Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Lecture #10: System of Linear Equations & Matrices Lecture #9: 1. 2. 3. 4. Linear Equations: y=mx +b Solution System: N.S., U.S., I.S. Augmented Matrix Solving a System of Linear Equations Today: 1. 2. 3. 4. Echelon Form, Reduced Echelon Form Gauss-Jordan Elimination Method Homogeneous Linear Equations Matrix Operations 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. System of linear equations • The general form: A11x1+A12x2+A13x3+…..A1nxn=B1 A21x1+A22x2+A23x3+…..A2nxn=B2 A31x1+A32x2+A33x3+…..A3nxn=B3 . . . . . . . . . . Am1x1+Am2x2+Am3x3+…Amnxn=Bm Rx=0 Ry=0 Rz=0 Matrix Form: • Coefficient Matrix ROW # a11 a21 a31 am1 a12 a22 a32 am 2 a13 ....... a23 ....... a33 ....... am3 ....... Column # a1n a2 n a3n amn Augmented Matrix: • System of linear eqns. • Augmented Matrix: (array of numbers of the system of eqns) 1x + y + 2z = 9 2x + 4y – 3z = 1 3x + 6y –5z = 0 Remember: Rx=0 Ry=0 Rz=0 1 1 2 9 2 4 3 1 3 6 5 0 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. 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. Gauss-Jordan Elimination • Goal: to reduce the augmented matrix into a form simple enough such that system of equation are solved by inspection. 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. IMPORTANT • Reduced Row echelom – Must have zeros above and below each leading 1. • Row-echelom form – Must have zeros below each leading 1. 1 0 0 9 0 1 0 1 1 4 0 9 0 1 0 1 0 0 1 0 0 0 1 0 1 0 0 9 0 1 0 1 1 10 12 9 0 0 1 7 0 0 0 0 0 0 0 1 0 0 0 1 0 1 7 1 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. Example #1 • For problem 3.22 find FAB, FAC, FAD using Gauss-Jordan method. F x 0 0 2 3 FAD FAC FAB 0 7 45 44 Fy 0 6 6 6 FAD FAC FAB 0 7 44 45 Fz 0 3 2 2 FAD FAC FAB 1250 7 44 45 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 Homogeneous System of Linear Equations • Non-homogeneous • Homogeneous A11x1+A12x2+A13x3+…..A1nxn=B1 A21x1+A22x2+A23x3+…..A2nxn=B2 A31x1+A32x2+A33x3+…..A3nxn=B3 . . . . . . . . Am1x1+Am2x2+Am3x3+…Amnxn=Bm A11x1+A12x2+A13x3+…..A1nxn= 0 A21x1+A22x2+A23x3+…..A2nxn= 0 A31x1+A32x2+A33x3+…..A3nxn= 0 . . . . . . . . Am1x1+Am2x2+Am3x3+…Amnxn= 0 The Constants B not equal to 0 The Constants B’s Equal to 0 Solutions in Homogeneous System • 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.