Chapter 1 Matrices and Systems of Linear Equations

1 And 1 Matrices and Systems of Linear Equations : Introduction to Matrices and Systems of Linear Equations : Echelon Form and Gauss-Jordan EliminationLectureLinear Algebra - Math 2568 Mon Friday, January 11, 2013 Oguz KurtMW Hrs:MWF 10:20-11:20 The Ohio State University and and and EquationsDefinitionAlinear equationin thenvariablesx1,x2, ,xnis an equation thatcan be written in the forma1x1+a2x2+ +anxn=bwhere thecoefficientsa1,a2, ,anand theconstant : 3x+4y+5z=12 is +y=1,siny+x=10 are a Linear equationa1x1+a2x2+ +anxn=bis a vector[s1,s2,..,sn]whose components satisfy the equation when wesubstitutex1=s1,x2=s2.

2 ,xn= : The Linear equationx1 x2+2x3=3 has[3,0,0],[0,1,2]and[6,1, 1].(In general,x1=3+s 2t,x2=s,x3=t.) and of Linear Equations (SLEs)DefinitionAsystem of Linear equationsis a finite set of Linear Equations , eachwith the same variables. Asolutionof system of Linear Equations is avector that issimultaneouslya solution of each equation in thesystem. Thesolution setof a system of Linear Equations is the set ofallsolutions of the system2x 3y=73x+y=5has[2, 1]as a solution. and and type of solutionsNote:A system of Linear Equations is calledconsistentif it has atleast one solution. A system with no solutions is system of Linear Equations with real coefficients has either1a unique solution (a consistent system ) or2infinitely many solutions (a consistent system ) or3no solutions (an inconsistent system ).

3 And a system of Linear EquationsExample:Solve the systemx y z=42y+z=53z=9 Note: To solve this system , we usually useback z=3z=3,2y+z=5 2y=2 y=1z=3,y=1,x y z=4 x=8[8,1,3]is the unique solution for this SLEs. and Augmented MatrixAssume we have the following SLEs withmequations andnunknowns:a1,1x1+a1, ,nxn=b1a2,1x1+a2, ,nxn= ,1x1+an, ,nxn=bmWe can represent it as ..xnEq. 1a1,1a1, ..a1,nb1Eq. 2a2,1a2, ..a2, mam,1am, ..am,nbm and MatrixThe following matrix is called theaugmented matrixof this SLEs: a1,1a1, ..a1,nb1a2,1a2, ..a2, ,1am, ..am,nbm Augmented Matrix can be used to solve SLEs. and :Find the augmented Matrices of the Linear y=02x+y=3= [1 10213] +5y= 1 x+y= 52x+4y=4= 15 1 1 1 5244.

4 And to solve SLEs using Aug. Matrix Changing the order of Equations Changing the order of variables (We will never use this method) Multiplying an equation by a non-zero constant Adding two will describe these as operations on the augmented matrix of thegiven SLEs. and MatrixThere are two important Matrices associated with a Linear matrixcontains the coefficients of the variables, andtheaugmented matrixis the coefficient matrix augmented by anextra column containing the constant terms:Assume we have the following SLEs withmequations andnunknowns:a1,1x1+a1, ,nxn=b1a2,1x1+a2, ,nxn= ,1x1+an, ,nxn=bm and and Augmented Matrices a1,1a1.

5 A1,na2,1a2, ..a2, ,1am, ..am,n Table :Coefficient Matrix a1,1a1, ..a1,nb1a2,1a2, ..a2, ,1am, ..am,nbm Table :Augmented MatrixRemark1:Note that if we denote the coefficient matrix of a linearsystem byAand the column vector of the constant terms byb, thenthe form of the augmented matrix is[A|b]. and the coefficient and augmented matricesAand[A|b]for thesystem2x+y z=3x+5z=1 x+3y 2z=0 The coefficient matrix isA= 21 1105 1 3 2 .The augmented matrix is[A|b]= 21 131051 1 3 20 and Echelon FormDefinitionA matrix is inrow echelon formif it satisfies the following properties:(1)All rows consisting entirely of zeros are at the bottom.

6 (2)In a nonzero row, the first nonzero entry (called theleading entry) is in a column to the left of any leading entriesbelow :Note that the textbook s definition of row echelon formrequires the leading terms to be 1. You are welcome to use following Matrices are in row echelon form: 2410 1 2000 1 0 10 1 50 0 4 1 1 2 10 0 1 30 0 0 0 0 201 1 30 0 1 1220 000400 00005 and that each of the Matrices in the previous example is anaugmented matrix, write out the corresponding Systems of linearequations and solve them. (Here, we will study the last matrix, and therest will be left as an exercise)Remark1:If we are asked to study a coefficient matrixAas theaugmented matrix[A|b], then we treatbas the zero [A|0]= x1x2x3x4x5x6c0201 13000 1122000004000000050 , then we have2x2+x4 x5+3x6=0 2x2+x4=0 x3+x4+2x5+2x6=0 x3+x4=04x5=0 x5=05x6=0 x6=0 Setx1=s,x4=t.

7 Then we have the solution set{x1=s,x2= t/2,x3=t=x4,x5=x6=0}as our solution set. and Row OperationsDefinitionThe followingelementary row operationscan be performed on amatrix:(1) Interchange two rows.(2) Multiply a row by a nonzero constant.(3) Add a multiple of a row to another row. and 2 33 45 6 R1 R2 3 42 35 6 Table :Operation (1) 2 33 45 6 R1:=2R1 4 63 45 6 Table :Operation (2) 2 33 45 6 R2:=R2+2R1 237 1056 Table :Operation (3) and the following matrix to row echelon form:A= 13201 0 1 1 110 104243 3133 1 0 0 Remark1:Note that the entry chosen to become a leading entry iscalled apivot, and this phase of the process is :=R4 R1 R2:=R2+R1 1 3 20100 2 11110 4 24330 0 1 1 1 0 R3.

8 =R3 2R2 1 3 20100 2 11110 0 02110 0 1 1 1 0 R4 R3 1 3 20100 2 11110 0 1 1 1 00 0 0211 and EquivalenceDefinitionMatricesAandBarerow equivalentif there is a sequence ofelementary row operations that A and B are row equivalent if and only if they can be reducedto the same row echelon :In fact, If A, B are row equivalent and B,C are rowequivalent, then A, C are row equivalent. The reason is simple: Wecan reverse the elementary row operations". and EliminationGaussian Elimination(1) Write the augmented matrix of the system of Linear Equations .(2) Use elementary row operations to reduce the augmented matrixto row echelon form.

9 (3) Using back substitution, solve the equivalent system thatcorresponds to the row-reduced :We have already applied all three steps in differentexamples. and : Gaussian EliminationSolve the following SLEs using Gaussian Elimination:3x1+x2 x3+2x4=1x1 x2+x3+2x4=22x1+2x2+x3+6x4=3[A|b](1)= 31 1 211 112222163 R3:=R3 2R2 R1:=R1 3R2 04 4 4 51 112204 12 1 R1 R2 1 112204 4 4 504 12 1 R3:=R3 R2 (2) 1 112204 4 4 500364 Remark1:The Gaussian Elimination does not ask us to use the rowoperations in a specific order. While there is a methodical way to dorow operations, it is sometimes faster not to use it. In fact, themethodological way can be summarized as repeating the followingtwo steps:1If needed, switch rows to get the next leading , ifRihas the next leading term then use theoperations of the formRj:=Rj Riwheneverj>ito cancel outnon-zero entries below this leading term.

10 And : Cont d (Step 3 in Gaussian Elimination)x1 x2+x3+2x4=24x2 4x3 4x4= 53x3+6x4=4 Setx4=t. thenx3=43 2x4=43 2t,x2= 54+x3+x4= 54+(43 2t)+t=112 t,x1=2+x2 x3 2x4=2+(112 t) (43 2t) 2t=912 tSolution set is{x1=912 t,x2=112 t,x3=43 2t,x4=t}.Remark1:Note that we assignedx4=t. WHY? and Row Echelon FormDefinitionA matrix is inreduced row echelon formif it satisfies the followingproperties:(1) It is in row echelon form.(2) The leading entry in each nonzero row is a 1 (called aleading 1).(3) Each column containing a leading 1 has zero everywhere :Reduced row echelon form can be seen as combining rowechelon form and back substitution in Gaussian : 1 0 30 120 00 , 1 2 0 0001 0000 1 , 1 1 0 011001 0 12000 12 2 and EliminationDefinition(1) Write the augmented matrix of the system of Linear Equations .