Review: Linear Equation and a Linear System Linear Algebra Lecture # 2: Systems of Linear Equations Important Concepts Gaussian Elimination and the Gauss-Jordan Method 1 A linear equation in n variables defines an( n − 1)-dimensional hyperplane in an n-dimensional space. 2 Monson H. Hayes A linear system is a set of m linear equations in n variables, and the system defines a set of m hyperplanes in an n-dimensional space. [email protected] A linear equation: a1x1 + a2x2 + ··· + anxn = b A system of m linear equations in n variables (unknowns): a11x1 + a12x2 + ··· + a1nxn = b1 a21x1 + a22x2 + ··· + a2nxn = b2 . This material is the property of the author and is for the sole and exclusive use of his students. It is not for publication, nor is it to be sold, reproduced, or generally distributed. am1x1 + am2x2 + ··· + amnxn = bm M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 1 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 2 / 36 Review: Systems of Linear Equations Solving a Linear System A number of different things can happen when working with systems Reading for Today's Class: Anton, pp. 11-24. of linear equations. Optional but recommended: The MATLAB tutorials written by Edward A system of linear equations may have: Neuman that are posted on the course web page. 1 No solution (inconsistent equations), 2 One and only one (unique) solution (consistent equations), or We now look at how to solve a linear system. 3 An infinite number of solutions (consistent equations). The first step will be the introduce to coefficient matrix and the An important thing to look at when dealing with systems of linear augmented matrix. These matrices will help facilitate the solution equations is procedure. 1 The number of unknowns (variables) and 2 The number of equations. Then we will define a set of operations that are performed on the equations to put them in a form where the solution is easily What, if anything, can you say if there are (think visually!): determined. 1 More unknowns than equations, 2 More equations than unknowns. These operations are known, in one form, as Gaussian Elimination and in another form as the Gauss-Jordan Method. M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 3 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 4 / 36 What is a Matrix? The Coefficient Matrix In mathematics, a matrix is a rectangular array of numbers, symbols, Consider the following linear system (two equations in two unknowns) or expressions. Page 1 of 1 The individual items in a matrix are called elements or entries. x1 − 2x2 = −1 In a general matrix with elements ai;j , the first index i is the row and −x1 + 3x2 = 3 the second index j is the column. A convenient and shorthand method for representing the coefficients that multiply the variables is to define a coefficient matrix, ai,j n columns j changes m x1 − 2x2 = −1 1 −2 rows a1,1 a1,2 a1,3 −x1 + 3x2 = 3 −1 3 (coefficient matrix) a2,1 a2,2 a2,3 a3,1 a3,2 a3,3 . M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 5 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 6 / 36 The Augmented Matrix Echelon Forms We may also include the parameters on the right-hand side of the A matrix is said to be in Row Echelon Form if linear system into the coefficient matrix, which results in the 1 If there are any rows containing only zeros, then they are grouped augmented coefficient matrix, together at the bottom of the matrix. 2 The first nonzero number in every nonzero row is equal to one (this is x1 − 2x2 = −1 1 −2 −1 called a leading 1). −x1 + 3x3 = 3 −1 3 3 (augmented matrix) 3 For any two successive nonzero rows, the leading 1 in the lower row Sometimes it may be useful to partition the augmented matrix to occurs farther to the right than the leading 1 in the higher row. emphasize/remind us that the right hand column of the matrix represents the parameters of the linear system. A matrix is said to be in Reduced Row Echelon Form if, in addition to the above, x − 2x = −1 1 −2 −1 1 2 5 Each column that contains a leading 1, has zeros everywhere else in −x1 + 3x3 = 3 −1 3 3 that column (not just below the leading 1). M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 7 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 8 / 36 file:///D:/LaTeX/Class/CAU%20Linear%20Algebra/Lectures/Lecture%2002/Matrixpng.svg 3/2/2012 Echelon Form Examples Algebraic Operations on a Linear System Solving a linear system involves performing algebraic operations on the system that do not alter the solution set. Row Echelon Form Reduced Row Echelon Form Eventually, one can determine whether or not the linear system is 21 ∗ ∗ ∗ ∗3 21 0 0 ∗ ∗3 consistent and, if it is, the solution set can be found. 60 1 ∗ ∗ ∗7 60 1 0 ∗ ∗7 6 7 6 7 A set of elementary algebraic operations on a linear system are 40 0 1 ∗ ∗5 40 0 1 ∗ ∗5 1 Interchange the order of two equations. 0 0 0 0 0 0 0 0 0 0 2 Multiply an equation by a (nonzero) constant. 3 Add one equation to another. 20 1 ∗ ∗ ∗ ∗ ∗ ∗ ∗3 20 1 ∗ 0 ∗ 0 0 0 ∗3 It is clear that each of these operations do not change the solution set 60 0 0 1 ∗ ∗ ∗ ∗ ∗7 60 0 0 1 ∗ 0 0 0 ∗7 of a linear system. 6 7 6 7 60 0 0 0 0 1 ∗ ∗ ∗7 60 0 0 0 0 1 0 0 ∗7 6 7 6 7 40 0 0 0 0 0 1 ∗ ∗5 40 0 0 0 0 0 1 0 ∗5 2x1 + x2 + 4x3 = 1 0 0 0 0 0 0 0 1 ∗ 0 0 0 0 0 0 0 1 ∗ 3x1 − x2 − 4x3 = 6 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 9 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 10 / 36 Elementary Row Operations Example Since the rows of an augmented matrix correspond to equations in a linear 2 3 system, the elementary algebraic operations performed on the equations may x1 + x2 + 2x3 = 9 1 1 2 9 be performed on the augmented matrix. 2x1 + 4x2 − 3x3 = 1 4 24 −3 1 5 These correspond to a set of elementary row operations: 3x1 + 6x2 − 5x3 = 0 3 6 −5 0 We begin by making the leading term in the second equation equal to zero. Elementary Algebraic Operations Elementary Row Operations To do this, we add −2 times the first equation to the second equation. Interchange order of two equations Interchange the order of two rows 2 3 Multiply equation by a constant Multiply a row by a constant x1 + x2 + 2x3 = 9 1 1 2 9 Add one equation to another Add one row to another 2x2 − 7x3 = −17 4 0 2 −7 −17 5 3x1 + 6x2 − 5x3 = 0 36 −5 0 Note that the last two operations may be combined into a single operation of: Adding a constant times one row to another. Next, we would like to make the first term in the third equation equal to Solving a Linear System - Step 1: Perform elementary row operations on the zero, so add −3 times the first equation to the third equation. augmented matrix to put it into row echelon form. 2 3 x1 + x2 + 2x3 = 9 1 1 2 9 2x2 − 7x3 = −17 4 0 2 −7 −17 5 3x2 − 11x3 = −27 0 3 −11 −27 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 11 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 12 / 36 Example (cont.) Example (cont.) 2 3 2 3 x1 + x2 + 2x3 = 9 1 1 2 9 x1 + x2 + 2x3 = 9 1 1 2 9 7 17 7 17 2x2 − 7x3 = −17 4 02 −7 −17 5 x2 − 2 x3 = − 2 4 0 1 − 2 − 2 5 1 3 1 3 3x2 − 11x3 = −27 0 3 −11 −27 − 2 x3 = − 2 0 0 − 2 − 2 The leading term in the second equation should be equal to one, so Finally, to make the leading term in the last row equal to one, multiply the multiply the second equation by 1=2. third equation by −2 2 3 x1 + x2 + 2x3 = 9 1 1 2 9 x + x + 2x = 9 2 1 1 2 9 3 7 17 7 17 1 2 3 x2 − 2 x3 = − 2 4 0 1 − 2 − 2 5 7 17 7 17 x2 − 2 x3 = − 2 4 0 1 − 2 − 2 5 3x2 − 11x3 = −27 03 −11 −27 x3 = 3 0 0 1 3 To make the first term in the last row equal to one, add −3 times the Note: The matrix is now in Row Echelon Form. second equation to the third equation, 2 3 x1 + x2 + 2x3 = 9 1 1 2 9 The linear system may be easily solved by a step called back-substitution. 7 17 7 17 x2 − 2 x3 = − 2 4 0 1 − 2 − 2 5 1 3 1 3 − 2 x3 = − 2 0 0 − 2 − 2 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 13 / 36 M.Hayes (CAU-GT) Lecture # 2 March 6, 2014 14 / 36 Example (cont.) Example (cont.) x1 + x2 + 2x3 = 9 We may continue with row reductions, and put the matrix into 7 17 Reduced Row Echelon Form.