1. ENGG2013 Unit 2 Linear Equations Jan, 2011.

2. Linear Equation in n variables a 1 x 1 + a 2 x 2 + … + a n x n = c – a 1, a 2, …, a n are called coefficients (real numbers). – x 1, x 2,…, x n are variables (or indeterminates). – c is a constant term (real number). Example – 2x + 3y – 4z = 0.2 Non-example – x 2 +y 2 =1 kshumENGG20132

3. Geometry of a linear equation kshumENGG20133 Two variables: straight line ax + by = c Three variables: plane ax + by + cz = d

4. System of linear equations A system of linear equations (or linear system) is a collection of one or more linear equations. – for example: A solution is a list of numbers (s 1, s 2, …, s n ) which satisfies all equalities after substituting x i by s i, for i =1,2,…,n. The set of all solutions is called the solution set. kshumENGG20134

5. Nutrition problem Find a combination of food A, B, C and D in order to satisfy the nutrition requirement exactly. Let x A, x B, x C and x D be the amount of food A, B, C and D respectively. kshumENGG20135 Food AFood BFood CFood DRequirement Protein98335 Carbohydrate1511145 Vitamin A0.020.0030.010.0060.01 Vitamin C0.01 0.0050.050.01

6. Formal notation Given a system of m linear equations in n variables the solution set is defined as kshumENGG20136 Double subscripts

7. Review of set notation Set of Greek letters = { , , , , , , , , , ,, ,, , , , , , , , , , ,  } Set of prime numbers = {2,3,5,7,11,13,17,23,29,31,37,41, …} Sphere with radius r centered at origin = {(x,y,z): x 2 +y 2 +z 2 =r 2 } kshumENGG20137 finite Countably infinite Uncountably infinite

8. Examples of solution sets kshumENGG20138 { (x,y): ax + by = c } x y

9. Consistency A linear system is called consistent if there is at least one solution, in other words, if the solution set is non-empty. kshumENGG20139 x y Inconsistent, no solution x y Consistent

10. Classification kshumENGG201310 Linear System Inconsistent (no solution) Consistent Unique solution Infinitely many solutions Tasks: Determine whether a linear system is consistent. If yes, find all solutions.

11. Short-hand notation using matrix kshumENGG201311 (2 rows, 4 columns) (4 rows, 3 columns) Usually called the augmented matrix

12. The nutrition example kshumENGG201312

13. Elementary row operations 1.Interchange two rows 2.Multiply a row by a non-zero constant 3.Replace a row by the sum of itself and a constant multiple of another row kshumENGG201313 Facts: Elementary row operations do not change the solution(s). (There is no loss, and no gain, of information.)

14. Illustration – row interchange kshumENGG201314

15. Illustration – Multiply by constant kshumENGG201315 22 22

16. Illustration – Row replacement kshumENGG201316 (1)  (1) – (2)

17. How to solve? Idea: Apply the three kinds of information- lossless elementary row operations, and transform the linear system into one which is easier to solve. kshumENGG201317 Linear system in upper triangular matrix form can be easily solved by backward substitution

18. Carl Friedrich Gauss kshumENGG201318 (1777~1855) The old Deutsche 10-Mark note

19. Gaussian elimination Step 0: Write the linear system in matrix format Step 1: Try to transform the matrix into upper triangular form Step 2: Solve for the variables one by one, in backward order kshumENGG201319

20. Example 1 (row operations) kshumENGG201320 (2)  (2) – (1) (3) (2) (1) (3)  (3) + (2)/2 Solve

21. Example 1 (backward sub.) kshumENGG201321 Upper triangular (3)  z = 7/3 (2)  – 2y – (7/3) = 1  y = –5/3 (1)  x+(–5/3)+(7/3) = 1  x = 1/3 Verify: x+y+z = 1/3 – 5/3 + 7/3 = 3/3 = 1 x–y = (1/3) – (– 5/3) = 6/3 = 2 y+2z = (– 5/3)+2(7/3) = 9/3 = 3 Solution: x=1/3, y = –5/3, z = 7/3 (unique solution)

22. Example 2 (row operations) kshumENGG201322 (2)  (2) – (1) (3)  (3) – (1) (3)  (3) – 2  (2) Solve

23. Example 2 (backward sub.) kshumENGG201323 z can be taken as a free variable. Let z to be any real number. (1) (2) From (2), y = –1 – z From (1), x +(–1 – z)+3z = 1  x = 2 – 2z Solution: x= 2– 2z, y = –1–z, z = any real number. Solution set = {(2 – 2z, –1–z, z): z is any real no.} (Infinitely many solutions) Note: You can let y to be the free variable as well, and obtain the solutions in terms of y.

24. Example 2 (cont’d) Verification – x+y+3z =(2 – 2z) + (– 1 – z) + 3z = 1 – x+2y+4z =(2 – 2z) + 2(– 1 – z) + 3z = 0 – x+3y+5z =(2 – 2z) + 3(– 1 – z) + 5z = – 1 kshumENGG201324 Solution: x= 2– 2z, y = –1–z, z = any real number Solution set = {(2 – 2z, –1–z, z): z is any real no.}

25. Example 3 (row operations) kshumENGG201325 Solve (2)  (2) – (1) (3)  (3) – (1) (3)  (3) – 2  (2)

26. Example 3 (cont’d) kshumENGG201326 Contradiction, cannot be true Answer: the linear system is inconsistent

27. Example 3 (picture) kshumENGG201327 Cross-section No common intersection An infinitely long triangular tube is formed by the three planes

28. Key concepts Three kinds of elementary row operations – The solution set is invariant under any elementary row operation Gaussian elimination – Transform a linear system to upper triangular form – Backward substitution Three types of solutions – No solution – Unique solution – Infinitely many solutions kshumENGG201328