Section I.Solving Linear Systems 17 Another thing shown plainly is that setting both w and u to zero gives that this 0 4 w 0 2 0 is a particular solution of the linear system. 2.14 Example In the same way,this system x-y+z=1 3x +z=3 5x-2y+3z=5 reduces 1 -11 1 -1 1 3 0 1 3 -3p1+P2 0 3 P2+P 5 -2 3 5 -5p1+P3 0 -2 0 0 0 100 to a one-parameter solution set. -1/3 2/3 zz∈R} 1 Before the exercises,we pause to point out some things that we have yet to do. The first two subsections have been on the mechanics of Gauss'method. Except for one result,Theorem 1.4-without which developing the method doesn't make sense since it says that the method gives the right answers-we have not stopped to consider any of the interesting questions that arise. For example,can we always describe solution sets as above,with a particular solution vector added to an unrestricted linear combination of some other vec- tors?The solution sets we described with unrestricted parameters were easily seen to have infinitely many solutions so an answer to this question could tell us something about the size of solution sets.An answer to that question could also help us picture the solution sets,in R2,or in R3,etc. Many questions arise from the observation that Gauss'method can be done in more than one way (for instance,when swapping rows,we may have a choice of which row to swap with).Theorem 1.4 says that we must get the same solution set no matter how we proceed,but if we do Gauss'method in two different ways must we get the same number of free variables both times,so that any two solution set descriptions have the same number of parameters? Must those be the same variables (e.g.,is it impossible to solve a problem one way and get y and w free or solve it another way and get y and z free)?
Section I. Solving Linear Systems 17 Another thing shown plainly is that setting both w and u to zero gives that this x y z w u = 0 4 0 0 0 is a particular solution of the linear system. 2.14 Example In the same way, this system x − y + z = 1 3x + z = 3 5x − 2y + 3z = 5 reduces 1 −1 1 1 3 0 1 3 5 −2 3 5 −3ρ1+ρ2 −→ −5ρ1+ρ3 1 −1 1 1 0 3 −2 0 0 3 −2 0 −ρ2+ρ3 −→ 1 −1 1 1 0 3 −2 0 0 0 0 0 to a one-parameter solution set. { 1 0 0 + −1/3 2/3 1 z ¯ ¯ z ∈ R} Before the exercises, we pause to point out some things that we have yet to do. The first two subsections have been on the mechanics of Gauss’ method. Except for one result, Theorem 1.4 — without which developing the method doesn’t make sense since it says that the method gives the right answers— we have not stopped to consider any of the interesting questions that arise. For example, can we always describe solution sets as above, with a particular solution vector added to an unrestricted linear combination of some other vectors? The solution sets we described with unrestricted parameters were easily seen to have infinitely many solutions so an answer to this question could tell us something about the size of solution sets. An answer to that question could also help us picture the solution sets, in R 2 , or in R 3 , etc. Many questions arise from the observation that Gauss’ method can be done in more than one way (for instance, when swapping rows, we may have a choice of which row to swap with). Theorem 1.4 says that we must get the same solution set no matter how we proceed, but if we do Gauss’ method in two different ways must we get the same number of free variables both times, so that any two solution set descriptions have the same number of parameters? Must those be the same variables (e.g., is it impossible to solve a problem one way and get y and w free or solve it another way and get y and z free)?
8 Chapter One.Linear Systems In the rest of this chapter we answer these questions.The answer to each is 'yes'.The first question is answered in the last subsection of this section.In the second section we give a geometric description of solution sets.In the final section of this chapter we tackle the last set of questions.Consequently,by the end of the first chapter we will not only have a solid grounding in the practice of Gauss'method,we will also have a solid grounding in the theory.We will be sure of what can and cannot happen in a reduction. Exercises 2.15 Find the indicated entry of the matrix,if it is defined. A=G (a)a2,1(b)a1,2 (c）a2,2 (d)a3,1 2.16 Give the size of each matrix 2.17 Do the indicated vector operation,if it is defined (a) (日)o目-目 +目o0目+0 e)( 2.18 Solve each system using matrix notation. Express the solution using vec- tors. (a)3x+6y=18 (b)x+y=1(c)x1+x3=4 x+2y=6 x-y=-1 x1-x2+2x3=5 4x1-x2+5x3=17 (d)2a+b-c=2 (e)x+2y-z =3.(f)x+z+0=4 2a+c=3 2x+y+w=4 2x+y-w=2 a-b=0 x-y+之+w=1 3x+y+z=7 v2.19 Solve each system using matrix notation.Give each solution set in vector notation. (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-w=7 3x-2y+3z+0=0 -1 -0=0 (d)a+2b+3c+d-e=1 3a-6+c+d+e=3 2.20 The vector is in the set.What value of the parameters produces that vec- tor? (a) ()(ke ())月：
18 Chapter One. Linear Systems In the rest of this chapter we answer these questions. The answer to each is ‘yes’. The first question is answered in the last subsection of this section. In the second section we give a geometric description of solution sets. In the final section of this chapter we tackle the last set of questions. Consequently, by the end of the first chapter we will not only have a solid grounding in the practice of Gauss’ method, we will also have a solid grounding in the theory. We will be sure of what can and cannot happen in a reduction. Exercises X 2.15 Find the indicated entry of the matrix, if it is defined. A = µ 1 3 1 2 −1 4¶ (a) a2,1 (b) a1,2 (c) a2,2 (d) a3,1 X 2.16 Give the size of each matrix. (a) µ 1 0 4 2 1 5¶ (b) Ã 1 1 −1 1 3 −1 ! (c) µ 5 10 10 5 ¶ X 2.17 Do the indicated vector operation, if it is defined. (a) Ã2 1 1 ! + Ã3 0 4 ! (b) 5 µ 4 −1 ¶ (c) Ã1 5 1 ! − Ã3 1 1 ! (d) 7 µ 2 1 ¶ + 9 µ 3 5 ¶ (e) µ 1 2 ¶ + Ã1 2 3 ! (f) 6 Ã3 1 1 ! − 4 Ã2 0 3 ! + 2Ã1 1 5 ! X 2.18 Solve each system using matrix notation. Express the solution using vectors. (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 X 2.19 Solve each system using matrix notation. Give each solution set in vector notation. (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 2.20 The vector is in the set. What value of the parameters produces that vector? (a) µ 5 −5 ¶ , { µ 1 −1 ¶ k ¯ ¯ k ∈ R} (b) Ã−1 2 1 ! , { Ã−2 1 0 ! i + Ã3 0 1 ! j ¯ ¯ i, j ∈ R}
Section I.Solving Linear Systems 19 2 0 nm,n∈R} 2.21 Decide if the vector is in the set. 回((2）ke时 o)(目()e (目()（日 rr∈R} (d) Θ9+（月 kj,k∈R} 2.22 Parametrize the solution set of this one-equation system. c1+E2+··+xn=0 v 2.23 (a)Apply Gauss'method to the left-hand side to solve x+2y w=a 2x +z=b x+y +2w=c for z,y,z,and w,in terms of the constants a,b,and c. (b)Use your answer from the prior part to solve this. x+2y-w=3 2x+z =1 x+y+2w=-2 2.24 Why is the comma needed in the notation'ai;'for matrix entries? √2.25 Give the4×4 matrix whose i,j-th entry is (a)i+j;(b)-1 to the i+j power. 2.26 For any matrix A,the transpose of A,written Atrans,is the matrix whose columns are the rows of A.Find the transpose of each of these (d) 0 2.27 (a)Describe all functions f(x)=ax2 bz +c such that f(1)=2 and f(-1)=6. (b)Describe all functions f(r)=ax2+bx+c such that f(1)=2. 2.28 Show that any set of five points from the plane R2 lie on a common conic section,that is,they all satisfy some equation of the form ar2+by2+cry+dr+ ey+f =0 where some of a,...f are nonzero. 2.29 Make up a four equations/four unknowns system having (a)a one-parameter solution set; (b)a two-parameter solution set; (c)a three-parameter solution set 2.30 (a)Solve the system of equations. ax+y=a2 x+ay=1 For what values of a does the system fail to have solutions,and for what values of a are there infinitely many solutions?
Section I. Solving Linear Systems 19 (c) Ã 0 −4 2 ! , { Ã1 1 0 ! m + Ã2 0 1 ! n ¯ ¯ m, n ∈ R} 2.21 Decide if the vector is in the set. (a) µ 3 −1 ¶ , { µ −6 2 ¶ k ¯ ¯ k ∈ R} (b) µ 5 4 ¶ , { µ 5 −4 ¶ j ¯ ¯ j ∈ R} (c) Ã 2 1 −1 ! , { Ã 0 3 −7 ! + Ã 1 −1 3 ! r ¯ ¯ r ∈ R} (d) Ã1 0 1 ! , { Ã2 0 1 ! j + Ã−3 −1 1 ! k ¯ ¯ j, k ∈ R} 2.22 Parametrize the solution set of this one-equation system. x1 + x2 + · · · + xn = 0 X 2.23 (a) Apply Gauss’ method to the left-hand side to solve x + 2y − w = a 2x + z = b x + y + 2w = c for x, y, z, and w, in terms of the constants a, b, and c. (b) Use your answer from the prior part to solve this. x + 2y − w = 3 2x + z = 1 x + y + 2w = −2 X 2.24 Why is the comma needed in the notation ‘ai,j ’ for matrix entries? X 2.25 Give the 4×4 matrix whose i, j-th entry is (a) i + j; (b) −1 to the i + j power. 2.26 For any matrix A, the transpose of A, written A trans, is the matrix whose columns are the rows of A. Find the transpose of each of these. (a) µ 1 2 3 4 5 6¶ (b) µ 2 −3 1 1 ¶ (c) µ 5 10 10 5 ¶ (d) Ã1 1 0 ! X 2.27 (a) Describe all functions f(x) = ax2 + bx + c such that f(1) = 2 and f(−1) = 6. (b) Describe all functions f(x) = ax2 + bx + c such that f(1) = 2. 2.28 Show that any set of five points from the plane R 2 lie on a common conic section, that is, they all satisfy some equation of the form ax2 + by2 + cxy + dx + ey + f = 0 where some of a, . . . , f are nonzero. 2.29 Make up a four equations/four unknowns system having (a) a one-parameter solution set; (b) a two-parameter solution set; (c) a three-parameter solution set. ? 2.30 (a) Solve the system of equations. ax + y = a 2 x + ay = 1 For what values of a does the system fail to have solutions, and for what values of a are there infinitely many solutions?
20 Chapter One.Linear Systems (b)Answer the above question for the system. ax+y=a3 x+ay=1 [USSR.Olympiad no.174] 2.31 In air a gold-surfaced sphere weighs 7588 grams.It is known that it may contain one or more of the metals aluminum,copper,silver,or lead.When weighed successively under standard conditions in water,benzene,alcohol,and glycerine its respective weights are 6588,6688,6778,and 6328 grams.How much,if any. of the forenamed metals does it contain if the specific gravities of the designated substances are taken to be as follows? Aluminum 2.7 Alcohol 0.81 Copper 8.9 Benzene 0.90 Gold 19.3 Glycerine 1.26 Lead 11.3 Water 1.00 Silver 10.8 [Math.Mag.,Sept.1952] I.3 General Particular Homogeneous The prior subsection has many descriptions of solution sets.They all fit a pattern.They have a vector that is a particular solution of the system added to an unrestricted combination of some other vectors.The solution set from Example 2.13 illustrates. 0 1/2 -1 0 +w 3 u 1/2 w,u∈R} 0 0 0 particular unrestricted solution combination The combination is unrestricted in that w and u can be any real numbers- there is no condition like "such that 2w-u=0"that would restrict which pairs w,u can be used to form combinations. That example shows an infinite solution set conforming to the pattern.We can think of the other two kinds of solution sets as also fitting the same pat- tern.A one-element solution set fits in that it has a particular solution,and the unrestricted combination part is a trivial sum (that is,instead of being a combination of two vectors,as above,or a combination of one vector,it is a combination of no vectors).A zero-element solution set fits the pattern since there is no particular solution,and so the set of sums of that form is empty. We will show that the examples from the prior subsection are representative, in that the description pattern discussed above holds for every solution set
20 Chapter One. Linear Systems (b) Answer the above question for the system. ax + y = a 3 x + ay = 1 [USSR Olympiad no. 174] ? 2.31 In air a gold-surfaced sphere weighs 7588 grams. It is known that it may contain one or more of the metals aluminum, copper, silver, or lead. When weighed successively under standard conditions in water, benzene, alcohol, and glycerine its respective weights are 6588, 6688, 6778, and 6328 grams. How much, if any, of the forenamed metals does it contain if the specific gravities of the designated substances are taken to be as follows? Aluminum 2.7 Alcohol 0.81 Copper 8.9 Benzene 0.90 Gold 19.3 Glycerine 1.26 Lead 11.3 Water 1.00 Silver 10.8 [Math. Mag., Sept. 1952] I.3 General = Particular + Homogeneous The prior subsection has many descriptions of solution sets. They all fit a pattern. They have a vector that is a particular solution of the system added to an unrestricted combination of some other vectors. The solution set from Example 2.13 illustrates. { 0 4 0 0 0 | {z } particular solution + w 1 −1 3 1 0 + u 1/2 −1 1/2 0 1 | {z } unrestricted combination ¯ ¯ w, u ∈ R} The combination is unrestricted in that w and u can be any real numbers — there is no condition like “such that 2w−u = 0” that would restrict which pairs w, u can be used to form combinations. That example shows an infinite solution set conforming to the pattern. We can think of the other two kinds of solution sets as also fitting the same pattern. A one-element solution set fits in that it has a particular solution, and the unrestricted combination part is a trivial sum (that is, instead of being a combination of two vectors, as above, or a combination of one vector, it is a combination of no vectors). A zero-element solution set fits the pattern since there is no particular solution, and so the set of sums of that form is empty. We will show that the examples from the prior subsection are representative, in that the description pattern discussed above holds for every solution set
Section I.Solving Linear Systems 21 3.1 Theorem For any linear system there are vectors B,...,B such that the solution set can be described as {D+C+…+cc1,..,ck∈R} where p is any particular solution,and where the system has k free variables. This description has two parts,the particular solution p and also the un- restricted linear combination of the 's.We shall prove the theorem in two corresponding parts,with two lemmas. We will focus first on the unrestricted combination part.To do that,we consider systems that have the vector of zeroes as one of the particular solutions, so that p+c1B1+...+ckBk can be shortened to c18+...+ckBk. 3.2 Definition A linear equation is homogeneous if it has a constant of zero, that is,if it can be put in the form a11+a2x2+...+ann=0. (These are homogeneous'because all of the terms involve the same power of their variable-the first power-including a 'Ozo'that we can imagine is on the right side.) 3.3 Example With any linear system like 3x+4=3 2x-y=1 we associate a system of homogeneous equations by setting the right side to zeros. 3x+4y=0 2x-y=0 Our interest in the homogeneous system associated with a linear system can be understood by comparing the reduction of the system 3x+4y=3-(2/3)p1+p23x+ 4y=3 2x-y=1 -(11/3)y=-1 with the reduction of the associated homogeneous system. 3z+4y=0-(2/3)2:+p23x+ 4y=0 2x-y=0 -(11/3)y=0 Obviously the two reductions go in the same way.We can study how linear sys- tems are reduced by instead studying how the associated homogeneous systems are reduced. Studying the associated homogeneous system has a great advantage over studying the original system.Nonhomogeneous systems can be inconsistent. But a homogeneous system must be consistent since there is always at least one solution,the vector of zeros
Section I. Solving Linear Systems 21 3.1 Theorem For any linear system there are vectors β~ 1, . . . , β~ k such that the solution set can be described as {~p + c1β~ 1 + · · · + ckβ~ k ¯ ¯ c1, . . . , ck ∈ R} where ~p is any particular solution, and where the system has k free variables. This description has two parts, the particular solution ~p and also the unrestricted linear combination of the β~’s. We shall prove the theorem in two corresponding parts, with two lemmas. We will focus first on the unrestricted combination part. To do that, we consider systems that have the vector of zeroes as one of the particular solutions, so that ~p + c1β~ 1 + · · · + ckβ~ k can be shortened to c1β~ 1 + · · · + ckβ~ k. 3.2 Definition A linear equation is homogeneous if it has a constant of zero, that is, if it can be put in the form a1x1 + a2x2 + · · · + anxn = 0. (These are ‘homogeneous’ because all of the terms involve the same power of their variable— the first power— including a ‘0x0’ that we can imagine is on the right side.) 3.3 Example With any linear system like 3x + 4y = 3 2x − y = 1 we associate a system of homogeneous equations by setting the right side to zeros. 3x + 4y = 0 2x − y = 0 Our interest in the homogeneous system associated with a linear system can be understood by comparing the reduction of the system 3x + 4y = 3 2x − y = 1 −(2/3)ρ1+ρ2 −→ 3x + 4y = 3 −(11/3)y = −1 with the reduction of the associated homogeneous system. 3x + 4y = 0 2x − y = 0 −(2/3)ρ1+ρ2 −→ 3x + 4y = 0 −(11/3)y = 0 Obviously the two reductions go in the same way. We can study how linear systems are reduced by instead studying how the associated homogeneous systems are reduced. Studying the associated homogeneous system has a great advantage over studying the original system. Nonhomogeneous systems can be inconsistent. But a homogeneous system must be consistent since there is always at least one solution, the vector of zeros