Linear Algebra (2009 Fall) Chapter 1 Matrices and Systems of Equations

Transcription

1 (9 Fall) Chapter Matrices and Sstems of Equations Chih-Wei Yi Dept. of Computer Science National Chiao Tung Universit October 9, 9

2 Linear Sstems Linear Sstems

3 Linear Sstems Linear Equations De nition (Linear Equations) A linear equation in n unknowns (variables),,..., n : a a a n n b. Eample,,, n : variables ( : leading varialbes) a, a,, a n : constants and called coe cients (a : leading coe cient) b: constant term 6 is a linear equation in unknowns, but sin and are not linear equations.

4 Linear Sstems Eamples Which are linear equations? sin (sin ) e log

5 Linear Sstems Solutions of Linear Equations De nition (Solutions) Assume s, s,..., s n are n real numbers. s, s,..., n s n is called a solution of linear equation a a a n n b if a s a s a n s n b is satis ed. Eample Consider the linear equation:. and is a solution. For an real number t, t and (Here t is called a parameter.) t is a solution.

6 Linear Sstems Eercise Problem Assume u (u, u,..., u n ) and v (v, v,..., v n ) are two solutions of a a... a n n b. Prove that for an real number c, u c(u v) is a solution of a a... a n n b. Solution (Hints) Show that (u v) is a solution of a a... a n n. Then, ou can prove u c(u v) is a solution of a a... a n n b.

7 Linear Sstems Linear Sstems De nition (Linear Sstems) A linear sstem of m equations in n unknows is a collection of m linear equations in n common unknowns. 8 a a... a n n b >< a a... a n n b.. >: a m a m... a mn n b m It is called an m n sstem. A solution to an m n sstem is an ordered n-tuple of real numbers (,,, n ) that satis es all m equations of the sstem. The set of all solutions to a linear sstem is called the solution set of the sstem. Here an (ordered) n-tuple of real numbers (,,, n ) is the same as a vector in R n space.

8 Linear Sstems Eample (, ) is a solution of set of the linear sstem.. f(, )g is the solutions Eample (,, ), (,, ), and (,, ) are solutions of. Actuall, this sstem has in nite solution, and its solution set is f( t,, t) j t Rg. Eample 8 < The sstem : is?. has no solution, so its solution set

9 Linear Sstems The Number of Solutions of A Linear Sstem 6

10 Linear Sstems Problem Prove the number of solutions of a linear sstem must be one of the following cases Eactl solution In nite number of solutions No solution (Hint: If u (u, u,, u n ) and v (v, v,, v n ) are solutions of a linear sstem in n variables, then for an c R, u c (u v) is also a solution of the sstem.) De nition (Consistent and Inconsistent) A linear sstem is inconsistent if its solution set is empt, otherwise it is consistent. Note: A consistent linear sstem has either eactl one solution or otherwise in nite solutions.

11 Gaussian Elimination Gaussian Elimination

12 Gaussian Elimination Row-Echelon Sstems De nition (Row-Echelon Form) A sstem is in row-echelon form if it follows a stair-step pattern and has leading coe cients of. All variables are aligned. In an equation, the leading coe cient is the coe cient of the rst variables. Eample not row echelon form row echelon form

13 Gaussian Elimination Solve a Row-Echelon Sstem A row-echelon sstem can be solved b back-substitution. 9 row-echelon sstem 9 9 ( ) back-substitution

14 Gaussian Elimination Gaussian Elimination De nition (Equivalent Sstems) Two sstems of linear equations are called equivalent if the have precisel the same solution set. Operations producing equivalent sstems Interchange two equations. Multipl an equation b a nonero constant. Add a multiple of an equation to another equation. Gaussian elimination: rewrite a sstem to an equivalent row-echelon sstem b a sequence of these three operations.

15 Gaussian Elimination Solve a Sstem of Linear Equations After appling backsubstitution, we have,,.

16 Gaussian Elimination An mn Matri n columns m rows a a M a m a a a M m L L O L a a a n n M mn st row nd row m st row st column n st column

17 Gaussian Elimination Represent a Linear Sstem b a Matri sstem coefficient matri augmented matri

18 Gaussian Elimination Elementar Row Operations Interchange two rows. Multipl a row b a nonero constant. Add a multiple of a row to another row. Remark: Note the elementar row operation performed in each step.

19 Gaussian Elimination Eample Notation R R ( ) Notation R R R ( ) Notation R R

20 Gaussian Elimination Row-Echelon Form Row-echelon form All rows consisting entirel of eros occur at the bottom of the matri. For each row that does not consist entirel of eros, the rst nonero entr is (called a leading ). For two successive (nonero) rows, the leading in the higher row is farther to the left than the leading in the lower row. Reduced row-echelon form Ever column that has a leading has eros in ever position ecepting the leading.

21 Gaussian Elimination Gaussian Elimination with Back-Substitution R R! R R R! R R R! R

22 Gaussian Elimination Gaussian Elimination with Back-Substitution 9 9 R! R 9 9 ( ) 9 ; Back-Substitution

23 Gaussian Elimination Practice w w 7 w Hint: Echange the st and nd equations.

24 Gaussian Elimination Solution ! 7! 7! ! 7!

25 Gaussian Elimination The Number of Solutions Inconsistent sstem: a row with eros ecept for the last entr (eample) Consistent sstem: not inconsistent sstems One solution: the number of not-ero rows is equal to the number of variables In nite solutions: the number of not-ero rows is less than the number of variables Qui: Homogeneous sstems in which each of the constant term is ero are consistent.

26 Gauss-Jordan Elimination Gauss-Jordan Elimination

27 Gauss-Jordan Elimination Gauss-Jordan Elimination Continues the procedure of Gaussian elimination until a reduced row-echelon form is obtained. For eample, !! 7 9 9!!!

28 Gauss-Jordan Elimination Eample w w 7 w 9

29 Gauss-Jordan Elimination Solution 6!! ! 7 7!!

30 Gauss-Jordan Elimination More Eamples 7 w w w t w t w t w t w Then,. Let t t t t w : Ans

31 Gauss-Jordan Elimination Eample 6w Solution!...! 6 Let s and w t, then, Ans: 6 w 7 w t w s t s t 6 s 7 t t

32 Gauss-Jordan Elimination Algorithm for Gaussian Elimination Input an mn matri A i ; j ; while i < m and j < n if all entries A(i, j), A(i, j),, A(m, j) are ero then j j and continue else if A(i, j) then 9k > i s.t. A(k, j) 6 and switch row(i) and row(k) //Here we have A(i, j) 6 divide row(i) b A(i, j) for k i to m if A(k, j) 6 then minus A(k, j) times of row(i) from row(k) i i ; j j ;

33 Applications of Linear Sstems Applications of Linear Sstems

34 Applications of Linear Sstems Polnomial Curve Fitting Given n points: (, ), (, ),, ( n, n ). Find a polnomial equation p() a a... a n n passing through all n points. FFF n points give n equations and therefore can solve n variables, a, a,, a n. Insert Figure.

35 Applications of Linear Sstems Eample Find the polnomial p() a a a that passes through the points (, ), (, ), and (, ). Solution From (, ), we have a a a. From (, ), we have a a a. From (, ), we have a a 9a. 9!...! Answer: p()

36 Applications of Linear Sstems Eercise Find the polnomial p() a a a a a passing through the points (, ), (, ), (, ), (, ), (, ). 6 ( ) ( ) ( ) ( ) ( ) ( ) Find a polnomial p() a a a a passing through the points (, ), (, ), (, ), (, ). 7

37 Applications of Linear Sstems Unit Review Linear equations and sstems of linear equations Solutions Number of solutions Gaussian elimination Equivalent sstems elementar row operation Row-echelon form Back-substitution Gauss-Jordan elimination Reduced row-echelon form Parametric solutions Matri representation