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Lecture #10: System of Linear Equations & Matrices

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

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Lecture #10: System of Linear Equations & Matrices

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

  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: A11x1+A12x2+A13x3+…..A1nxn=B1 Rx=0 A21x1+A22x2+A23x3+…..A2nxn=B2 Ry=0 A31x1+A32x2+A33x3+…..A3nxn=B3 Rz=0 . . . . . . . . . . Am1x1+Am2x2+Am3x3+…Amnxn=Bm

  4. Matrix Form: • Coefficient Matrix ROW # Column #

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

  6. Solving a System of Linear Eqns. • GOAL • FIND the solution for x, y,z (TA, TB, TC, TD, TE) • 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 • If row does not consist entirely of zeros, then the first non-zero number in row is 1. • If a row consist of zeros, then they are moved to the bottom of matrix. • 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. • Each column that contains a leading 1 has zero everywhere.

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

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

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

  14. Non-homogeneous A11x1+A12x2+A13x3+…..A1nxn=B1 A21x1+A22x2+A23x3+…..A2nxn=B2 A31x1+A32x2+A33x3+…..A3nxn=B3 . . . . . . . . Am1x1+Am2x2+Am3x3+…Amnxn=Bm The Constants B not equal to 0 Homogeneous A11x1+A12x2+A13x3+…..A1nxn= 0 A21x1+A22x2+A23x3+…..A2nxn= 0 A31x1+A32x2+A33x3+…..A3nxn= 0 . . . . . . . . Am1x1+Am2x2+Am3x3+…Amnxn= 0 The Constants B’s Equal to 0 Homogeneous System of Linear Equations

  15. Trivial Solution X1 = 0 X2 = 0 X3 = 0 …. …. Xn = 0 For same # equations and same # unknowns Non trivial solution: X1 = C1 X2 = C2 X3 = C3 …. …. Xn = C4 When there is more unknowns than equations . Solutions in Homogeneous System

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