −𝟏 = 𝟏 𝟒 ( 𝟕 𝟏 − 𝟑 − 𝟏𝟗 − 𝟏 𝟏𝟏 − 𝟏𝟏 − 𝟏 𝟕 ) 𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟕 − 𝟓 𝟏𝟐 ) 𝑿=𝑨−𝟏 𝑩 = = = = 3
- 40. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 . 𝑨= ( 𝟐 − 𝟑 𝟓 𝟑 𝟐 − 𝟒 𝟏 𝟏 − 𝟐) |𝑨|= | 𝟐 −𝟑 𝟓 𝟑 𝟐 −𝟒 𝟏 𝟏 −𝟐| ¿ 𝟐 ×|𝟐 − 𝟒 𝟏 − 𝟐|−(−𝟑)×|𝟑 −𝟒 𝟏 −𝟐|+𝟓 ×|𝟑 𝟐 𝟏 𝟏| = 2 (-4 + 4) + 3 (-6 + 4) + 5 ( 3 – 2) = 0 - 6 + 5 = -1 -1
- 41. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝑨−𝟏 = -4 + 4 = 0 = - (-6 + 4) = 2 = 3 – 2 = 1 = -(6 – 5) = - 1 = -4 – 5 = - 9 = -(2 + 3) = - 5 = 12 – 10 = 2 = -(-8 – 15) = 23 = 4 + 9 = 13 𝑨= ( 𝟐 − 𝟑 𝟓 𝟑 𝟐 − 𝟒 𝟏 𝟏 − 𝟐)
- 42. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 0 2 1 = -1 - 9 -5 2 23 13 𝑨𝒊𝒋= ( 𝟎 𝟐 𝟏 −𝟏 −𝟗 −𝟓 𝟐 𝟐𝟑 𝟏𝟑 )𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( 𝟎 −𝟏 𝟐 𝟐 −𝟗 𝟐𝟑 𝟏 −𝟓 𝟏𝟑) 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋( 𝑨) 𝑨 −𝟏 = 𝟏 −𝟏 ( 𝟎 −𝟏 𝟐 𝟐 −𝟗 𝟐𝟑 𝟏 −𝟓 𝟏𝟑) 𝑨 −𝟏 = (− 𝟎 𝟏 − 𝟐 𝟐 𝟗 −𝟐𝟑 − 𝟏 𝟓 −𝟏𝟑)
- 43. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝟐 𝒙−𝟑 𝒚 +𝟓 𝒛=𝟏𝟏;𝟑 𝒙+𝟐 𝒚 −𝟒 𝒛=−𝟓; 𝒙+ 𝒚 −𝟐 𝒛=−𝟑 𝑨= ( 𝟐 − 𝟑 𝟓 𝟑 𝟐 − 𝟒 𝟏 𝟏 − 𝟐)𝑿= ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟏𝟏 − 𝟓 − 𝟑) Then the system of equations can be written as AX = B 𝑿=𝑨−𝟏 𝑩 𝑿=𝑨−𝟏 𝑩 = = = 3
- 44. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 16.The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method. Let the cost of onions, wheat, and rice per kg be Rs., Rs. and Rs. respectively. Then, the given situation can be represented by a system of equations as:
- 45. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 90 6 𝑨= ( 𝟒 𝟑 𝟐 𝟐 𝟒 𝟔 𝟔 𝟐 𝟑)𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟔 𝟎 𝟗𝟎 𝟕𝟎 ) Then the system of equations can be written as AX = B 𝑿=𝑨−𝟏 𝑩 |𝑨|= | 𝟒 𝟑 𝟐 𝟐 𝟒 𝟔 𝟔 𝟐 𝟑| ¿ 𝟒 ×|𝟒 𝟔 𝟐 𝟑|−𝟑 ×|𝟐 𝟔 𝟔 𝟑|+𝟐 ×|𝟐 𝟒 𝟔 𝟐| = 4 (12 – 12) - 3 (6 – 36) + 2 (4 – 24) = 0 + 90 - 40 = 50 |𝑨|=𝟓𝟎
- 46. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝑨−𝟏 = 12 – 12 = 0 = - (6 – 36) = 30 = 4 – 24 = - 20 = -(9 – 4) = -5 = 12 – 12 = 0 = -(8 - 18) = 10 = 18 – 8 = 10 = -(24 – 4) = - 20 = 16 – 6 = 10 𝑨= ( 𝟒 𝟑 𝟐 𝟐 𝟒 𝟔 𝟔 𝟐 𝟑)
- 47. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 0 30 - 20 = -5 0 10 10 -20 10 𝑨𝒊𝒋= ( 𝟎 𝟑𝟎 −𝟐𝟎 −𝟓 𝟎 𝟏𝟎 𝟏𝟎 −𝟐𝟎 𝟏𝟎 )𝒂𝒅𝒋 ( 𝑨)=( 𝑨𝒊𝒋) ′ = ( 𝟎 −𝟓 𝟏𝟎 𝟑𝟎 𝟎 −𝟐𝟎 −𝟐𝟎 𝟏𝟎 𝟏𝟎 ) 𝑨 −𝟏 = 𝟏 |𝑨| 𝒂𝒅𝒋( 𝑨)𝑨 −𝟏 = 𝟏 𝟓𝟎 ( 𝟎 − 𝟓 𝟏𝟎 𝟑𝟎 𝟎 − 𝟐𝟎 − 𝟐𝟎 𝟏𝟎 𝟏𝟎 )
- 48. SOLUTIONS TO CLASS XII NCERT TEXT BOOK QUESTIONS EXERCISE 4.6 𝑨 −𝟏 = 𝟏 𝟓𝟎 ( 𝟎 − 𝟓 𝟏𝟎 𝟑𝟎 𝟎 − 𝟐𝟎 − 𝟐𝟎 𝟏𝟎 𝟏𝟎 ) 𝑿 = ( 𝒙 𝒚 𝒛 )𝑩= ( 𝟔 𝟎 𝟗𝟎 𝟕𝟎 ) 𝑿=𝑨−𝟏 𝑩 = = = = 8 Cost of onion = Rs.5 per kg Cost of Wheat = Rs.8 per kg Cost of Rice = Rs.8 per kg