5.7 Cramer's Rule 1. Using Determinants to Solve Systems Assumes the system of two equations in two unknowns

Transcription

1 5.7 Cramer's Rule 1. Using Determinants to Solve Systems Assumes the system of two equations in two unknowns (1) possesses the solution and provided that.. The numerators and denominators are recognized as determinants. That is the system has the unique solution provided that the (2) determinant of the matrix of coefficients. For a system of linear equations in unknowns it is convenient to define a special matrix th column In other words is the same as the matrix that the th column of

2 has been replaced by the entries of the column matrix The generalization of (2) known as Cramer's Rule is given in the next theorem. 2. Cramer's Rule Let be the coefficient matrix of the system(1). If det then the solution of 1. is given by where is defined in (3). Proof We first write the system 1. as. Since det exists and so Now the entry in the But where have th row of the last matrix is is the cofactor expansion of det is the matrix given in (3) along the for. Example : Use Carmer's rule to solve the system th column. Hence we

3 Solution) Thus 2. gives. 5.8 System of Linear Algebraic Equations 1. Linear Equations Recall that any equation of the form ax+by=c where a b and c are real numbers is said to be a linear equation in the variable x and y. The graph of a linear equation in two variable is a straight line. For real numbers a b c and d ax+by+cz=d is a linear equation in the variables x y and z and is the equation of a plain. In general an equation of the form where the variables and are real numbers is a linear equation in. 2. General Form A system of linear equations in unknowns has the general form...(1) The coefficients of the unknowns in the linear system (1) can be

4 abbreviated as where denotes the row and denotes the column in which the coefficient appears. For example is the coefficient of the unknown in the second row and third column ( that is ). Thus and. The numbers are called the constants of the system. If all the constants are zero the system(1) is said to be homogeneous otherwise it is non-homogeneous. For example This system is homogeneous This system is nonhomogeneous 3. Solution A solution of a linear system(1) is a set of n numbers that satisfies each equation in the system. For example is a solution of the system To see this we replace by 3 and by -1 in each equations: 3(3) + 6(-1)9-6 =3 and 3-4(-1) = 3+4 = 7 A linear system of equations is said to be consistent if it has at least one solution and inconsistent if it has no solutions. If a linear system is consistent it has either a unique solution (that is precisely one solution) or infinitely many solutions. 4. Solving Systems We can transform a system of linear equations into an equivalent system (that is one having the same solutions) by the following elementary operations : Multiply an equation by a non-zero constant. Interchange the positions of equations in the system Add a non-zero multiple of one equation to any other equations.

5 5. Augmented Matrix Reflecting on the solution of the linear system should convince you that the solution of the system does not depend on what symbols are used as variables. Thus the systems and have the same solutions as the system. In other words in the solution of a linear system used to denote the variables are immaterial; it is the coefficients of the variables and the constants that determine the solution of the system. In fact we can solve a system of form(1) by dropping the variables entirely and performing operations on the rows of the array of coefficients and constants : This array is called the augmented matrix of the system or simply the matrix of the system(1). Example linear system the matrix of the system

6 6. Elementary Row Operations Since the rows of an augmented matrix represent the equations in a linear system the three elementary operations on a linear system listed above are equivalent to the following elementary row operations on a matrix: Multiply an row by a non-zero constant. Interchange any two rows. Add a non-zero multiple of one row to any other row. Of course when we add a multiple of one row to another we add the corresponding entries in the rows. We say that two matrices are row equivalent if one can be obtained from the other through a sequence of elementary row operations. The procedure of carrying out elementary row operations on a matrix to obtain a row equivalent matrix is called row reduction. 7. Gauss Elimination Method To solve a system such as (1) using an augmented matrix we shall use either Gaussian elimination or the Gauss-Jordan elimination method. In the former method we row reduce the augmented matrix of the system until we arrive at a row-equivalent augmented matrix in row-echelon form : The first non-zero entry in a non-zero row is a 1. In consecutive non-zero rows the first entry 1 in the lower row appears to the right of the 1 in the higher row. Rows consisting of all zeros are at the bottom of the matrix. In the Gauss-Jordan method the row operations are continued until we obtain an augmented matrix that is reduced row-echelon form. A reduced row-echelon matrix has the same three properties listed above but in addition : A column containing a first entry 1 has zeros everywhere else. Example : Echelon Forms (a) row-echelon form

7 and (b) reduced row-echelon form and It should be noted that in Gaussian elimination we stop when we have obtained an augmented matrix in row-echelon form. In other words by using different sequences of row operations we may arrive at different row-echelon forms. This method then requires the use of back-substitution. The solution of the system will be apparent by inspection of the final matrix. In terms of the equations of the original system our goal in both methods is simply to make the coefficient of in the first equations 1) equal to one and then use multiples of that equations to eliminate from other equations. The process is repeated for the other variables. To keep track of the row operations used on an augmented matrix we shall utilize the follow notation: Symbol Meaning Interchange rows and Multiply the th row by the nonzero constant Multiply the th row by and add to the th row Example : solve the linear system using Gaussian elimination Solution) 1) We can always interchange equations so that the first equation contains the variable

8 The last matrix is in row-echelon form. 8. Gauss-Jordan Elimination Method In Gauss-Jordan elimination we stop when we have obtained the augmented matrix in reduced row-echelon form. Any sequence of row operations will lead to the same augmented matrix in reduced row-echelon form. This method does not require back-substitution; Example : solve the linear system in above Example using Gauss-Jordan elimination Solution) The last matrix is in reduced row-echelon form. Example : Use Gauss-Jordan elimination to solve Solution) linearly dependent

9 In this case the last matrix in reduced row-echelon form implies that the original system of three equations in three unknowns is really equivalent to two equations in the unknowns. Since only is common to both equations (the non-zero rows). 9. LU Decomposition a) Introduction A practical direct method for equation solving especially when several RHS's are needed and/or when all RHS's are not known at the start of the problem. b) Basic Approach (i) [A]{x} = {C} (ii) Decompose the matrix [A] into [L][U] where either [L] or [U] has 1's on diagonal. It should be noted that [L] and [U] represent Lower and Upper triangular matrices respectively (iii) [A]{x} = [L][U]{x} = C (iv) Let {d} = [U]{x} and substitute [L]{d} = {C} (v) Solve [L]{d} = {C} for {d} by forward substitution. (vi) With {d} solve [U}{x} = {d} for {x} by back substitution.

10 10. Iterative Solution Technique a) Impetus for Iterative Schemes (i) Direct methods may be inadequate for round-off-error prone problems. (ii) It is self-correcting. (iii) It provides a means for controlling level of error (iv) It may be more rapid if coefficient matrix is sparse. (v) It may be more economical with respect to memory. (vi) It may also be applied to solve nonlinear systems. b) Disadvantages (i) It may not converge or may be slowly convergent. (ii) It may appropriate for only diagonally dominant systems. c) Basic Mechanics (i) Equations (ii) Solve each equation for one variable. (iii) Start with an initial estimate of substitute this into the RHS of all of the above equations and generate a new approximation. (iv) Note the analogy to one-point iteration. (v) Repeat until the maximum number of iterations is reached or

11 ε with ε being a tiny number. 11. Jacobi Iterative Method a) Matrix Derivation To solve [A]{x} = {C} separate [A] into [A] = [ ] + [D] + [ ] where [D] = [A]'s diagonal [ ] is a lower triangular matrix with zeros on diagonal and [ ] is an upper triangular with zeros on diagonal. b) Rewrite the given system. note that (i) If the system is diagonally dominant (ii) c) The Final Form is d) 초기값과수렴성 Jacobi 반복법이대각선지배성 을만족하면초기값에관계없이수렴된해를구할수있다. 하지만이조 건은수렴하기위한충분조건이지필요조건은아니다. 따라서 이조건을 만족하지않더라도때때로수렴된해를얻을수있다. 12. Gauss - Seidel Iterative Method

12 a) In most cases using the newest values within RHS equations will provide better estimates of the next value. b) The basic procedure c) If either method is going to converge Gauss-Seidel Iterative Method converges faster than Jacobi Iterative Method. d) Why use the Jacobi Iterative Method? Because we can separate the n-equations into n independent tasks it is very well suited computers with parallel processors.