Matrix notation. A nm : n m : size of the matrix. m : no of columns, n: no of rows. Row matrix n=1 [b 1, b 2, b 3,. b m ] Column matrix m=1

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1 Matrix notation A nm : n m : size of the matrix m : no of columns, n: no of rows Row matrix n=1 [b 1, b 2, b 3,. b m ] Column matrix m=1 n = m square matrix Symmetric matrix Upper triangular matrix: matrix with elements below main diagonal are zero Diagonal matrix : square matrix with all off diagonal elements are zeros. lower triangular matrix: matrix with elements above main diagonal are zero Identity matrix: diagonal matrix with diagonal elements set to 1 A banded matrix : all elements are zero except of a band centered on the main diagonal Tri-diagonal matrix Matrix operations Addition/subtraction [C]=[A] [B]

2 [A]+[B]= [B]+[A] ([A]+[B])+[C]= [A]+([B]+[C]), g: is scalar Product of two matrices [C]=[A][B] Where n: the column dimension of [A] and row dimension of [B] [A] m n [B] n l = [C] m l No of columns of first matrix should be equal to the No of rows of second matrix. [A][B] [B][A] Division is not defined [A][A] -1 = [A] -1 [A]=I [A] -1 : the inverse of A. Transpose of a matrix Eigen values, NOT Required [A]{ x} = {b} [A]{ x}= { x} ([A]- [I] ){ x}=0 Det ([A]- [I] )=0 n th degree polynomial that is called the characteristic polynomial.

3 Find the eigen-values and eigenvectors of the following matrix To find eigen vector solve for [ ]{x}=0 For Solving, let, For Solving, Let, Linear Algebraic Equations.. =

4 [A]{X}={B}, 9.2 Naïve Gauss elimination a. Elimination step b. back substitution.(i).(ii)....(iii) Forward elimination Multiply (i) by ( subtract it from (ii) to give Prime indicates that the elements have been changed from their original values. Procedure is then repeated for the remaining equations.

5 Back substitution Example (3x3) Consider the simple example : x 1 + x 2 + x 3 = 6 2x 1 + 3x 2 + 4x 3 = 20 3x 1 + 4x 2 + 2x 3 = 17 Augmented matrix is A b =, and the unknown vector X = (x 1, x 2, x 3) T [R2-(2/1)R1]: Multiply 1 st row by (-2/1) and added to 2 nd row [R3-(3/1)R1] : Multiply 1 st row by (-3/1) and added to 3 rd row to get [R3-(1/1)R1] : Multiply 2 nd row by (-1/1) and added to 3 rd row to get

6 Back Substitution : From the last row -3x 3 = -9 x 3 = 3 From the second row x 2 + 2x 3 = 8 x 2 = 8-2x 3 = 2 Now from the first row x 1+ x 2 + x 3 = 6 x 1 = =1 x = (1, 2, 3) T The same procedure can be extended to the system of any order n provided the system has an unique solution. 9.3 Pitfalls of elimination Methods Division by Zero 2 x x 3 = 8 4 x x x 3 = -3 2 x 1 + x x 3 = Round-Off errors Error depends on the number of significant figures used ill-conditioned systems x x 2 = x x 2 = 10.4 Solve the same system with x x 2 = x x 2 = 10.4,, Ill-conditioned systems have near zero determinants

7 Effect of Scale on the determinant x x 2 = x x 2 = 10.4 D = 1(2)-2(1.1) = x 1 + x 2 = x 1 + x 2 = 5.2 D = 0.5 (1)-0.55(1) = (x x 2 = 10) *(10) (1.1 x x 2 = 10.4) *(10) D = 10(20)-20(11)= Techniques for improving solutions Use of more significant figures Pivoting If the diagonal element is zero or a vanishingly very small then the elements of the rows below this diagonal become very large in magnitude and difficult to handle because of the finite storage capacity of the computers. Alternative is to convert the system such that the element which has large magnitude in that column comes at the pivotal position i.e., the diagonal position. Partial Pivoting : If only row interchanging is used to bring the element of large magnitude of the pivotal column to the pivotal position at each step of diagonalization then such a process is called partial pivoting Complete Pivoting : In this process the largest element(in magnitude) of the whole coefficient matrix A is first brought at 1x 1 position of the coefficient matrix and then leaving the first row and first column, the largest among the remaining elements is brought to the pivotal 2 x2 position and so on by using both row and column transformations, is called complete pivoting Example x 1+ x 2 + x 3 = 6, 3x 1 + 3x 2 + 4x 3 = 20, 2x 1 + x 2 + 3x 3 = 13 Partial pivoting : Since the largest element in the first column is at 3x 1 which is not in the pivotal position, perform the row transformation R 1 «R 2. Now the system is 3x 1 + 3x 2 + 4x 3 = 20 x 1 + x 2 + x 3 = 6 2x 1 + x 2 + 3x 3 = 13 R 2 R 2-1/3 R 1 R 3 R 3-2/3 R 1 Augmented matrix is =

8 A b = 0 0-1/3-2/ /3-1/3 Since this is a '0' at the pivotal position i.e., at second row second column apply R 2«R 3 (interchange rows tow and three) A b = 0-1 1/3-1/ /3-2/3 Now the augmented matrix is in diagonal form(the part of coefficient matrix A). Back substitution : From the last row : -1/3x 3 = -2/3 x 3 = 2 Now from the second row -x 2 = -1/3-1/3x 2 x 2 =1 and from the first row 3x 1 = 20-3x 1-4x(2) x 1 =3 Hence the solution is X = (3 1 2) T Complete Pivoting : The given system is 3x 1 + 3x 2 + 4x 3 = 20 x 1 + x 2 + x 3 = 6 2x 1 + x 2 + 3x 3 = 13 Since the largest element in magnitude is at first row third column perform the column transformation C 1 C 3 (interchange first and third columns) then the augmented matrix is Augmented matrix is (please note the order of the individual elements of the unknown vector x is now (x 3 x 2 x 1) T Perform R 2 R 2-1/4R 1 R 3 R 3-3/4R A b = 0 1/4 1/ /4-1/4-2 Now the element with the largest magnitude is in the third row(leaving the first row aside) Perform R 2 R 3

9 R 3 R 3 - (-1/5) R 2 Back substitution A b = A b = /4-1/ /4 1/ /4-1/ /5 3/5 From the last row 1/5x 1 = 3/5 x 1 =3 From the second row -5/4x 2 = -2 +1/4 x 3 x 2=1 4x 3 = 20-3x 2-3x 1 = = 8 x 3 = 2 Ill-conditioned system Use Gaussian elimination to solve the system of linear equations. x + y= 0, x + (401/400) y= 20, Round each intermediate calculation to four significant digits. SOLUTION Using Gaussian elimination with rational arithmetic, you can find the exact solution to be y= 8000 and x= and But rounding 401/400 = to four significant digits introduces a large rounding error, as follows. (R2 - R1) So, y = 10,000, and back-substitution produces x = -y = - 10,000 This solution represents a percentage error of 25% for both the -value and the -value. Note that this error was caused by a rounding error of only (when you rounded to 1.002).

10 Gausss Jordan Method

11 Example 2- solve using Gauss Jordan Method 2x 1 + x 2 - x 3 = 1 5x 1 + 2x 2 + 2x 3 = -4 3x 1 + x 2 + x 3 = 5