Cramer s Rule and Gauss Elimination

Transcription

1 Outlines September 28, 2004

2 Outlines Part I: Review of Previous Lecture Part II: Review of Previous Lecture

3 Outlines Part I: Review of Previous Lecture Part II: Cramer s Rule Introduction Matrix Version and

4 Part I Review of Previous Lecture

5 Review of Previous Lecture Graphical interpretation Solvable and unsolvable problems Linear dependence and independence Ill-conditioning

6 Part II

7 Introduction Cramer s Rule (1750) A linear system of equations can be solved by using Cramer s rule, which for a system of 2 equations [ ] { } { } a11 a 12 x1 b1 = a 21 a 22 x 2 b 2 yields where x 1 = [A 1] [A], x 2 = [A 2] [A] [ ] b1 a A 1 = 12 b 2 a 22 [ ] a11 b A 2 = 1 a 21 b 2

8 Introduction Cramer s Rule Details For a system of n equations, Cramer s rule requires that you calculate n + 1 determinants of n n matrices. In the general case for a system of equations [A]{x} = {b}, the matrix [A i ] is obtained by replacing the ith column of the original [A] matrix with the contents of the {b} vector. Each unknown variable x i is found by dividing the determinant [A i ] by the determinant of the original coefficient matrix [A].

9 Introduction Cramer s Rule Solve the following system of equations using Cramer s rule: x 1 x 2 + x 3 = 3 2x 1 + x 2 x 3 = 0 3x 1 + 2x 2 + 2x 3 = 15 Convert the system of equations into matrix form: x x 2 = x 3 15

10 Introduction Cramer s Rule (continued) [A] = , {x} = Define matrices [A 1 ], [A 2 ], and [A 3 ] as [A 1 ] = [A 3 ] = x 1 x 2 x 3, [A 2 ] =, {b} = ,

11 Introduction Cramer s Rule (continued) Calculate determinants of [A], [A 1 ], [A 2 ], and [A 3 ]: [A] = 12 [A 1 ] = 12 [A 2 ] = 24 [A 3 ] = 48 Unknowns x 1, x 2, and x 3 are then calculated as x 1 = [A 1] [A] = = 1, x 2 = [A 2] = 24 [A] 12 = 2, x 3 = [A 3] = 48 [A] 12 = 4

12 Introduction Cramer s Rule (continued) Be sure to double-check your answers by substituting them into the original equations: x 1 x 2 + x 3 = = 3 2x 1 + x 2 x 3 = = 0 3x 1 + 2x 2 + 2x 3 = = 15

13 Introduction Cramer s Rule Advantages/Disadvantages Advantages Easy to remember steps Disadvantages Computationally intensive compared to other methods: the most efficient ways of calculating the determinant of an n n matrix require (n 1)(n!) operations. So Cramer s rule would require (n 1)((n + 1)!) total operations. For 8 equations, that works out to 7(9!) = operations, or around 700 hours if you can perform one operation per second. Roundoff error may become significant on large problems with non-integer coefficients.

14 Matrix Version and Recall the scaffolding problem from the beginning of Chapter 3. Its matrix form was T A P T B 5P T C P = 2 T D P T E P T F P 3 Notice that its coefficient matrix contains nothing but zeroes below the diagonal. This is an example of an upper triangular matrix, and these systems of equations are very easy to solve.

15 Matrix Version and Introduction (continued) The original system of equations on the scaffolding problem was T A +T B T C T D T F = P 1 9T B +T C +4T D +7T F = 5P 1 T C +T D T E = P 2 3T D +2T E = P 2 T E +T F = P 3 4T F = P 3 Notice that we can solve for T F using only the sixth equation in the system. That is, T F = P 3 4. After solving for T F, we can solve for T E using only the fifth equation. The pattern continues, back-substituting through the system of equations until finally we solve for T A using the first equation..

16 Matrix Version and Introduction (continued) The goal of Gauss elimination is to convert any given system of equations into an equivalent upper triangular form. Once converted, we can back-substitute through the equations, solving for the unknowns algebraically.

17 Matrix Version and Rules The operations used in converting a system of equations to upper triangular form are known as elementary operations and are: Any equation may be multiplied by a nonzero scalar. Any equation may be added to (or subtracted from) another equation. The positions of any two equations in the system may be swapped.

18 Matrix Version and 2x 1 x 2 + x 3 = 4 (1) 4x 1 + 3x 2 x 3 = 6 (2) 3x 1 + 2x 2 + 2x 3 = 15 (3) To eliminate the 4x 1 term in Equation 2, multiply Equation 1 by 2 and subtract it from Equation 2. To eliminate the 3x 1 term in Equation 3, multiply Equation 1 by 3 2 and subtract it from Equation 3. This gives a system of equations 2x 1 x 2 + x 3 = 4 (4) 5x 2 3x 3 = 2 (5) 7 2 x x 3 = 9 (6)

19 Matrix Version and (continued) To eliminate the 7 2 x 2 term from Equation 6, multiply Equation 5 by 7 10 and subtract it from Equation 6. This gives a system of equations 2x 1 x 2 + x 3 = 4 (7) 5x 2 3x 3 = 2 (8) 13 5 x 3 = 52 5 (9)

20 Matrix Version and (continued) Equation 9 can easily be solved for x 3. x 3 = 52 ( ) 5 = Equation 8 can easily be solved for x 2, once x 3 is known. x 2 = 1 5 ( 2 + 3x 3) = 1 ( 2 + 3(4)) = 2 5 Equation 7 can easily be solved for x 1, once both x 2 and x 3 are known. x 1 = 1 2 (4 + x 2 x 3 ) = 1 ( ) = 1 2

21 Matrix Version and Matrix Version of The Gauss elimination method can be applied to a system of equations in matrix form. Instead of eliminating terms from equations, we ll be replacing certain elements of the coefficient matrix with zeroes.

22 Matrix Version and Matrix Version (Step 0) Start by defining the augmented matrix [C (0) ] for the problem: [C (0) ] = n 1,n n 2,n n1 n2 a nn (0) n,n+1 where the first n columns are the elements of the original [A] matrix, and the last column is the elements of the original {b} matrix.

23 Matrix Version and Matrix Version (Step 1) Zero out the first column of the [C] matrix, rows 2 n. To turn a 21 to a zero, multiply row 1 by a 21 a 11, then subtract the numbers on row 1 from row 2. To turn a 31 to a zero, multiply row 1 by a 31 a 11, then subtract the numbers on row 1 from row 3. Repeat for rows 4 n. [C (1) ] = n 1,n+1 0 a (1) 22 a (1) 2n a (1) 2,n a (1) n2 a nn (1) a (1) n,n+1

24 Matrix Version and Matrix Version (Step 2) Zero out the second column of the [C] matrix, rows 3 n. To turn a 32 to a zero, multiply row 2 by a 32 a 22, then subtract the numbers on row 2 from row 3. To turn a 42 to a zero, multiply row 2 by a 42 a 22, then subtract the numbers on row 2 from row 4. Repeat for rows 5 n. [C (1) ] = n 1,n+1 0 a (1) 22 a (1) 13 a (1) 2n a (1) 2,n a (2) 33 a (2) 3n a (2) 3,n a (2) n3 a nn (2) a (2) n,n+1

25 Matrix Version and Matrix Version (Step n-1) Zero out the (n 1)th column of the [C] matrix, row n. To turn a n,n 1 to a zero, multiply row n 1 by a n,n 1 a n 1,n 1, then subtract the numbers on row n 1 from row n n 1,n+1 [C (1) ] = 0 a (1) 22 a (1) 13 a (1) 2n a (1) 2,n a (2) 33 a (2) 3n a (2) 3,n a nn (n 1). a (n 1) n,n+1

26 Matrix Version and Solve the following system of equations with Gauss elimination: x x 2 = x 3 15 First, set up the augmented matrix [C (0) ]: [C (0) ] =

27 Matrix Version and (continued) Step 1a: eliminate the 4 on row 2, column 1. Multiply all the elements of row 1 by a 21 a 11 = 4 2 = 2, then subtract them from the elements of row [C] = 4 (2)(2) 3 (2)( 1) 1 (2)(1) 6 (2)(4) =

28 Matrix Version and (continued) Step 1b: eliminate the 3 on row 3, column 1. Multiply all the elements of row 1 by a 31 a 11 = 3 2 = 1.5, then subtract them from the elements of row [C (1) ] = (1.5)(2) 2 (1.5)( 1) 2 (1.5)(1) 15 (1.5)(4) = This completes the first elimination step.

29 Matrix Version and (continued) Step 2: eliminate the 3.5 on row 3, column 2. Multiply all the elements of row 2 by a 32 a 22 = = 0.7, then subtract them from the elements of row [C (2) ] = (0.7)(5) 0.5 (0.7)( 3) 9 (0.7)( 2) = This completes the second elimination step.

30 Matrix Version and (continued) We ve now converted the original system of equations x x 2 = x 3 15 into an equivalent upper-triangular system of equations x x 2 = x

31 Matrix Version and (continued) The new system of equations can be converted back to algebraic form as: 2x 1 x 2 + x 3 = 4 (10) 5x 2 3x 3 = 2 (11) 2.6x 3 = 10.4 (12) Solve Equation 12 for x 3 : x 3 = = 4. Then solve Equation 11 for x 2 : x 2 = 1 5 ( 2 + 3x 3) = 2. Then solve Equation 10 for x 1 : x 1 = 1 2 (4 + x 2 x 3 ) = 1.

32 Matrix Version and (continued) Double-check the solution by substituting the values of x 1, x 2, and x 3 into the original equations: 2x 1 x 2 + x 3 = 2(1) = 4 4x 1 3x 2 x 3 = 4(1) + 3(2) 4 = 6 3x 1 + 2x 2 + 2x 3 = 3(1) + 2(2) + 2(4) = 15

33 Matrix Version and Advantages/Disadvantages Advantages Much less computation required for larger problems. Gauss elimination requires n3 3 multiplications to solve a system of n equations. For 8 equations, this works out to around 170 operations, versus the roughly 2.5 million operations for Cramer s rule. Disadvantages Not quite as easy to remember the procedure for hand solutions. Roundoff error may become significant, but can be partially mitigated by using more advanced techniques such as pivoting or scaling.

34 Solve Problem 3.4 using: Cramer s rule, Gauss elimination, and MATLAB s \ operator. Double-check your answers by substituting them back into the original equations.