12 Chapter One.Linear Systems can also be described as {(,y,z)2x+z=3 and-y-3z/2=-1/2).How- ever,this second description is not much of an improvement.It has two equa- tions instead of three,but it still involves some hard-to-understand interaction among the variables. To get a description that is free of any such interaction,we take the vari- able that does not lead any equation,2,and use it to describe the variables that do lead,x and y.The second equation gives y =(1/2)-(3/2)z and the first equation gives x =(3/2)-(1/2)z.Thus,the solution set can be de- scribed as {(,y,z)=((3/2)-(1/2)z,(1/2)-(3/2)z,z)zER).For instance, (1/2,-5/2,2)is a solution because taking z=2 gives a first component of 1/2 and a second component of-5/2. The advantage of this description over the ones above is that the only variable appearing,z,is unrestricted-it can be any real number. 2.2 Definition The non-leading variables in an echelon-form linear system are free variables. In the echelon form system derived in the above example,x and y are leading variables and z is free. 2.3 Example A linear system can end with more than one variable free.This row reduction x+y+z-w=1 x+y+z-0=1 y-z+w=-1-3p1+P3 y-2+0=-1 3x +6z-60=6 -3y+3z-3w=3 -y+2-w=1 -y+z-w=1 x+y+z-0=1 3p2+P3 y-2+=-1 P2+P4 0=0 0=0 ends with x and y leading,and with both z and w free.To get the description that we prefer we will start at the bottom.We first express y in terms of the free variables z and w with y =-1+z-w.Next,moving up to the top equation,substituting for y in the first equation x+(-1+z-w)+z- w 1 and solving for x yields r =2-2z+2w.Thus,the solution set is {2-2z+2,-1+z-w,z,w)|2,w∈R} We prefer this description because the only variables that appear,z and w, are unrestricted.This makes the job of deciding which four-tuples are system solutions into an easy one.For instance,taking z=1 and w =2 gives the solution (4,-2,1,2).In contrast,(3,-2,1,2)is not a solution,since the first component of any solution must be 2 minus twice the third component plus twice the fourth
12 Chapter One. Linear Systems can also be described as {(x, y, z) ¯ ¯ 2x + z = 3 and −y − 3z/2 = −1/2}. However, this second description is not much of an improvement. It has two equations instead of three, but it still involves some hard-to-understand interaction among the variables. To get a description that is free of any such interaction, we take the variable that does not lead any equation, z, and use it to describe the variables that do lead, x and y. The second equation gives y = (1/2) − (3/2)z and the first equation gives x = (3/2) − (1/2)z. Thus, the solution set can be described as {(x, y, z) = ((3/2) − (1/2)z,(1/2) − (3/2)z, z) ¯ ¯ z ∈ R}. For instance, (1/2, −5/2, 2) is a solution because taking z = 2 gives a first component of 1/2 and a second component of −5/2. The advantage of this description over the ones above is that the only variable appearing, z, is unrestricted — it can be any real number. 2.2 Definition The non-leading variables in an echelon-form linear system are free variables. In the echelon form system derived in the above example, x and y are leading variables and z is free. 2.3 Example A linear system can end with more than one variable free. This row reduction x + y + z − w = 1 y − z + w = −1 3x + 6z − 6w = 6 −y + z − w = 1 −3ρ1+ρ3 −→ x + y + z − w = 1 y − z + w = −1 −3y + 3z − 3w = 3 −y + z − w = 1 3ρ2+ρ3 −→ρ2+ρ4 x + y + z − w = 1 y − z + w = −1 0 = 0 0 = 0 ends with x and y leading, and with both z and w free. To get the description that we prefer we will start at the bottom. We first express y in terms of the free variables z and w with y = −1 + z − w. Next, moving up to the top equation, substituting for y in the first equation x + (−1 + z − w) + z − w = 1 and solving for x yields x = 2 − 2z + 2w. Thus, the solution set is {2 − 2z + 2w, −1 + z − w, z, w) ¯ ¯ z, w ∈ R}. We prefer this description because the only variables that appear, z and w, are unrestricted. This makes the job of deciding which four-tuples are system solutions into an easy one. For instance, taking z = 1 and w = 2 gives the solution (4, −2, 1, 2). In contrast, (3, −2, 1, 2) is not a solution, since the first component of any solution must be 2 minus twice the third component plus twice the fourth
Section I.Solving Linear Systems 13 2.4 Example After this reduction 2x-2y =0 2x-2y =0 z+3w=2 -(3/2)P1+P3 z+3w=2 3x-3y =0 -(1/2)p1+p4 0=0 x-y+2z+6w=4 2z+6w=4 2x-2y =0 -2p2+P4 z+3w=2 0=0 0=0 x and z lead,y and w are free.The solution set is {(y,y,2-3w,w)y,w ER). For instance,(1,1,2,0)satisfies the system-take y=1 and w=0.The four- tuple (1,0,5,4)is not a solution since its first coordinate does not equal its second. We refer to a variable used to describe a family of solutions as a parameter and we say that the set above is parametrized with y and w.(The terms parameter'and free variable'do not mean the same thing.Above,y and w are free because in the echelon form system they do not lead any row.They are parameters because they are used in the solution set description.We could have instead parametrized with y and z by rewriting the second equation as w =2/3-(1/3)z.In that case,the free variables are still y and w,but the parameters are y and z.Notice that we could not have parametrized with r and y,so there is sometimes a restriction on the choice of parameters.The terms 'parameter'and 'free'are related because,as we shall show later in this chapter, the solution set of a system can always be parametrized with the free variables. Consequently,we shall parametrize all of our descriptions in this way.) 2.5 Example This is another system with infinitely many solutions. T+2y =1 =1 2x +z =2 -2p+p%工+2 -4y+z=0 3x+2y+z-w=4 -3P1+P3 -4g+2-0=1 I+2y =1 -2tP% -4y+z=0 -w=1 The leading variables are r,y,and w.The variable z is free.(Notice here that, although there are infinitely many solutions,the value of one of the variables is fixed-w =-1.)Write w in terms of z with w=-1+0z.Then y=(1/4)z. To express x in terms of z,substitute for y into the first equation to get x= 1-(1/2)z.The solution set is {(1-(1/2)z,(1/4)z,2,-1)2ER) We finish this subsection by developing the notation for linear systems and their solution sets that we shall use in the rest of this book. 2.6 Definition An mxn matrir is a rectangular array of numbers with m rows and n columns.Each number in the matrix is an entry
Section I. Solving Linear Systems 13 2.4 Example After this reduction 2x − 2y = 0 z + 3w = 2 3x − 3y = 0 x − y + 2z + 6w = 4 −(3/2)ρ1+ρ3 −→ −(1/2)ρ1+ρ4 2x − 2y = 0 z + 3w = 2 0 = 0 2z + 6w = 4 −2ρ2+ρ4 −→ 2x − 2y = 0 z + 3w = 2 0 = 0 0 = 0 x and z lead, y and w are free. The solution set is {(y, y, 2 − 3w, w) ¯ ¯ y, w ∈ R}. For instance, (1, 1, 2, 0) satisfies the system — take y = 1 and w = 0. The fourtuple (1, 0, 5, 4) is not a solution since its first coordinate does not equal its second. We refer to a variable used to describe a family of solutions as a parameter and we say that the set above is parametrized with y and w. (The terms ‘parameter’ and ‘free variable’ do not mean the same thing. Above, y and w are free because in the echelon form system they do not lead any row. They are parameters because they are used in the solution set description. We could have instead parametrized with y and z by rewriting the second equation as w = 2/3 − (1/3)z. In that case, the free variables are still y and w, but the parameters are y and z. Notice that we could not have parametrized with x and y, so there is sometimes a restriction on the choice of parameters. The terms ‘parameter’ and ‘free’ are related because, as we shall show later in this chapter, the solution set of a system can always be parametrized with the free variables. Consequently, we shall parametrize all of our descriptions in this way.) 2.5 Example This is another system with infinitely many solutions. x + 2y = 1 2x + z = 2 3x + 2y + z − w = 4 −2ρ1+ρ2 −→ −3ρ1+ρ3 x + 2y = 1 −4y + z = 0 −4y + z − w = 1 −ρ2+ρ3 −→ x + 2y = 1 −4y + z = 0 −w = 1 The leading variables are x, y, and w. The variable z is free. (Notice here that, although there are infinitely many solutions, the value of one of the variables is fixed— w = −1.) Write w in terms of z with w = −1 + 0z. Then y = (1/4)z. To express x in terms of z, substitute for y into the first equation to get x = 1 − (1/2)z. The solution set is {(1 − (1/2)z,(1/4)z, z, −1) ¯ ¯ z ∈ R}. We finish this subsection by developing the notation for linear systems and their solution sets that we shall use in the rest of this book. 2.6 Definition An m×n matrix is a rectangular array of numbers with m rows and n columns. Each number in the matrix is an entry
14 Chapter One.Linear Systems Matrices are usually named by upper case roman letters,e.g.A.Each entry is denoted by the corresponding lower-case letter,e.g.aij is the number in row i and column j of the array.For instance, 12.25 A=(34-7) has two rows and three columns,and so is a 2x3 matrix.(Read that "two- by-three";the number of rows is always stated first.)The entry in the second row and first column is a2.1=3.Note that the order of the subscripts matters: a1.2 a2.1 since a1.2 2.2.(The parentheses around the array are a typo- graphic device so that when two matrices are side by side we can tell where one ends and the other starts. Matrices occur throughout this book.We shall use Mnxom to denote the collection of nxm matrices 2.7 Example We can abbreviate this linear system 工1+2x2 =4 2-t3=0 工1 +2x3=4 with this matrix. 2 0 02 The vertical bar just reminds a reader of the difference between the coefficients on the systems's left hand side and the constants on the right.When a bar is used to divide a matrix into parts,we call it an augmented matrix.In this notation,Gauss'method goes this way. 1 2 0 4 1 2 0 4 /12 0 4 01 0 P1+P3 0 个y -1 0 2p2+P3 0 1 -1 0 10 2 0 -2 0 0 0 0 The second row stands for y-z=0 and the first row stands for x+2y=4 so the solution set is {(4-2z,z,z)2ER).One advantage of the new notation is that the clerical load of Gauss'method-the copying of variables,the writing of +'s and ='s,etc.-is lighter. We will also use the array notation to clarify the descriptions of solution sets.A description like {(2-2z+2w,-1+z-w,z,w)2,w E R}from Ex- ample 2.3 is hard to read.We will rewrite it to group all the constants together, all the coefficients of z together,and all the coefficients of w together.We will write them vertically,in one-column wide matrices. w|z,w∈R}
14 Chapter One. Linear Systems Matrices are usually named by upper case roman letters, e.g. A. Each entry is denoted by the corresponding lower-case letter, e.g. ai,j is the number in row i and column j of the array. For instance, A = µ 1 2.2 5 3 4 −7 ¶ has two rows and three columns, and so is a 2×3 matrix. (Read that “twoby-three”; the number of rows is always stated first.) The entry in the second row and first column is a2,1 = 3. Note that the order of the subscripts matters: a1,2 6= a2,1 since a1,2 = 2.2. (The parentheses around the array are a typographic device so that when two matrices are side by side we can tell where one ends and the other starts.) Matrices occur throughout this book. We shall use Mn×m to denote the collection of n×m matrices. 2.7 Example We can abbreviate this linear system x1 + 2x2 = 4 x2 − x3 = 0 x1 + 2x3 = 4 with this matrix. 1 2 0 4 0 1 −1 0 1 0 2 4 The vertical bar just reminds a reader of the difference between the coefficients on the systems’s left hand side and the constants on the right. When a bar is used to divide a matrix into parts, we call it an augmented matrix. In this notation, Gauss’ method goes this way. 1 2 0 4 0 1 −1 0 1 0 2 4 −ρ1+ρ3 −→ 1 2 0 4 0 1 −1 0 0 −2 2 0 2ρ2+ρ3 −→ 1 2 0 4 0 1 −1 0 0 0 0 0 The second row stands for y − z = 0 and the first row stands for x + 2y = 4 so the solution set is {(4 − 2z, z, z) ¯ ¯ z ∈ R}. One advantage of the new notation is that the clerical load of Gauss’ method— the copying of variables, the writing of +’s and =’s, etc. — is lighter. We will also use the array notation to clarify the descriptions of solution sets. A description like {(2 − 2z + 2w, −1 + z − w, z, w) ¯ ¯ z, w ∈ R} from Example 2.3 is hard to read. We will rewrite it to group all the constants together, all the coefficients of z together, and all the coefficients of w together. We will write them vertically, in one-column wide matrices. { 2 −1 0 0 + −2 1 1 0 · z + 2 −1 0 1 · w ¯ ¯ z, w ∈ R}
Section I.Solving Linear Systems 15 For instance,the top line says that x =2-2z+2w.The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way. 2.8 Definition A vector (or column vector)is a matrix with a single column. A matrix with a single row is a row vector.The entries of a vector are its components. Vectors are an exception to the convention of representing matrices with capital roman letters.We use lower-case roman or greek letters overlined with an arrow:d,b,...or a,B,...(boldface is also common:a or a).For instance, this is a column vector with a third component of 7. 3 7 2.9 Definition The linear equation ax1+a2x2+...+anzn d with unknowns 1,...,In is satisfied by S1 Sn if ais1+a2s2+...+ansn d.A vector satisfies a linear system if it satisfies each equation in the system. The style of description of solution sets that we use involves adding the vectors,and also multiplying them by real numbers,such as the z and w.We need to define these operations. 2.10 Definition The vector sum of u and is this. 1十1 + In general,two matrices with the same number of rows and the same number of columns add in this way,entry-by-entry. 2.11 Definition The scalar multiplication of the real number r and the vector U is this. In general,any matrix is multiplied by a real number in this entry-by-entry way
Section I. Solving Linear Systems 15 For instance, the top line says that x = 2 − 2z + 2w. The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way. 2.8 Definition A vector (or column vector ) is a matrix with a single column. A matrix with a single row is a row vector . The entries of a vector are its components. Vectors are an exception to the convention of representing matrices with capital roman letters. We use lower-case roman or greek letters overlined with an arrow: ~a, ~b, . . . or ~α, β~, . . . (boldface is also common: a or α). For instance, this is a column vector with a third component of 7. ~v = 1 3 7 2.9 Definition The linear equation a1x1 + a2x2 + · · · + anxn = d with unknowns x1, . . . , xn is satisfied by ~s = s1 . . . sn if a1s1 + a2s2 + · · · + ansn = d. A vector satisfies a linear system if it satisfies each equation in the system. The style of description of solution sets that we use involves adding the vectors, and also multiplying them by real numbers, such as the z and w. We need to define these operations. 2.10 Definition The vector sum of ~u and ~v is this. ~u + ~v = u1 . . . un + v1 . . . vn = u1 + v1 . . . un + vn In general, two matrices with the same number of rows and the same number of columns add in this way, entry-by-entry. 2.11 Definition The scalar multiplication of the real number r and the vector ~v is this. r · ~v = r · v1 . . . vn = rv1 . . . rvn In general, any matrix is multiplied by a real number in this entry-by-entry way
16 Chapter One.Linear Systems Scalar multiplication can be written in either order:r.or .r,or without the .'symbol:r.(Do not refer to scalar multiplication as 'scalar product' because that name is used for a different operation. 2.12 Example 28 5 -7 -21 Notice that the definitions of vector addition and scalar multiplication agree where they overlap,for instance,+=20. With the notation defined,we can now solve systems in the way that we will use throughout this book. 2.13 Example This system 2x+y -0=4 y+w+u=4 -z+2w =0 reduces in this way. 21 0 -104 2 1 0 -1 0 4 0 10 11 4 -(1/2)P+p3 0 1 0 11 4 .0-1200/ 0 -1/2 -15/20 -2 /21 0 -1 0 4 (1/2)P2+P3 0 1 0 1 1 4 0 0-1 3 1/20 The solution set is {(w+(1/2)u,4-w-u,3w+(1/2)u,w,u)w,u ER).We write that in vector form 0 1/2 y -1 0 1/2 u|w,u∈R} 0 0 Note again how well vector notation sets off the coefficients of each parameter. For instance,the third row of the vector form shows plainly that if u is held fixed then z increases three times as fast as w. That format also shows plainly that there are infinitely many solutions.For example,we can fix u as 0,let w range over the real numbers,and consider the first component x.We get infinitely many first components and hence infinitely many solutions
16 Chapter One. Linear Systems Scalar multiplication can be written in either order: r · ~v or ~v · r, or without the ‘·’ symbol: r~v. (Do not refer to scalar multiplication as ‘scalar product’ because that name is used for a different operation.) 2.12 Example 2 3 1 + 3 −1 4 = 2 + 3 3 − 1 1 + 4 = 5 2 5 7 · 1 4 −1 −3 = 7 28 −7 −21 Notice that the definitions of vector addition and scalar multiplication agree where they overlap, for instance, ~v + ~v = 2~v. With the notation defined, we can now solve systems in the way that we will use throughout this book. 2.13 Example This system 2x + y − w = 4 y + w + u = 4 x − z + 2w = 0 reduces in this way. 2 1 0 −1 0 4 0 1 0 1 1 4 1 0 −1 2 0 0 −(1/2)ρ1+ρ3 −→ 2 1 0 −1 0 4 0 1 0 1 1 4 0 −1/2 −1 5/2 0 −2 (1/2)ρ2+ρ3 −→ 2 1 0 −1 0 4 0 1 0 1 1 4 0 0 −1 3 1/2 0 The solution set is {(w + (1/2)u, 4 − w − u, 3w + (1/2)u, w, u) ¯ ¯ w, u ∈ R}. We write that in vector form. { x y z w u = 0 4 0 0 0 + 1 −1 3 1 0 w + 1/2 −1 1/2 0 1 u ¯ ¯ w, u ∈ R} Note again how well vector notation sets off the coefficients of each parameter. For instance, the third row of the vector form shows plainly that if u is held fixed then z increases three times as fast as w. That format also shows plainly that there are infinitely many solutions. For example, we can fix u as 0, let w range over the real numbers, and consider the first component x. We get infinitely many first components and hence infinitely many solutions