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system linear equations
- 1. Chapter 02 SYSTEM OF LINEAR EQUATIONS BY SANAULLAH MEMON
- 2. SYSTEM OF LINEAR EQUATIONS “Non-homogeneous System of Linear Equations”. n.The systemof m linearequationsinnvariables/unknownsisgivenby: a11 x1 + a12 x2 + . .. + a1n xn = b1 a21 x1 + a22 x2 + . . .+ a2n xn = b2 ………………………………. ………………………………. am1 x1 + am2 x2 + .. . + amn xn = bm ai j and bj fori = 1, 2, …, m andj = 1, 2, … ,n are constants. Thenumbersaij arecalledthecoefficientsofthesystem 11
- 3. Homogeneous System of Linear Equations a11 x1 + a12 x2 + . . .+ a1n xn = 0 a21 x1 + a22 x2 + . . .+ a2n xn = 0 ………………………………. ( ………………………………. am1 x1 + am2 x2 + .. . + amn xn = 0 2
- 4. SOLUTION OF Non- Homogeneous System of Linear Equations
- 5. SOLUTION OF Non-homogeneous System of Linear Equations A sequence of n numbers s1, s2, s3, …sn, for which (1) is satisfied when we substitute x1 = s1, x2 = s2, …,xn = sn, is called a solution of (1). The set of all such solutions is called solution set or the general solution of (1). A system of equations that has no solution is said to be inconsistent. If there is at least one solution of the system, the system is called consistent system.
- 6. SOLUTION OF Homogeneous System of Linear Equations
- 7. SOLUTION OF Homogeneous System of Linear Equations • The system (2) always possesses one solution, namely the zero solution that is x1 = 0, x2 = 0, … xn = 0. The zero solution is also known as trivial solution. Any other solution, if it exists, is called a non-zero or non –trivial solution. In coming sections we shall learn under what conditions the solution of systems (1) and (2) exist.
- 9. Matrix form of homo 0R 2
- 10. NOTE (i) In place of variables x1, x2, …, xn if we use other variables say u1, u2, …, un there will be no effect on the solution. These variables are dummy variables and are immaterial. However, if coefficient aij or bi is changed the solution of the system will be different. (ii) If m = n the system is called square system.
- 13. METHODS OF SOLVING SYSTEM OF LINEAR EQUATIONS This Photo by Unknown Author is licensed under CC BY
- 14. Ax = b Ax = b
- 15. A x = 0. A x = 0.
- 16. • Non-Homogeneous System of Linear Equations ➢(i) Gauss Elimination Method ➢ (ii) Gauss-Jordan Method
- 17. METHODS ➢(i) Gauss Elimination Method ➢ (ii) Gauss-Jordan Method.
- 18. Gauss Elimination Method Non-Homogeneous System of Linear Equations
- 19. Augmented matrix 𝐴 𝑏= Non-Homogeneous System of Linear Equations
- 20. Non- Homogeneous System of Linear Equations Step1.Changethe system oflinearequationstotheformA x=b. Step2.Form the augmented coefficient matrix Ab by including the elements of b as an extracolumninthe matrixA. Step3.Convert the augmented matrix into echelon form by using elementary row operations. Step4.Convert the echelon matrix into matrix form and find x by using “Backward Substitution”.
- 21. Note •A system A x = b is called over-determined if it has more equations than unknowns i.e. m > n. It is called determined if the number of equations is equal to the number of unknowns, i.e. m = n and underdetermined if m < n.
- 23. Use Gauss’s elimination method to solve the system of linear equations (m = n). x1 + 5 x2 + 2 x 3 = 9 x1 + x2 + 7 x 3 = 6 - 3 x2 + 4 x 3 = -2 Solution: Step1. We change the system of linear equations in matrix form A x = b. − = − 2 6 9 3x 2x 1x 430 711 251 Step2. The augmented coefficient matrix bA is: −− = 2430 6711 9251 bA Step3. Now we convert this augmented coefficient matrix into an echelon form using elementary row operations as follows: ( ) 1R12R 2430 6711 9251 bA −+ −− = ( ) 2R13R 2430 3540 9251 −+ −− −− 3R42R 1110 3540 9251 + − −− 23R 1110 1100 9251 − − 1100 1110 9251 Step4. Convert the echelon matrix into equation form =− 1 1 9 3x 2x 1x 100 110 251 This is equivalent to: x1 +5 x2 + 2 x 3 = 9 (i) x2 - x 3 = 1 (ii) x 3 = 1 (iii) From (iii), we get 13x = . Substituting 13x = into (ii), we get, = 22x Now put 13x = and 22x = in (i), we get −= 31x Thus, required solution of given system of equations is: ==−= 13x,22x,31x
- 25. Gauss-Jordan Method we reduces the augmented matrix into “Reduced Echelon” form
- 26. Method ➢ Step1. Change the system of linear equations to the form Ax = b. ➢Step2. Form the augmented coefficient matrix Ab by including the column vector b. ➢Step3. Convert the augmented matrix into “Reduced Echelon” form by using elementary row operations. ➢Step4. Convert the reduced echelon matrix into system of equations and find x directly without using backward substitution.
- 27. Use Gauss-Jordan method to solve the system of linear equations. 234x23x 637x2x1x 932x25x1x −=+− =++ =++ Step1. We change the system of linear equations in matrix form = bAx − = − 2 6 9 3x 2x 1x 430 711 251 Step2. The augmented matrix is −− = 2430 6711 9251 A b . Step3. We convert this augmented coefficient matrix into reduced echelon form elementary row operations. ( ) 1R12R 2430 6711 9251 bA −+ −− = ( ) 2R13R 2430 3540 9251 −+ −− −− 3R42R 1110 3540 9251 + − −− 23R 1110 1100 9251 − ( ) 2R51R 1100 1110 9251 −+− ( ) 3R2R,3R71R 1100 1110 4701 +−+− − 1100 2010 3001 Step4. Writing this matrix into equation form, we have: ( ) ( ) ( )iii13x2x01x0 ii23x02x1x0 i33x02x01x =++ =++ −=++ ==−= 13x,22x,31x
- 28. ( ) ( ) 3a3xa12x1x 2a3x2xa11x a3x2x1x =+++ =+++ =++ Solution: Step1. We change the system of linear equations in matrix form = bAx = + + a3 a2 a 3x 2x 1x a111 1a11 111 Step2. The augmented coefficient matrix Ab is + += a3a111 a21a11 a111 bA Step3. Convert augmented coefficient matrix into reduced echelon form using ele row operations. ( ) ( ) 1R13R,1R12R a3a111 a21a11 a111 bA −+−+ + += 3R a 1 ,1R a 1 a2a00 a0a0 a111 ( ) 2R11R 2100 1010 a111 −+ ( ) 3R11R 2100 1010 1a101 −+ − − 2100 1010 3a001 Step4. Writing the above matrix in equation form, we have: − = 2 1 3a 3x 2x 1x 100 010 001 Or, ( ) ( ) ( )iii23x2x01x0 ii13x02x1x0 i3a3x02x01x =++ =++ −=++
- 29. INVERSE OF MATRIX BY ROW OPERATION REPEAT WITH NEW EXAMPLE
- 30. creative
- 32. − −− −− − = 32032 122361 22101 50121 A − − −− − −−−−− − += 32 51 22 01 02 11 32 21 121 51 21 01 31 11 61 21 21 51 21 01 11 11 01 21 Rank1ARank . 722-1 7224- 7222- Rank1ARank − − − += This reduced non – zero matrix is of order 3 × 4. Hence, − −−− − − − − −− −− − − ++= 71 72 21 22 21 22 74 72 24 22 24 22 Rank11ARank − − += 722 1444 Rank2ARank This reduced non – zero matrix is of order 2 × 3. Thus − − += 72 144 22 44 Rank3ARank 00Rank3ARank += . This last matrix is of order 1 × 2; and is a zero matrix, hence its rank is zero. Thus, Rank (A ) = 3 + 0 = 3. An echelon matrix equivalent to A = − −− − 00000 144400 72220 50121