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linear algebraic systems i

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18 diapositives · 524 Views · Sep 15, 2011

Fundamental Theorem. Linear algebraic systemConsistencyThe system has solutions if and only if the matrices A and have the same rank rUniquenessThe system has a single solution if and only if both matrices have rank r = nInfinitely many solutionsThe system has infinitely many solutions if and only if both matrices have rank r < n.

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    1. Linear Algebraic Systems I Existence and uniqueness of solutions Determinants and matrix inverses Gauss-Jordan elimination Ill-conditioned matrices

    3. Implications Homogeneous system Trivial solution: x = 0 Nontrivial solutions exist if and only if rank(A) < n Nontrivial solutions are said to be contained in the null space of A Nonhomogeneous system If the system is consistent, then all solutions can be represented as x = x0+xh x0 is a particular solution of the nonhomogeneous system xh is any solution of the homogeneous system

    4. Fundamental Theorem Examples Unique solution Infinitely many solutions No solutions

    5. Determinants Only applicable to square matrices Notation: det(A), |A| 2x2 matrix 3x3 matrix More general formulas based on cofactors are presented in the text

    6. Determinant Examples By hand Using Matlab >> A=[1 2; 3 4]; >> det(A) ans = -2 >> A=[1 2 3;4 5 6;7 8 9]; >> det(A) ans = 0

    7. Properties of Determinants |A| = |AT| Diagonal & triangular matrices Products: |AB| = |A||B| A zero column or row produces a zero determinant Linearly dependent rows or columns produce a zero determinant A square matrix A has full rank n if and only if |A| is non-zero

    8. Matrix Inverse Definition Assume A is a nxn matrix The inverse of A is denoted A-1 The inverse satisfies the equations: Existence & uniqueness The inverse exists if and only if: If A has an inverse, then the inverse is unique Concepts Singular matrix: A-1 does not exist, det(A) = 0, rank(A) < n Nonsingular matrix: A-1 exists, det(A) non-zero, rank(A) = n If rank(A) < n, the matrix is said to rank deficient

    9. Special Cases 2x2 matrix Diagonal matrix Product of square matrices

    10. Gauss-Jordan Elimination Method to compute A-1 using row operations Form augmented matrix Eliminate first entry in last two rows

    11. Gauss-Jordan Elimination Eliminate x2 entry from third row Make the diagonal elements unity

    12. Gauss-Jordan Elimination cont. Eliminate first two entries in third column Obtain identity matrix Matrix inverse

    13. Gauss-Jordan Elimination cont. Verify result

    14. Using the Matrix Inverse Linear algebraic equation system: Ax = b Assume A is a non-singular matrix Solution Example

    15. Matlab Examples >> A=[-1 1 2; 3 -1 1; -1 3 4]; >> inv(A) ans = -0.7000 0.2000 0.3000 -1.3000 -0.2000 0.7000 0.8000 0.2000 -0.2000 >> A=[1 2; 3 5]; >> b=[1; 2]; >> x=inv(A)*b x = -1.0000 1.0000

    16. Ill-Conditioned Matrices Matrix inversion: Ax = b ? x = A-1b Assume A is a perfectly known matrix Consider b to be obtained from measurement with some uncertainty Terminology Well-conditioned problem: “small” changes in the data b produce “small” changes in the solution x Ill-conditioned problem: “small” changes in the data b produce “large” changes in the solution x Ill-conditioned matrices Caused by nearly linearly dependent equations Characterized by nearly singular A matrix Solution is not reliable Common problem for large algebraic systems Ill-conditioning quantified by the condition number (covered later)

    17. Ill-Conditioned Matrix Example Example e represents measurement error in b2 Two rows (columns) are nearly linearly dependent Analytical solution 10% error (e = 0.1)

    18. Matlab Example >> A=[1 2; 3 5]; >> cond(A) ans = 38.9743 (well conditioned) >> A=[0.9999 -1.0001; 1 -1]; >> cond(A) ans = 2.0000e+004 (poorly conditioned) >> b=[1; 1.1] >> x=inv(A)*b x = 500.5500 499.4500