Section I.Solving Linear Systems 27 Now,apply Lemma 3.8 to conclude that a solution set p+h h solves the associated homogeneous system} is either empty(if there is no particular solution p),or has one element(if there is a p and the homogeneous system has the unique solution 0),or is infinite(if there is a p and the homogeneous system has a non-0 solution,and thus by the prior paragraph has infinitely many solutions) QED This table summarizes the factors affecting the size of a general solution. number of solutions of the associated homogeneous system one infinitely many unique infinitely many particular yes solution solutions solution exists？ no no solutions solutions The factor on the top of the table is the simpler one.When we perform Gauss'method on a linear system,ignoring the constants on the right side and so paying attention only to the coefficients on the left-hand side.we either end with every variable leading some row or else we find that some variable does not lead a row,that is,that some variable is free.(Of course,"ignoring the constants on the right"is formalized by considering the associated homogeneous system. We are simply putting aside for the moment the possibility of a contradictory equation. A nice insight into the factor on the top of this table at work comes from con- sidering the case of a system having the same number of equations as variables This system will have a solution,and the solution will be unique,if and only if it reduces to an echelon form system where every variable leads its row,which will happen if and only if the associated homogeneous system has a unique solution. Thus,the question of uniqueness of solution is especially interesting when the system has the same number of equations as variables. 3.12 Definition A square matrix is nonsingular if it is the matrix of coeffi- cients of a homogeneous system with a unique solution.It is singular otherwise, that is,if it is the matrix of coefficients of a homogeneous system with infinitely many solutions. 3.13 Example The systems from Example 3.3,Example 3.5,and Example 3.9 each have an associated homogeneous system with a unique solution.Thus these matrices are nonsingular
Section I. Solving Linear Systems 27 Now, apply Lemma 3.8 to conclude that a solution set {~p + ~h ¯ ¯ ~h solves the associated homogeneous system} is either empty (if there is no particular solution ~p), or has one element (if there is a ~p and the homogeneous system has the unique solution ~0), or is infinite (if there is a ~p and the homogeneous system has a non-~0 solution, and thus by the prior paragraph has infinitely many solutions). QED This table summarizes the factors affecting the size of a general solution. number of solutions of the associated homogeneous system particular solution exists? one infinitely many yes unique solution infinitely many solutions no no solutions no solutions The factor on the top of the table is the simpler one. When we perform Gauss’ method on a linear system, ignoring the constants on the right side and so paying attention only to the coefficients on the left-hand side, we either end with every variable leading some row or else we find that some variable does not lead a row, that is, that some variable is free. (Of course, “ignoring the constants on the right” is formalized by considering the associated homogeneous system. We are simply putting aside for the moment the possibility of a contradictory equation.) A nice insight into the factor on the top of this table at work comes from considering the case of a system having the same number of equations as variables. This system will have a solution, and the solution will be unique, if and only if it reduces to an echelon form system where every variable leads its row, which will happen if and only if the associated homogeneous system has a unique solution. Thus, the question of uniqueness of solution is especially interesting when the system has the same number of equations as variables. 3.12 Definition A square matrix is nonsingular if it is the matrix of coeffi- cients of a homogeneous system with a unique solution. It is singular otherwise, that is, if it is the matrix of coefficients of a homogeneous system with infinitely many solutions. 3.13 Example The systems from Example 3.3, Example 3.5, and Example 3.9 each have an associated homogeneous system with a unique solution. Thus these matrices are nonsingular. µ 3 4 2 −1 ¶ 3 2 1 6 −4 0 0 1 1 1 2 −1 2 4 0 0 1 −3
28 Chapter One.Linear Systems The Chemistry problem from Example 3.6 is a homogeneous system with more than one solution so its matrix is singular. 7 0 -7 0 8 2 0 1 -3 0 0 3 -6 3.14 Example The first of these matrices is nonsingular while the second is singular 1 36 because the first of these homogeneous systems has a unique solution while the second has infinitely many solutions. x+2y=0 x+2y=0 3x+4y=0 3x+6y=0 We have made the distinction in the definition because a system(with the same number of equations as variables)behaves in one of two ways,depending on whether its matrix of coefficients is nonsingular or singular.A system where the matrix of coefficients is nonsingular has a unique solution for any constants on the right side:for instance,Gauss'method shows that this system x+2y=a 3x+4y=b has the unique solution x=6-2a and y=(3a-6)/2.On the other hand,a system where the matrix of coefficients is singular never has a unique solution- it has either no solutions or else has infinitely many,as with these. x+2y=1x+2y=1 3x+6y=23x+6y=3 Thus,'singular'can be thought of as connoting "troublesome",or at least "not ideal". The above table has two factors.We have already considered the factor along the top:we can tell which column a given linear system goes in solely by considering the system's left-hand side-the constants on the right-hand side play no role in this factor.The table's other factor,determining whether a particular solution exists,is tougher.Consider these two 3x+2y=53x+2y=5 3x+2y=5 3x+2y=4 with the same left sides but different right sides.Obviously,the first has a solution while the second does not,so here the constants on the right side
28 Chapter One. Linear Systems The Chemistry problem from Example 3.6 is a homogeneous system with more than one solution so its matrix is singular. 7 0 −7 0 8 1 −5 −2 0 1 −3 0 0 3 −6 −1 3.14 Example The first of these matrices is nonsingular while the second is singular µ 1 2 3 4¶ µ1 2 3 6¶ because the first of these homogeneous systems has a unique solution while the second has infinitely many solutions. x + 2y = 0 3x + 4y = 0 x + 2y = 0 3x + 6y = 0 We have made the distinction in the definition because a system (with the same number of equations as variables) behaves in one of two ways, depending on whether its matrix of coefficients is nonsingular or singular. A system where the matrix of coefficients is nonsingular has a unique solution for any constants on the right side: for instance, Gauss’ method shows that this system x + 2y = a 3x + 4y = b has the unique solution x = b − 2a and y = (3a − b)/2. On the other hand, a system where the matrix of coefficients is singular never has a unique solution — it has either no solutions or else has infinitely many, as with these. x + 2y = 1 3x + 6y = 2 x + 2y = 1 3x + 6y = 3 Thus, ‘singular’ can be thought of as connoting “troublesome”, or at least “not ideal”. The above table has two factors. We have already considered the factor along the top: we can tell which column a given linear system goes in solely by considering the system’s left-hand side — the constants on the right-hand side play no role in this factor. The table’s other factor, determining whether a particular solution exists, is tougher. Consider these two 3x + 2y = 5 3x + 2y = 5 3x + 2y = 5 3x + 2y = 4 with the same left sides but different right sides. Obviously, the first has a solution while the second does not, so here the constants on the right side
Section I.Solving Linear Systems 29 decide if the system has a solution.We could conjecture that the left side of a linear system determines the number of solutions while the right side determines if solutions exist,but that guess is not correct.Compare these two systems 3x+2y=53x+2y=5 4z+2则=4 3x+2y=4 with the same right sides but different left sides.The first has a solution but the second does not.Thus the constants on the right side of a system don't decide alone whether a solution exists:rather,it depends on some interaction between the left and right sides. For some intuition about that interaction,consider this system with one of the coefficients left as the parameter c. x+2y+3z=1 x+y+z=1 cx+3y+4z=0 If c=2 this system has no solution because the left-hand side has the third row as a sum of the first two,while the right-hand does not.If c2 this system has a unique solution (try it with c=1).For a system to have a solution,if one row of the matrix of coefficients on the left is a linear combination of other rows, then on the right the constant from that row must be the same combination of constants from the same rows. More intuition about the interaction comes from studying linear combina- tions.That will be our focus in the second chapter,after we finish the study of Gauss'method itself in the rest of this chapter. Exercises 3.15 Solve each system.Express the solution set using vectors.Identify the par- ticular solution and the solution set of the homogeneous system. (a)3x+6y=18(b)x+y=1(c)x1+x3=4 x+2y=6 x-y=-1 T1-x2+2x3=5 4r1-x2+5x3=17 (d)2a+b-c=2(e)x+2y-z=3()x+z+w=4 2a+c=3 2x+y+w=4 2x+y-w=2 a-b=0 E-y+2+0=1 3c+y+z=7 3.16 Solve each system,giving the solution set in vector notation.Identify the particular solution and the solution of the homogeneous system. (a)2x+y-z=1(b)x-z=1(c)x-y+z=0 4x-y=3 y+2z-w=3 y+w=0 x+2y+3z-0=7 3x-2y+3z+0=0 -1 -w=0 (d)a+2b+3c+d-e=1 3a-6+c+d+e=3 √3.17 For the system 2x-y-w=3 y+z+2w=2 x-2y-z=-1
Section I. Solving Linear Systems 29 decide if the system has a solution. We could conjecture that the left side of a linear system determines the number of solutions while the right side determines if solutions exist, but that guess is not correct. Compare these two systems 3x + 2y = 5 4x + 2y = 4 3x + 2y = 5 3x + 2y = 4 with the same right sides but different left sides. The first has a solution but the second does not. Thus the constants on the right side of a system don’t decide alone whether a solution exists; rather, it depends on some interaction between the left and right sides. For some intuition about that interaction, consider this system with one of the coefficients left as the parameter c. x + 2y + 3z = 1 x + y + z = 1 cx + 3y + 4z = 0 If c = 2 this system has no solution because the left-hand side has the third row as a sum of the first two, while the right-hand does not. If c 6= 2 this system has a unique solution (try it with c = 1). For a system to have a solution, if one row of the matrix of coefficients on the left is a linear combination of other rows, then on the right the constant from that row must be the same combination of constants from the same rows. More intuition about the interaction comes from studying linear combinations. That will be our focus in the second chapter, after we finish the study of Gauss’ method itself in the rest of this chapter. Exercises X 3.15 Solve each system. Express the solution set using vectors. Identify the particular solution and the solution set of the homogeneous system. (a) 3x + 6y = 18 x + 2y = 6 (b) x + y = 1 x − y = −1 (c) x1 + x3 = 4 x1 − x2 + 2x3 = 5 4x1 − x2 + 5x3 = 17 (d) 2a + b − c = 2 2a + c = 3 a − b = 0 (e) x + 2y − z = 3 2x + y + w = 4 x − y + z + w = 1 (f) x + z + w = 4 2x + y − w = 2 3x + y + z = 7 3.16 Solve each system, giving the solution set in vector notation. Identify the particular solution and the solution of the homogeneous system. (a) 2x + y − z = 1 4x − y = 3 (b) x − z = 1 y + 2z − w = 3 x + 2y + 3z − w = 7 (c) x − y + z = 0 y + w = 0 3x − 2y + 3z + w = 0 −y − w = 0 (d) a + 2b + 3c + d − e = 1 3a − b + c + d + e = 3 X 3.17 For the system 2x − y − w = 3 y + z + 2w = 2 x − 2y − z = −1
30 Chapter One.Linear Systems which of these can be used as the particular solution part of some general solu- tion? 0 2 1 (a) (b) 1 8 0 0 3.18 Lemma 3.8 says that any particular solution may be used for p.Find,if possible,a general solution to this system x-y+0=4 2x+3y-2 =0 y+之+D=4 that uses the given vector as its particular solution 0 5 2 (a) 0 (b c 10 3.19 One of these is nonsingular while the other is singular.Which is which? a（432) 3 3.20 Singular or nonsingular? 1 (a) 13 回() 1 (Careful!) 3.21 Is the given vector in the set generated by the given set? (a) 食Θ)) (Θ日Θ ( 目Θ目目目 0 (d) 300 3.22 Prove that any linear system with a nonsingular matrix of coefficients has a solution,and that the solution is unique. 3.23 To tell the whole truth,there is another tricky point to the proof of Lemma 3.7. What happens if there are no non-'0=0'equations?(There aren't any more tricky points after this one.) 3.24 Prove that if s and t satisfy a homogeneous system then so do these vec- tors. (a)+(b)35 (c)ks+mt'for k,mER What's wrong with:"These three show that if a homogeneous system has one solution then it has many solutions-any multiple of a solution is another solution
30 Chapter One. Linear Systems which of these can be used as the particular solution part of some general solution? (a) 0 −3 5 0 (b) 2 1 1 0 (c) −1 −4 8 −1 X 3.18 Lemma 3.8 says that any particular solution may be used for ~p. Find, if possible, a general solution to this system x − y + w = 4 2x + 3y − z = 0 y + z + w = 4 that uses the given vector as its particular solution. (a) 0 0 0 4 (b) −5 1 −7 10 (c) 2 −1 1 1 3.19 One of these is nonsingular while the other is singular. Which is which? (a) µ 1 3 4 −12¶ (b) µ 1 3 4 12¶ X 3.20 Singular or nonsingular? (a) µ 1 2 1 3¶ (b) µ 1 2 −3 −6 ¶ (c) µ 1 2 1 1 3 1¶ (Careful!) (d) Ã1 2 1 1 1 3 3 4 7! (e) Ã 2 2 1 1 0 5 −1 1 4! X 3.21 Is the given vector in the set generated by the given set? (a) µ 2 3 ¶ , { µ 1 4 ¶ , µ 1 5 ¶ } (b) Ã−1 0 1 ! , { Ã2 1 0 ! , Ã1 0 1 ! } (c) Ã1 3 0 ! , { Ã1 0 4 ! , Ã2 1 5 ! , Ã3 3 0 ! , Ã4 2 1 ! } (d) 1 0 1 1 , { 2 1 0 1 , 3 0 0 2 } 3.22 Prove that any linear system with a nonsingular matrix of coefficients has a solution, and that the solution is unique. 3.23 To tell the whole truth, there is another tricky point to the proof of Lemma 3.7. What happens if there are no non-‘0 = 0’ equations? (There aren’t any more tricky points after this one.) X 3.24 Prove that if ~s and ~t satisfy a homogeneous system then so do these vectors. (a) ~s + ~t (b) 3~s (c) k~s + m~t for k, m ∈ R What’s wrong with: “These three show that if a homogeneous system has one solution then it has many solutions— any multiple of a solution is another solution
Section I.Solving Linear Systems 31 and any sum of solutions is a solution also-so there are no homogeneous systems with exactly one solution."? 3.25 Prove that if a system with only rational coefficients and constants has a solution then it has at least one all-rational solution.Must it have infinitely many?
Section I. Solving Linear Systems 31 and any sum of solutions is a solution also — so there are no homogeneous systems with exactly one solution.”? 3.25 Prove that if a system with only rational coefficients and constants has a solution then it has at least one all-rational solution. Must it have infinitely many?