* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Gaussian_elimination_V2 - Ms
Survey
Document related concepts
Linear least squares (mathematics) wikipedia , lookup
Rotation matrix wikipedia , lookup
Jordan normal form wikipedia , lookup
Eigenvalues and eigenvectors wikipedia , lookup
Determinant wikipedia , lookup
Singular-value decomposition wikipedia , lookup
Four-vector wikipedia , lookup
Matrix (mathematics) wikipedia , lookup
Non-negative matrix factorization wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
Orthogonal matrix wikipedia , lookup
Cayley–Hamilton theorem wikipedia , lookup
Matrix calculus wikipedia , lookup
System of linear equations wikipedia , lookup
Transcript
Digital Lesson Gaussian Elimination Gaussian elimination with back-substitution is a procedure for solving linear systems by applying a particular sequence of elementary row operations to the augmented matrix of the system. A unique augmented matrix is associated with each system of linear equations. System 2 x 5 y 15 3 x 2 y 13 Augmented matrix 2 5 3 - 2 15 13 Elementary row operations are of three types. 1. Interchange two rows of a matrix. 2. Multiply a row of a matrix by a nonzero constant. 3. Add a multiple of one row of a matrix to another. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 A row of a matrix is a zero row if it consists entirely of zeros. A row which is not a zero row is a nonzero row. A unit column of a matrix is a column with one entry a 1 and all other entries 0. non-zero row zero row 0 0 1 0 2 0 6 0 1 0 1 0 1 0 0 0 0 0 2 0 3 0 0 0 unit column Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 A matrix is in row-echelon form if and only if: 1. All zero rows occur below all nonzero rows. 2. The first nonzero entry of any nonzero row is a one. This is called the leading 1 of the row. 3. For any two nonzero rows, the leading 1 of the higher row is to the left of the leading 1 of the lower row. 1 0 6 3 3. 0 1 - 1 4 2. 0 0 1 2 1. 0 0 0 0 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 A matrix is in reduced row-echelon form if and only if: 1. The matrix is in row echelon form. 2. Every column containing a leading one is a unit column. unit columns 0 0 0 0 1 0 0 0 3 0 0 0 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 0 1 0 0 0 0 1 0 2 5 leading 1s 2 0 5 Example: Tell if each matrix is in row-echelon form. not a unit column 1 2 3 A 0 1 7 0 0 0 0 B 0 1 C 0 0 1 0 A is in row-echelon form. A fails to be in reduced row-echelon form because the second column contains a leading 1, but is not a unit column. leading 1 0 B is in both row-echelon form and reduced row-echelon form. 1 2 1 5 0 0 1 0 1 0 C fails to be in row-echelon form because the leading 1 in the second row is to the right of the leading 1 in the third row. Since C is not in row-echelon form, it cannot be in reduced row-echelon form. leading 1s Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 To solve a system of linear equations using Gaussian elimination with back substitution, 1. Write the augmented matrix of the system. 2. Use elementary row operations to transform the augmented matrix into row-echelon form. 3.Write the system of equations corresponding to the augmented matrix in row-echelon form 4. Use back substitution to find the solution. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 Example: Use Gaussian elimination with back substitution to solve y 3z 3 the system 10 2 x 4 y 2 x 4 y 3z 16 0 2 2 1 R 1 2 1 4 4 3 0 3 1 2 0 0 1 3 2 4 3 3 10 16 5 3 16 –2R1 + R3 R2 2 R1 0 2 4 1 4 0 3 3 10 3 16 1 0 0 2 1 0 0 3 3 5 3 6 Example continued Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Example continued 1 0 0 2 0 1 3 0 3 5 3 1 R3 6 3 1 0 0 2 0 1 3 0 1 5 3 2 Row-echelon form 5 x 2 y The system associated with this augmented matrix is: y 3z 3 z2 Substitute z = 2 back into the second equation and solve for y . y + 3(2) = 3 y = –3. Substitute y = –3 back into the first equation and solve for x . x + 2(–3) = 5 x = –1. The unique solution is (–1, –3, 2). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 Example: Use Gaussian elimination with back substitution to solve the system x y 4 . 2 x 2 y 6 1 1 2 2 4 6 –2R1 + R2 1 1 0 0 4 - 2 Row-echelon form The system corresponding to this augmented matrix is: x y 4 0 2 Since the second equation is false, the system has no solutions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10