1. What is the iterative method in system of linear equations?2. Give atleast two examples of each iterative method.Answer question number 1 only.These is the reference that might help to answer the question. The solution is X1 = 1.0013, X2 = 0.9945, X3 = 0.9983.The exact solution is x = 1, X2 = 1, X3 = 1.Example3x1 + x2 - x3 = 32x1 + 4x2 + x3 = 7X1 - x2 + 4x3 = 4Consider the initial approximation, x1(0) = 0, x2(0) = 0, x3) = 0. Apply Gauss-Seidel method until the last iteration has a difference that is less than 0.005.References:Kharab, A. & Guenther R. (2019). An introduction t(4 ed.). Boca Raton: CRC Press.Gupta, R. (2019). Numerical methods: Fundamentals and applications. India: CambridgeUniversity Press,numerical methods: A MATLAB approachStep 1. Rewrite the system.X1 =,(3 – x2 + x3)x2 = (7 – 2x1 – x3)x3 = (4 - x1 + x2)Step 2. Get the first approximation.x,(1)=(3 – x2(0) + x3(0)) = 1X2(1)=-(7 – 2x,- x3(0)) = 1.25x3(1) = (4 – x,(1) + x2(1)) = 1.0625%3DStep 3. Perform approximation until desired accuracy is obtained.1=-(3-x," + x,"3(2)(") 0.9375(7–2x,2) – x,"))=1.01564(2) =(2) -(2)(4-x,4 +x,2))= 1.01954As required, the difference of the last iteration should be less than 0.005,which is after the third iteration (faster than Jacobi).1.0000| 1.06251.01950.9983Iteration 11.2500Iteration 20.93751.0156Iteration 31.00130.9945 Systems of Linear Equations (Iterative Methods)Iterative methods are based on successive improvement of initial guessesfor the solution.The first approximation [x1(1), x2(1), ..., Xn(1)] is used to solve for the seconditeration. The process is repeated until desired accuracy is obtained. The(k+1)th iteration can be obtained from kth iteration by the following formula:x,(k+1) =b,-(a,x,4) + a,zx,«) +•.+a„x„")]%3DIterative Method 1:Jacobi Method/Method ofSimultaneousDisplacement1x, (k).(k)(k) +...+azX(k+1)x,The linear system of equations can be rewritten as:-[b, - (a,,x, +a3X, +…·+a„X,(k+1)%3D(k).k = 0,1,2,...a1x, =-b, - (a,x, +a,x, + +a„X,The Jacobi iteration formula can be written as:1(k+1)%3D(k)1sisn k=0,1,2,...j=1X =-b, - (a,x, +aX2 ++a,n-1\,-1aExampleInitial approximation is required to get the next approximation. Let the initialapproximation be [x1©, x2), .., Xn]. These values are used to get the nextapproximation [x,("), x2(1), ..., xn")] of the Jacobi method.3x1 + x2 – x3 = 32x1 + 4x2 + x3 = 7X1 - x2 + 4x3 = 4(1)%3D- b, - (a,,x2 + a,3x, +.+a,„X„")](0)(0)(0)Consider the initial approximation, x,0) = 0, x20) = 0, x30) = 0. Apply Jacobimethod until the last iteration has a difference that is less than 0.005.(1)(0),- (a,x +azxStep 1. Rewrite the system.1X1 = (3 – x2 + x3)(1)%3Dx2 = (7 – 2x1 – x3)(0)b,-(a,x,® +a„zx," +…+a,n-*n-")|annX3 =7(4 – x1 + x2)06 Handout 1E student.feedback@sti.edu*Property of STIPage 1 of 3ASTIIT1903Step 2. Get the first approximation.Solve for the approximation of variable x1:1(1)(0)(1)(0)(0)(0)(3-x, +x,'3)=1Co"x"lv+..+ o*xlv)– 'q ]x,"(1)=-(7-2x,0 – x,)=1.75(0)To solve for x2(1), use x1(1) instead of x1(0) (as used in Jacobi):41b,-(a,x," +a,x,(1)(0)(0)X,+..+a,„x,") |x, ),-(4-x, + x,")=14%3D(0Solve for the first approximation to the ith variable:Step 3. Perform approximation until desired accuracy is obtained.+a* +.+ax,%3D1(2)(1)(3-x,"+x,")=0.753X" =.The last variable is calculated as:6, - (a,,x," +a„Xa(1)(1)(1)+...+a,n-1Xn-1(2)(1)(7-2x," - x,)=14X, =.The first approximation [x1("), x2"), ..., x,(")] is used to solve for the seconditeration. The (k+1)th iteration can be obtained from kth iteration by thefollowing formula:(2)(1)x, =.(4-x,+x,")=1.1875As required, the difference of the last iteration should be less than 0.005,which is after the sixth iteration.(k) +...+ aX,%3DaIteration 1Iteration 21.0000 1.7500| 1.00001.18751.06251.00390.98960.99621.00001.0625 1.07810.7500-(Iteration 3Iteration 4Iteration 50.9948 0.95311.0169 1.00160.9960 | 0.9941Iteration 6(k+1) + a,X2(k+1) =*+1) +.+a,x**)) k= 0,1,2,...(k+1)The solution is x1 = 0.9960, x2 = 0.9941, X3 = 0.9962.The exact solution is x1 = 1, X2 = 1, X3 = 1.aThe Gauss-Seidel iteration formula can be written as:Iterative Method 2: Gauss-Seidel Method/Method of Successive1(k+1)(k)1sisn k=0,1,2,..(k+1)Displacement/Liebmann Methoda,j=lLet the initial approximation be [x,(0), x20), ..., x,]. The latest availablevalues are used to get the next approximation. Let the next approximationbe (x"), x2"), ..., Xn()].

Question

1. What is the iterative method in system of linear equations?2. Give atleast two examples of each iterative method.Answer question number 1 only.These is the reference that might help to answer the question. The solution is X1 = 1.0013, X2 = 0.9945, X3 = 0.9983.The exact solution is x = 1, X2 = 1, X3 = 1.Example3x1 + x2 - x3 = 32x1 + 4x2 + x3 = 7X1 - x2 + 4x3 = 4Consider the initial approximation, x1(0) = 0, x2(0) = 0, x3) = 0. Apply Gauss-Seidel method until the last iteration has a difference that is less than 0.005.References:Kharab, A. & Guenther R. (2019). An introduction t(4 ed.). Boca Raton: CRC Press.Gupta, R. (2019). Numerical methods: Fundamentals and applications. India: CambridgeUniversity Press,numerical methods: A MATLAB approachStep 1. Rewrite the system.X1 =,(3 – x2 + x3)x2 = (7 – 2x1 – x3)x3 = (4 - x1 + x2)Step 2. Get the first approximation.x,(1)=(3 – x2(0) + x3(0)) = 1X2(1)=-(7 – 2x,- x3(0)) = 1.25x3(1) = (4 – x,(1) + x2(1)) = 1.0625%3DStep 3. Perform approximation until desired accuracy is obtained.1=-(3-x,&#34; + x,&#34;3(2)(&#34;) 0.9375(7–2x,2) – x,&#34;))=1.01564(2) =(2) -(2)(4-x,4 +x,2))= 1.01954As required, the difference of the last iteration should be less than 0.005,which is after the third iteration (faster than Jacobi).1.0000| 1.06251.01950.9983Iteration 11.2500Iteration 20.93751.0156Iteration 31.00130.9945 Systems of Linear Equations (Iterative Methods)Iterative methods are based on successive improvement of initial guessesfor the solution.The first approximation [x1(1), x2(1), ..., Xn(1)] is used to solve for the seconditeration. The process is repeated until desired accuracy is obtained. The(k+1)th iteration can be obtained from kth iteration by the following formula:x,(k+1) =b,-(a,x,4) + a,zx,«) +•.+a„x„&#34;)]%3DIterative Method 1:Jacobi Method/Method ofSimultaneousDisplacement1x, (k).(k)(k) +...+azX(k+1)x,The linear system of equations can be rewritten as:-[b, - (a,,x, +a3X, +…·+a„X,(k+1)%3D(k).k = 0,1,2,...a1x, =-b, - (a,x, +a,x, + +a„X,The Jacobi iteration formula can be written as:1(k+1)%3D(k)1sisn k=0,1,2,...j=1X =-b, - (a,x, +aX2 ++a,n-1\,-1aExampleInitial approximation is required to get the next approximation. Let the initialapproximation be [x1©, x2), .., Xn]. These values are used to get the nextapproximation [x,(&#34;), x2(1), ..., xn&#34;)] of the Jacobi method.3x1 + x2 – x3 = 32x1 + 4x2 + x3 = 7X1 - x2 + 4x3 = 4(1)%3D- b, - (a,,x2 + a,3x, +.+a,„X„&#34;)](0)(0)(0)Consider the initial approximation, x,0) = 0, x20) = 0, x30) = 0. Apply Jacobimethod until the last iteration has a difference that is less than 0.005.(1)(0),- (a,x +azxStep 1. Rewrite the system.1X1 = (3 – x2 + x3)(1)%3Dx2 = (7 – 2x1 – x3)(0)b,-(a,x,® +a„zx,&#34; +…+a,n-*n-&#34;)|annX3 =7(4 – x1 + x2)06 Handout 1E student.feedback@sti.edu*Property of STIPage 1 of 3ASTIIT1903Step 2. Get the first approximation.Solve for the approximation of variable x1:1(1)(0)(1)(0)(0)(0)(3-x, +x,'3)=1Co&#34;x&#34;lv+..+ o*xlv)– 'q ]x,&#34;(1)=-(7-2x,0 – x,)=1.75(0)To solve for x2(1), use x1(1) instead of x1(0) (as used in Jacobi):41b,-(a,x,&#34; +a,x,(1)(0)(0)X,+..+a,„x,&#34;) |x, ),-(4-x, + x,&#34;)=14%3D(0Solve for the first approximation to the ith variable:Step 3. Perform approximation until desired accuracy is obtained.+a* +.+ax,%3D1(2)(1)(3-x,&#34;+x,&#34;)=0.753X&#34; =.The last variable is calculated as:6, - (a,,x,&#34; +a„Xa(1)(1)(1)+...+a,n-1Xn-1(2)(1)(7-2x,&#34; - x,)=14X, =.The first approximation [x1(&#34;), x2&#34;), ..., x,(&#34;)] is used to solve for the seconditeration. The (k+1)th iteration can be obtained from kth iteration by thefollowing formula:(2)(1)x, =.(4-x,+x,&#34;)=1.1875As required, the difference of the last iteration should be less than 0.005,which is after the sixth iteration.(k) +...+ aX,%3DaIteration 1Iteration 21.0000 1.7500| 1.00001.18751.06251.00390.98960.99621.00001.0625 1.07810.7500-(Iteration 3Iteration 4Iteration 50.9948 0.95311.0169 1.00160.9960 | 0.9941Iteration 6(k+1) + a,X2(k+1) =*+1) +.+a,x**)) k= 0,1,2,...(k+1)The solution is x1 = 0.9960, x2 = 0.9941, X3 = 0.9962.The exact solution is x1 = 1, X2 = 1, X3 = 1.aThe Gauss-Seidel iteration formula can be written as:Iterative Method 2: Gauss-Seidel Method/Method of Successive1(k+1)(k)1sisn k=0,1,2,..(k+1)Displacement/Liebmann Methoda,j=lLet the initial approximation be [x,(0), x20), ..., x,]. The latest availablevalues are used to get the next approximation. Let the next approximationbe (x&#34;), x2&#34;), ..., Xn()].

Accepted Answer

We can solve the 2 examples using jacobi formula

What is the iterative method in system of linear equations? 2. Give atleast two examples of each iterative method. Answer question number 1 only. These is the reference that might help to answer the question.