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# Lesson 7

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### Lesson 7

1. 1. Last modified:January 1, 1970 GMT An Introduction to Matlab(tm): Lesson 7 SOLVING SIMULTANEOUS LINEAR EQUATIONS One of the most common applications of matrix algebra occurs in the solution of linear simultaneous equations. Consider a set of n equations for which the unknowns are x1, x2, ....... xn. a11x1 + a12x2 + a13x3 + ............... a1nxn = b1 a21x1 + a22x2 + a23x3 + ............... a2nxn = b2 a31x1 + a32x2 + a33x3 + ............... a3nxn = b3 . . . . . . . . . . . . . . . . . . . . an1x1 + an2x2 + an3x3 + ............... annxn = bn The matrix format for these equations is [a][x] = [b] where a11 a12 a13 .... a1n x1 b1 a21 a22 a23 .....a2n x2 b2 [a] = a31 a32 a33 .... a3n . . . . . . . . . . . . . . . . . . . . an1 an2 an3 .... ann [x] = x3 . . . . xn [b] = b3 bn Note that only when the equations are linear in the unknown xi's is the matrix formulation possible. The classical matrix solution to this equation is based on the definition of the inverse of a matrix. The inverse of a matrix is that matrix which when multiplied by the original matrix, results in the identity matrix. Then a-1 a = I
2. 2. where a-1 denotes the 'inverse of matrix a' and I denotes the identity matrix of the same order as matrix 'a'. If 'a' is a known coefficient matrix and if 'b' is a column vector of known terms, the problem is one of finding the n vales of x1, x2, .... xn that satisfy these simultaneous equations. Pre-multiplying both sides of the equation by the inverse of 'a' gives [a-1][a][x] = [a-1][b] or [x] = [a-1][b] Thus if we find the inverse of the coefficient matrix, the product of the inverse times the matrix of non-homogeneous terms, b, yields the unknown matrix, x. To observe the simple property of the inverse, define the 4x4 matrix, C C = [1 -4 3 2; 3 1 -2 1; 2 1 1 -1; 2 -1 3 1] calculate the inverse using the 'matlab' command, inv(). C_inverse = inv(C) and note that C_inverse * C = [1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1] where the right hand side is the 4x4 identity matrix. We leave to a study of linear algebra, the method and steps required to calculate the inverse of a matrix. Suffice it to say here that for systems of equations numbering in the hundreds or thousands, which often occur in complex engineering problems, the calculation of the matrix inverse is a time consuming operation even for modern computers. As an example of solving a system of equations using the matrix inverse, consider the following system of three equations. x1 - 4x2 + 3x3 = -7 3x1 + x2 - 2x3 = 14 2x1 + x2 + x3 = 5 Using 'matlab', define the two matrices for [a] and [b].
3. 3. a = [ 1 -4 3; 3 1 -2; 2 1 1]; b = [ -7; 14; 5]; Find the solution vector, x, by x = inv(a)*b giving x=[3 1 -2 ] LEFT AND RIGHT MATRIX 'DIVISION' IN MATLAB In contrast to matrix inversion, the preferred methods of solution for large-scale systems of simultaneous equations are methods of back substitution whereby the equations can be rearranged so that first xn is solved, then xn-1 and so on until finally x1 is obtained in order. Thus in back substitution we never solve the equations simultaneously, rather we solve them sequentially. For example, the above set of equations can be rearranged through appropriate combination into the below equations. (You should carry out this algebra to ensure you understand the steps matlab takes in solving with back substitution. In the general case, the steps or algorithm is much more complex than that required for 3 equations.) x1 - 4x2 + 3x3 = -7 13x2 - 11x3 = 35 (34/13)x3 = (68/13) In this format, we see from the third equation the solution for x3 = 2 and knowing this, the second equation yields x2 = 1, and knowing both x3 and x2, the first equation gives x1 = 1. This is an application of back substitution, whereby each unknown is obtained by simple sequential solution to a single equation. Methods of back substitution are employed by 'matlab' when we invoke the 'right division' and 'left division' commands. / left division right division Understand that matrices cannot be divided. The operation is not
4. 4. defined. The syntax of 'right division' and 'left division' simply invoke back substitution methods for obtaining a solution vector from a system of simultaneous equations. Whether 'right division' (/) or 'left division' () is appropriate depend on how the matrix equation is posed. Left Division. If the matrix equation is [a][x] = [b] then we have a nxn matrix [a] pre-multiplying a nx1 matrix [x], resulting in a nx1 matrix [b]. The solution for x is obtained by using the left division operation. [x] = [a][b] If the matrix equation is cast in this format, the use of right division to obtain the solution vector x will result in an error message. Returning to 'matlab', recall we have defined the matrices [a] and [b] in the format of the matrix equation appropriate for left division. a = [ 1 -4 3; 3 1 -2; 2 1 1]; b = [ -7; 14; 5]; Now enter the command x1 = ab and obtain the result x1 = [ 3 1 -2 ] which i s the same result found earlier for x when solving by matrix inversion. Right Division. The matrix format in which right division is employed is less common but no less successful in solving the problem of simultaneous equations. Right Division is invoked when the equations are written in the form
5. 5. [x][A] = [B] Note that in this form [x] and [B] are row vectors rather than column vectors as above. This form represent a 1xn matrix (x) premultiplying a nxn matrix (A), resulting an a 1xn matrix (B). Again, recalling the set of equations, x1 - 4x2 + 3x3 = -7 3x1 + x2 - 2x3 = 14 2x1 + x2 + x3 = 5 These equations can be written in matrix format [x][A] =[B] if x = [x1 x2 x3] B = [ -7 14 5] and 1 3 2 [A] = -4 1 1 3 -2 1 Note that [A] is equal to the transpose of [a]. A = a' Having so defined A and B, the solution for x can be obtained by right division. x = B/A results in x=[3 1 -2 ] PRACTICE PROBLEMS
6. 6. 1. Find the solution to r + s + t + w = 4 2r - s + w = 2 3r + s - t - w = 2 r - 2s - 3t + w = -3 using the matrix inverse and left and right division. 2. Find the solution to 2x1 + x2 - 4x3 + 6x4 + 3x5 - 2x6 = 16 -x1 + 2x2 + 3x3 + 5x4 - 2x5 = -7 x1 - 2x2 - 5x3 + 3x4 + 2x5 + x6 = 1 4x1 + 3x2 - 2x3 + 2x4 + x6 = -1 3x1 + x2 - x3 + 4x4 + 3x5 + 6x6 = -11 5x1 + 2x2 - 2x3 + 3x4 + x5 + x6 = 5 using the matrix inverse and left and right division. 3. Back substitution is facilitated by recognizing that any nonsingular symmetric matrix can be decomposed or factored into the product of two matrices, called here L and U. A=LU This is called LU factorization. U is an upper triangular matrix (all 0's below the diagonal). L is called a permutation of a lower triangular matrix in which 1's occur at every element which is a permutation of the two numbers defining the size of the matrix. These matrices can be obtained in 'matlab' using the command [L,U] = lu(A) Using the matrix A defined in Prob 2 above, find the upper and lower triangular factor matrices for the matrix of coefficients, A. If these matrices are called AU and AL, show that
7. 7. AL * AU * x = B where x [ 2 -1 1 0 3 -4 ]' and B = [16 -7 1 -1 -11 5]' Back to Matlab In-House Tutorials