Problem 1.1.31(4ed) 33(5ed) Find all the polynomials f(t) of - PowerPoint PPT Presentation

9/5/13 ¡ Problem 1.1.31(4ed) 33(5ed)  Find all the polynomials f(t) of degree ≤ 2 whose graph runs through the points (1,3) and (2,6), such that f’(1)=1. MAT 211 Linear Equations Note to students: Before reading these instructions, solve a few linear systems with pencil and paper. Definition Solving linear equations by elimination i th step A m x n matrix is a If there is not variables in the system formed the i th equation to the 1. bottom, then the process ends. rectangular array of If there is an equation of the form 0 = number not zero, then the 2. m.n numbers arranged system has no solution (and the process is complete). In the system formed the i th equation to the bottom, find the first in m horizontal rows 3. occurrence of the variable with smallest subindex, say x k . and n vertical columns If x k does not appear in the i th equation, swap the i th equation with 4. the first equation below that contains x k . Divide the i th equation, by the coefficient of x k in the i th equation. 5. (Side note: the image is Eliminate x k from all the other equations by subtracting to each of 6. from wikipedia) the other equations the appropriate multiple of the i th equation. (If the coefficient of x k in the j th equation is c, replace the i th equation by (j th eq.- c .i th eq) 1 ¡

9/5/13 ¡ A matrix is in reduced row- A system of equations has an Problem 1.2.45 echelon form if it satisfies the “obvious” or “very easy to three following conditions. compute” solution when:  If a row has non-zero  The coefficient of the leading entries, then the first non-  Consider the system of zero entry (from left to variable is 1. (The leading equations below, where k is right) is 1. This entry is variable is the variable with called the leading 1 or x + 2y + 3z = 4 an arbitrary constant. smallest subindex in the pivot . For which values of the equation) a. x + ky + 4z = 6  If a column contains a constant k does the system leading 1, then all other  The leading variable in each entries in that column are x + 2y + (k+2)z = 6 have a unique solution. equation does not appear in 0. When is there no solution? any other equation. b.  If a row contains a  The leading variables When are there infinitely leading 1, then each row c. above it contains a appear in natural order from many solutions? leading 1 further to the top to bottom (that is from left. the smallest subindex to the largest. Decide whether each of the matrices below is Decide whether each of the matrices below is in reduced row echelon form. in reduced row echelon form. 0 1 0 0 0 0 0 0 0 0 0 0 8 0 1 0 0 0 0 0 0 0 0 0 0 8 0 1 0 0 0 0 0 0 0 0 0 0 8 0 1 0 0 0 0 0 0 0 0 0 0 8 0 0 0 1 0 0 0 0 0 0 0 0 9 0 0 0 1 0 0 0 0 0 0 0 0 9 0 0 0 1 0 0 0 0 0 0 8 0 9 0 0 0 1 0 0 0 0 0 8 0 0 9 0 0 0 0 1 0 0 0 0 0 0 0 π 0 0 0 0 0 0 0 0 0 0 0 0 π 0 0 0 0 1 0 0 0 0 0 0 0 π 0 0 0 0 1 0 0 0 0 0 0 0 π 0 0 0 0 0 0 0 0 1 0 0 0 e 0 0 0 0 0 0 0 0 1 0 0 0 e 0 0 0 0 0 0 0 0 1 0 0 0 e 0 0 0 0 0 0 0 0 1 0 0 0 e 0 0 0 0 0 0 0 0 0 1 0 0 3 0 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 1 0 0 3 0 0 0 0 0 0 0 0 0 1 0 0 3 0 0 0 0 0 0 0 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 ¡

9/5/13 ¡ Decide whether each of the matrices below is in reduced row echelon form. Elementary row operations.  Interchange the positions of two rows in the matrix.  Multiply a row by a non-zero constant. 0 1 0 0 0 0 0 0 0 0 0 0 8 0 1 0 0 0 0 0 0 0 0 0 0 8  Replace a row with the sum of that row and 0 0 0 1 0 0 0 0 0 8 0 0 9 0 0 0 1 0 0 0 0 0 0 8 0 9 a multiple of any other row. 0 0 0 0 1 0 0 0 0 0 0 0 π 0 0 0 0 2 0 0 0 0 0 0 0 π 0 0 0 0 0 0 0 0 1 0 0 0 e 0 0 0 0 0 0 0 0 1 0 0 0 e 0 0 0 0 0 0 0 0 0 1 0 0 3 • Any matrix can be reduced to a matrix in reduced 0 0 0 0 0 0 0 0 0 1 0 0 3 row echelon form by elementary row operations. 0 0 0 0 0 0 0 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 1 4 (One way to do it is to “copy” the recipe or 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 algorithm to solve a linear system we saw before.) 0 0 0 0 0 0 0 0 0 0 0 0 0 • If a matrix is associated to a linear system then by 0 0 0 0 0 0 0 0 0 0 0 0 0 reducing it to a reduced row echelon form, we can 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 find the solution of the system. A linear system is consistent , if it has one or The rank of a matrix more solutions. Otherwise, (that is, if it has no solutions) is inconsistent . Theorem: Consider a system of n linear equations with m variables. The coefficient matrix of the system A has size n x m. The rank of a matrix A, (denoted by The augmented matrix has size n x (m +1). Moreover, rank(A)) is the number of leading 1’s the linear system is inconsistent if and only if the associated in its reduced row echelon form. 1. augmented matrix in reduced row echelon form has a row of the form [0 0 0…. 0 1] Theorem: Each matrix can be transformed elementary the linear system is consistent if and only the associated augmented 2. matrix in reduced row echelon form does NOT have a row of the form row operations in an unique matrix in reduced row [0 0 0…. 0 1]. echelon form. If the linear system is consistent then, either 3. It has infinitely many solutions. In this case, the number of leading a. variables is strictly smaller than the total number of variables. This implies that the definition of rank makes sense. It has exactly one solution. This holds if and only all variables are b. leading . 3 ¡

9/5/13 ¡ Theorem: A system with the The rank of a matrix A is A linear system is consistent , if it has one or same number of equations more solutions. Otherwise, (that is, if it has no the number of leading 1’s than variables has a unique solutions) is inconsistent . in its reduced row echelon solution if and only if … form. Theorem: A system of n linear equations with m variables is Theorem The coefficient matrix A of a consistent if and only if either Theorem: A system of n linear linear system of n linear equations with m equations with m variables is variables is an n x m matrix, which satisfies It has infinitely many solutions. In this case, the rank of A is a. consistent if and only if either the following statements: strictly less than m. OR It has infinitely many a. • rank(A) ≤ n It has exactly one solution. This holds if and only all solutions. In this case, the b. • rank(A) ≤ m. rank of A is strictly less variables are leading . In this case, the rank of A is m. • If rank(A)=n then the system is consistent than m. OR (and n ≤ m) It has exactly one solution. b. • If rank(A)=m then the system has at most This holds if and only all one solution. variables are leading . In • If rank(A)<m then the system has either this case, the rank of A is infinitely many solutions or none. m. Definition of a linear system. Characterization of an “easy to solve”   system. Relation to reduced row Vectors and Matrices Geometric interpretation (2 or 3 variables)  echelon form of a matrix. Associated matrices  The number of solutions of a linear  Augmented system is either zero OR exactly one  Coefficient. OR infinite  Gauss Jordan elimination Variables on a (easy to solve) linear   system Linear systems summary Elementary row operations in matrices.  Leading Reduced row echelon form of a matrix.  continues   Vectors Free   Geometric interpretation Linear systems summary  Vector spaces  Linear combinations  A linear system is inconsistent if it has no solutions. Otherwise, it is consistent .  Dot product  Theorem:  Matrices  A linear system is inconsistent if and only if the augmented associated matrix in  Rank How does the reduced row reduced row echelon form has a row of the form [0 0 0 … 0 1]. echelon form look in each  Algebraic operations:  If a linear system is consistent then either case?  Sum  All the variables are leading (Thus the system has exactly one solution.)  Multiplication by a scalar.  There is at least one no leading variable. (Thus the system has infinitely many solutions. )  Product of a matrix and a vector  Relation between the number of solutions and the rank of the associated  Algebraic rules. matrix.  Matrix form of a linear system .  Consequences of the relations between the number of equations and the number of unknown. 4 ¡