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Lecture 2. Contents

Lecture 2. Contents. Solution of Linear Systems of Equations Consistency Rank Geometric Interpretation Gaussian Elimination. Systems of linear equations. In Linear equations the unknowns appear raised to the power one.

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Lecture 2. Contents

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  1. Lecture 2. Contents • Solution of Linear Systems of Equations • Consistency • Rank • Geometric Interpretation • Gaussian Elimination Lecture 2

  2. Systems of linear equations In Linearequations the unknowns appear raised to the power one. If the variables in the equation are ordered as x1, … xn and the missing variables xi are written as … + 0·xi + … , then the linear equation can be written in the matrix notation, according to the matrix multiplication rules. Solution of linear equations was one of the reasons, for matrix notation invention. Lecture 2

  3. Linear Systems in Matrix Form (1)  Lecture 2

  4. Solution of Linear Systems Each side of the equation (2) Can be multiplied by A-1 : Due to the definition of A-1: Therefore the solution of (2) is: Lecture 2

  5. Consistency (Solvability) • A-1 does not exist for every A • The linear system of equations A·x=b has a solution, or said to be consistent IFF Rank{A}=Rank{A|b} • A system is inconsistent when Rank{A}<Rank{A|b} Rank{A} is the maximum number of linearly independent columns or rows of A. Rank can be found by using ERO (Elementary Row Oparations) or ECO (Elementary column operations). Lecture 2

  6. Elementary row and column operations • The following operations applied to the augmented matrix [A|b], yield an equivalent linear system • Interchanges: The order of two rows/columns can be changed • Scaling: Multiplying a row/column by a nonzero constant • Sum: The row can be replaced by the sum of that row and a nonzero multiple of any other row. One can use ERO and ECO to find the Rank as follows: EROminimum # of rows with at least one nonzero entry or ECOminimum # of columns with at least one nonzero entry Lecture 2

  7. An inconsistent example: Geometric interpretation ERO:Multiply the first row with -2 and add to the second row Rank{A}=1 Rank{A|b}=2 > Rank{A} Lecture 2

  8. Uniqueness of solutions • The system has a unique solution IFF Rank{A}=Rank{A|b}=n n is the order of the system • Such systems are called full-rank systems Lecture 2

  9. Full-rank systems • If Rank{A}=n Det{A}  0  A-1 exists  Unique solution Lecture 2

  10. Rank deficient matrices • If Rank{A}=m<n Det{A} = 0  A is singular so not invertible infinite number of solutions (n-m free variables) under-determined system Rank{A}=Rank{A|b}=1 Consistent so solvable Lecture 2

  11. Ill-conditioned system of equations • A small deviation in the entries of A matrix, causes a large deviation in the solution.   Lecture 2

  12. Ill-conditioned continued..... • A linear system of equations is said to be “ill-conditioned” if the coefficient matrix tends to be singular Lecture 2

  13. Back substitution ERO Gaussian Elimination • By using ERO, matrix A is transformed into an upper triangular matrix (all elements below diagonal 0) • Back substitution is used to solve the upper-triangular system Lecture 2

  14. Pivotal element Pivotal Element The first coefficient of the first row (pivot) is used to zero out first coefficients of Other rows. Lecture 2

  15. First step of elimination First row, multiplied by appropriate factor is subtracted from other rows. Lecture 2

  16. Pivotal element Second step of elimination The second coefficient (pivot) of the second row is used to zero out second coefficients of other rows. Lecture 2

  17. Second step of elimination Second row, multiplied by appropriate factor is subtracted from other rows (ERO). Lecture 2

  18. Gaussion elimination algorithm For c=p+1 to n Lecture 2

  19. Back substitution algorithm Finally, the following system is obtained: Lecture 2

  20. Back substitution algorithm The answer is obtained as following: Lecture 2

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