Jacobian Method

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Linear equation systems are linked to many engineering and scientific topics, as well as quantitative industry, statistics, and economic problems. It is possible and easy to solve a large number of symmetric, linear algebraic equations after the invention of computers. The solution of a very large set of simultaneous equations by numerical methods in time is an important factor in the practical application of the results. If the equations are solved in considerable time, we can increase productivity significantly. 


We have mainly two numerical methods: the direct method and the iterative method for solving linear equation systems. For a big set of linear equations, particularly for sparse and structured coefficient equations, the iterative methods are preferable as they are largely unaffected by round-off errors. The well known classical iterative methods are the Jacobian and Gauss-Seidel methods. Here, we are going to discuss the Jacobi or Jacobi Method. 


Jacobi Iteration Method

Matrices in the form of "AX=b" can easily represent a large linear system, where "A" represents a square matrix containing the ordered coefficients of our system of linear equations, "X" contains all of our various variables, and "B" represents the constants equal to each linear equation. For our unknown x-values, we wish to solve, and we can do so by using the Jacobian Method. As discussed, we can summarize the Jacobi Iterative Method with the equation "AX=B."  The "a" variables indicate the elements of the "A" coefficient matrix, the "x" variables give us the unknown X-values which we are solving for, and the constants of each equation are represented by "b".

Now, AX=B is a system of linear equations, where

A = \[\begin{bmatrix} a_{11} & a_{12} & \cdots  & a_{1n}\\ a_{21} & a_{22} & \cdots  & a_{2n}\\ \vdots  & \vdots & \ddots  & \vdots \\  a_{n1} & a_{n2} & \cdots  & a_{nn} \end{bmatrix}\], X = \[\begin{bmatrix} x_{1}\\ x_{2}  \\ \vdots \\ x_{n} \end{bmatrix}\], B = \[\begin{bmatrix} b_{1}\\ b_{2}  \\ \vdots \\ b_{n} \end{bmatrix}\]

Suppose that none of the diagonal entries are zero without loss of generality; otherwise, swap them in rows, and the matrix A can be broken down as,

A=(D+U+L) , 

where D is the Diagonal matrix of A, U denotes the elements above the diagonal of matrix A, and L denotes the elements below the diagonal of matrix A.

Where D = \[\begin{bmatrix} a_{11} & 0 & \cdots  & 0\\0 & a_{22} & \cdots  & 0\\ \vdots  & \vdots & \ddots  & \vdots \\ 0 & 0 & \cdots  & a_{nn} \end{bmatrix}\]  and L + U is  \[\begin{bmatrix} 0 & a_{12} & \cdots  & a_{1n}\\ a_{21} & 0 & \cdots  & a_{2n}\\ \vdots  & \vdots & \ddots  & \vdots \\  a_{n1} & a_{n2} & \cdots  & 0 \end{bmatrix}\]

The solution can be obtained iteratively via using the following relation:

X\[^{(k+1)}\] = D\[^{-1}\] (B - (L + U)X\[^{(k)}\]) 

Where Xk and X(k+1) are kth and (k+1)th iteration of X. The elements of X(k+1) can be calculated by using element based formula that is given below: 

X\[_{i}\]\[^{(k+1)}\] = \[\frac{1}{a_{ii}}\] \[\sum_{j\neq i}^{}\] (b\[_{i}\] - a\[_{ij}\] - x\[_{j}^{k}\]), i = 1, 2, 3, ………, n

Therefore, after placing the previous iterative value of X in the equation above, the new X value is determined until the necessary precision is achieved. The vital point is that the method should converge in order to find a solution. The sufficient but not possible condition for the method to converge is that the matrix should be strictly diagonally dominant. The Jacobi Method is also known as the simultaneous displacement method. 


Gauss-Seidel and Jacobi Methods

The difference between Gauss-Seidel and Jacobi methods is that, Gauss Jacobi method takes the values obtained from the previous step, while the Gauss–Seidel method always uses the new version values in the iterative procedures. The reason why the Gauss-Seidel method is commonly referred to as the successive displacement method is that the second unknown is calculated by the first unknown in the current iteration, the third unknown is calculated from the 1st and 2nd unknown, etc.


Jacobi Method Example

Question: Solve the below using the Jacobian method, which is a system of linear equations in the form AX = B. 

A = \[\begin{bmatrix} 2 & 5\\ 1 & 7 \end{bmatrix}\], b = \[\begin{bmatrix} 13 \\ 11 \end{bmatrix}\], x\[^{0}\] = \[\begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Ans: 

We know that X(k+1) = D-1(B – RX(k)) is the formula that is used to estimate X.

Now, we’ll rewrite the formula as D-1(B – RX(k)) = TX(k) + C for our convenience. 

In this, T = -D-1R, C = D-1B, R = L + U

We’ll now split the matrix A as a diagonal matrix and remainder.

D = \[\begin{bmatrix} 2 & 0\\ 0 & 7 \end{bmatrix}\] ⇒ D\[^{-1}\] = \[\begin{bmatrix} \frac{1}{2} & 0\\ 0 & frac{1}{7} \end{bmatrix}\]

The lower and upper parts of the reminder of A are as follows:

R = \[\begin{bmatrix} 0 & 1\\ 5 & 0 \end{bmatrix}\], L = \[\begin{bmatrix} 0 & 0\\ 5 & 0 \end{bmatrix}\], U = \[\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\]  

Here,

R = Remainder matrix

L = Lower part of R

U = Upper part of R

T = -D-1(L + U) = D-1[-L + (-U)]

T = \[\begin{bmatrix} \frac{1}{2} & 0\\ 0 & \frac{1}{7} \end{bmatrix}\]  {\[\begin{bmatrix} 0 & 0\\ -5 & 0 \end{bmatrix}\] + \[\begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix}\]} = \[\begin{bmatrix} 0 & - \frac{1}{2}\\ - \frac{5}{7} & 0 \end{bmatrix}\]

C = \[\begin{bmatrix} \frac{1}{2} & 0\\ 0 & \frac{1}{7} \end{bmatrix}\] \[\begin{bmatrix} 13 \\ 11 \end{bmatrix}\] = \[\begin{bmatrix} \frac{13}{2} \\ \frac{11}{7} \end{bmatrix}\] 

x\[^{1}\] = \[\begin{bmatrix} 0 & - \frac{1}{2}\\ - \frac{5}{7} & 0 \end{bmatrix}\] \[\begin{bmatrix} 1 \\ 1 \end{bmatrix}\]  + \[\begin{bmatrix} \frac{13}{2} \\ \frac{11}{7} \end{bmatrix}\] = \[\begin{bmatrix} \frac{12}{2} \\ \frac{6}{7} \end{bmatrix}\] ≈ \[\begin{bmatrix} 6 \\ 0.857 \end{bmatrix}\]

x\[^{2}\] = \[\begin{bmatrix} 0 & - \frac{1}{2}\\ - \frac{5}{7} & 0 \end{bmatrix}\] \[\begin{bmatrix} 6 \\ \frac{6}{7} \end{bmatrix}\]  + \[\begin{bmatrix} \frac{13}{2} \\ \frac{11}{7} \end{bmatrix}\] = \[\begin{bmatrix} \frac{85}{14} \\ -\frac{19}{7} \end{bmatrix}\] ≈ \[\begin{bmatrix} 6.071 \\ -2.714 \end{bmatrix}\]

We’ll repeat the process until it converges. 


What is the “T” Matrix?

In the Jacobi Method example problem we discussed the “T” Matrix. Let’s now understand what it is about.   While the application of the Jacobi iteration is very easy, the method may not always converge on the set of solutions. As a result, a convergence test must be carried out prior to the implementation of the Jacobi Iteration. This convergence test completely depends on a special matrix called our "T" matrix. We'll re-write this system of equations in a way that the whole system is split into the form "Xn+1 = TXn+c." In simple words, the matrix on the RHS of the equation can be split into the matrix of coefficients and the matrix of constants. 


Conclusion

The Jacobian method, one of the most basic methods to find solutions of linear systems of equations, is studied. Jacobian problems and solutions have many significant disadvantages, such as low numerical stability and incorrect solutions (in many instances), particularly if downstream diagonal entries are small. We can see while solving a variety of problems, that this method yields very accurate results when the entries are high. This method has applications in Engineering also as it is one of the efficient methods for solving systems of linear equations, when approximate solutions are known. This significantly reduces the number of computations required.

FAQ (Frequently Asked Questions)

1. What is the Jacobian Method?

Ans: The method of finding solutions to the diagonally dominant linear equation system is called Gauss Jacobi Iterative Method. Reduce the matrix to the diagonal matrix and iterate until it converges. Use this calculator to solve the linear system of equations for matrix variables.

2. What is the Use of the Jacobi Method?

Ans: In linear algebra, the Jacobian method is an iterative algorithm used to determine the solutions for a diagonally dominant system of linear equations. Each diagonal element is fixed, and an approximate value is plugged in.

3. Why Do We Use the Gauss-Seidel Method?

Ans: Gauss-Seidel Method solves the linear equations of the system.  It is the method of iteration for solving the linear equation with unknown variables. This method is very simple and is used for computing on digital computers. In this method, the value of unknown is immediately reduced to the number of iterations, and the calculated value replaces the earlier value at the end of the iteration.