Lecture 12 Simultaneous Linear Equations Gaussian Elimination (1) Dr.Qi Ying

Transcription

1 Lecture 12 Simultaneous Linear Equations Gaussian Elimination (1) Dr.Qi Ying

2 Objectives Understanding forward elimination and back substitution in Gaussian elimination method Understanding the concept of singularity and ill-condition

3 Introduction Linear equation: each term in each equation contains only one unknown, and each unknown appears to the first power.

4 Determinant and Cramer s Rule The determinant of a matrix [A] is written as: det(a), deta or A Cramer s Rule: For a linear system [A]{x}={b}, the solution is given by x i = det( A ) i det( A) A i is the matrix formed by replacing the i-th column of [A] by the column vector {b}

5 Determinant for 2x2 and 3x3 matrices 2x2 det( ) 3x3 a a A = = a11a22 a12a21 a21 a22 a a a det( A) = a a a a a a a a a a a a = a a + a a a a a a a i+ j = ( 1) aa i=1 or 2 or 3 j= 1 ij ij minor of the original matrix 22 32

6 General solution for two equations a x + a x = b a x + a x = b and two unknowns x b a b a, a b = x = a b a a a a a a a a

7 Elimination method For systems with more than 3 equations, Cramer s rule becomes impractical. e.g. for 20x20 matrix, needs 4.62x10 19 multiplications Will take 15,000 years for a computer that does 100 million floating-point operations per second. Some more efficient methods, such as elimination methods are needed to solve linear systems of equations.

8 Example problem 1 x y = 1 2x+ y = 4 Eliminate x x y = 1 3y = 6 Solve for y x =1 Back substitute to solve for x y = 2

9 Example problem 1 y (x=1, y=2) x-y=-1 2x+y=4 X

10 Example problem 2 x+y+z=0 x+3y+9z=2 x + y + z = 0 x + 3y + 9z = 2 x y z = 0 Solution x=0,y=-1/3,z=1/3 x+y+z=0

11 Example problem 2 x + y + z = 0 x + 3y + 9z = 2 x y z = 0 Eliminate x x + y + z = 0 2y+ 8z = 2 2y 2z = 0 Eliminate y x + y + z = 0 2y+ 8z = 2 6z = 2 Solve for z x = 0 y = 1 z = x + y + z = 0 1 y = 3 1 z = 3 Back substitute x + y + z = 0 2y+ 8z = 2 1 z = 3 Back substitute

12 Example problem 2 Forward elimination in matrix form using row operations

13 Gauss Elimination a a... a x + a x + a x a x = b n n 1 x + a x + a x a x = b n x + a x + a x a x = b n1 1 n2 2 n3 3 nn n n n 2 (1) (2) (n)

14 Gauss Elimination Forward Elimination To eliminate x1 in equation 2: 1) Multiply (a21/a11) with all the terms in equation (1), then 2) Add the resulting equation to equation (2) +) a x a x a a a a x... a a nx a n = b a a a a a a x + a x + a x a nxn = b a 21 a21 a21 a21 a21 a21 a11 1 a22 a12 2 a a2n 1n n 2 1 a x 11 a x 11 a a x a x = b b 11 a11 a11 (2 ) 0 a 22 a 23 a 2n b 2

15 Gauss Elimination Forward Elimination a x + a x + a x a x = n n 1 a a x + a x a x = b n... x + a x a x = b n2 2 n3 3 nn n n n b 2 (1) (2 ) (n )

16 a Gauss Elimination Forward Elimination x + a x + a x a x = b n n 1 a x + a x a x = b n a x a x = b 33 3 n3 3 nn n 2 3n n 3... a x a x = b n n

17 a Gauss Elimination Forward Elimination x + a x + a x a x = b n n 1 a x + a x a x = b n a x a x = b n n 3 n... 2 ( n 1) = ( n 1) nn n n a x b

18 Gauss Elimination Forward Elimination a11 a12 a13... a1 n a11 a12 a13... a1 n a21 a22 a23... a 2n 0 a 22 a a 2n a31 a32 a33... a3n 0 a 32 a a 3n a a a... a 0 a a... a n1 n2 n3 nn n2 n3 nn a11 a12 a13... a1 n x1 b1 0 a 22 a a 2n x 2 b a a 3n x3 = b n 1 ( n 1) ann xn bn

19 a a... a a a... a a a... a Gauss Elimination Forward ii i, i+ 1 i, n i+ 1, i i+ 1, i+ 1 i+ 1, n n, i n, i+ 1 n, n Elimination Current row: i Goal: eliminate a i+1,i ;a i+2,i ; ;a n,i % current row is i % forward elimination for rows i+1 to n for m=i+1:n fac=a(m,i)/a(i,i); % evaluate all coefficients from column i to n for j=i:n A(m,j)=A(m,j)-fac*A(i,j); end end

20 Gauss Elimination Forward Elimination Now x n can be solved from the last equation: x n = ( n 1) b n ( n 1) ann Plug this into the second last equation: a x + a x = b x ( n 2) ( n 2) ( n 2) n 1, n 1 n 1 n 1, n n n 1 x n-1 can be solved: ( n 2) ( n 2) b n 1 a n 1, n xn n 1 = ( n 2) an 1, n 1

21 Gauss Elimination Forward Elimination Now x n-2 can be solved from the third last equation: a x + a x + a x = b ( n 3) ( n 3) ( n 3) ( n 3) n 2, n 1 n 2 n 2, n 1 n 1 n 2, n n n 2 x = ( a x a x ) ( n 3) ( n 3) ( n 3) b n 2 n 2, n 1 n 1 + n 2, n n n 2 ( n 3) an 2, n 2 Repeat the back substitution steps, x i can be solved as: x i = b n ( i 1) ( i 1) i aij j= i+ 1 a ( i 1 ii x j ) for i=n-1, n-2,..., 1

22 function x = gauss_simple (A, b) % gauss_simple: simple Gauss elimination % input: A = coefficient matrix, b = right hand side of the equations n=size(a,1); % forward elimination for i=1:n-1 % loop over all the rows (no need for last row) for m=i+1:n % loop over rows i+1 to n fac=a(m,i)/a(i,i); for j=i:n % calculate coefficients in row m A(m,j)=A(m,j)-fac*A(i,j); end b(m)=b(m)-fac*b(i); % calculate the right hand side(rhs) end end % back substitution x(n)=b(n)/a(n,n); for i=n-1:-1:1 sum=0; for j=i+1:n sum=sum+a(i,j)*x(j); end x(i)=(b(i)-sum)/a(i,i); end end

23 Singularity No solution or infinite number of solutions for the linear equations. No Solution x-y+10=0 x-y=0

24 Singularity Infinite number of solutions

25 Ill-conditioned systems A system is either singular or it is not if the operations can be carried out in infinite precision. In real computer systems, a system can be almost singular, leading to a solution that has little reliability.

26 Ill-conditioned systems Example Compare the solution of the two systems of equations: x1+ 2x2 = x1+ 2x2 = 10.4 x1+ 2x2 = x1+ 2x2 = 10.4

27 Ill-conditioned systems Example 5x+ 7 y = 12 7x+ 10y = 17 x = 1, y = 1 Now consider: (2.415,0) x = 2.415, y = 0 5x+ 7 y = x+ 10y =

28 Gauss Elimination Improvements Examine the factor used in the forward elimination Fac=A(m,i)/A(i,i); 1. A(i,i) = 0 -- leads to divide by zero error 2. A(i,i) is close to zero lead to round-off error. Solution: rearrange the rows so that the ith diagonal element is as large as possible.