1.  Lecture #9: 1.Linear Equations: y=mx +b 2.Solution System: N.S., U.S., I.S. 3.Augmented Matrix 4.Solving a System of Linear Equations  Today: 1.Echelon Form, Reduced Echelon Form 2.Gauss-Jordan Elimination Method 3.Homogeneous Linear Equations 4.Matrix Operations Lecture #10: System of Linear Equations & Matrices

2. Announcements: Review Class on Tuesday 28: –Room: Ricketts 203 –Time: 6:30-8:00pm Exam #1 Next Wednesday: 9/29 Unit Vectors, Cartesian Vector Form.

3. System of linear equations The general form: A 11 x 1 +A 12 x 2 +A 13 x 3 +…..A 1n x n =B 1 Rx=0 A 21 x 1 +A 22 x 2 +A 23 x 3 +…..A 2n x n =B 2 Ry=0 A 31 x 1 +A 32 x 2 +A 33 x 3 +…..A 3n x n =B 3 Rz=0..... A m1 x 1 +A m2 x 2 +A m3 x 3 +…A mn x n =B m

4. Matrix Form: Coefficient Matrix ROW # Column #

5. Augmented Matrix: System of linear eqns. 1x + y + 2z = 9 2x + 4y – 3z = 1 3x + 6y –5z = 0 Remember: R x =0 R y =0 R z =0 Augmented Matrix: (array of numbers of the system of eqns)

6. Solving a System of Linear Eqns. GOAL –FIND the solution for x, y,z (T A, T B, T C, T D, T E ) The idea is to replace a given system by a system which has the same solution set, but it is easier to solve.

7. Basic Operations to find Unknown Multiply a row by a nonzero constant. (the row you multiply by a number after adding the two rows will not change) Interchange two rows. Add a multiple of one row to another row.

8. Gauss-Jordan Elimination Goal: to reduce the augmented matrix into a form simple enough such that system of equation are solved by inspection.

9. Reduced row-echelom form 1.If row does not consist entirely of zeros, then the first non-zero number in row is 1. 2.If a row consist of zeros, then they are moved to the bottom of matrix. 3.In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occur farther to the right of above row. 4.Each column that contains a leading 1 has zero everywhere.

10. IMPORTANT Reduced Row echelom –Must have zeros above and below each leading 1. Row-echelom form –Must have zeros below each leading 1.

11. Gauss-Jordan Elimination Method Step1: Locate the leftmost column that does not consist entirely of zero. Step 2: Interchange the top row with another row, if necessary, to bring a nonzero entry to the top from step 1. Step 3: If the entry that is now at the top is a constant, divide entire row by it. Step 4: Add multiples to top row to the rows below such that all entries have 1 as leading term. Step 5: Cover top row and begin with step 1 applied to submatrix.

12. For problem 3.22 find FAB, FAC, FAD using Gauss-Jordan method. Example #1

13. Activity:#1 For Problem in example #1 solve using Gauss-Jordan Method. -T A (0.766) + T B (0.866) = 1699 T A (0.643) + T B (0.500) = 2943

14. Homogeneous System of Linear Equations Non-homogeneous A 11 x 1 +A 12 x 2 +A 13 x 3 +…..A 1n x n =B 1 A 21 x 1 +A 22 x 2 +A 23 x 3 +…..A 2n x n =B 2 A 31 x 1 +A 32 x 2 +A 33 x 3 +…..A 3n x n =B 3.. A m1 x 1 +A m2 x 2 +A m3 x 3 +…A mn x n =B m The Constants B not equal to 0 Homogeneous A 11 x 1 +A 12 x 2 +A 13 x 3 +…..A 1n x n = 0 A 21 x 1 +A 22 x 2 +A 23 x 3 +…..A 2n x n = 0 A 31 x 1 +A 32 x 2 +A 33 x 3 +…..A 3n x n = 0.. A m1 x 1 +A m2 x 2 +A m3 x 3 +…A mn x n = 0 The Constants B’s Equal to 0

15. Solutions in Homogeneous System Trivial Solution X 1 = 0 X 2 = 0 X 3 = 0 …. X n = 0 For same # equations and same # unknowns Non trivial solution: X 1 = C 1 X 2 = C 2 X 3 = C 3 …. X n = C 4 When there is more unknowns than equations.