Chapter 7 Systems and Matrices

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1 Section 7. Solving Sstems of wo Equations Chapter 7 Sstems and Matrices Section 7. Solving Sstems of wo Equations Eploration. In #, there ma be more than one good wa to choose the variable for which the substitution will be made. One approach is given. In most cases, the solution is onl shown up to the point where the value of the first variable is found.. (, )=(9, ): Since =, we have -4=, so =9.. (, )= a 0 : =0-, 7, b 0 so - (0-)=0, or 7=0, so = 7.. [0, 0] b [, ] 7. (, )= a - : = ( - 7)/,, b so (-7)+=8, or =, so =. 9. No solution: =+6, so (+6)+6=4, or =4 not true.. (, )= (;, 9); he second equation gives =9, so, =9, or = ;. [0, 0] b [, ] [0, 0] b [, ]. (, )= a - or (, )= a :,, 7 b b. he function ln is onl defined for 7 0, so all solutions must be positive. As approaches infinit, -4+ is going to infinit much more quickl than ln is; and hence will alwas be larger than ln for -values greater than 4. Quick Review 7.. =- = -. (+)(-)=0 +=0 or -=0 = = =. -4=0 ( -4)=0 (-)(+)=0 =0, =, = 7. m = - 4, - = - 4 ( + ) = = 9. (+)= () 4-6= 0 Section 7. Eercises. (a) No: (0) - (4) Z 8. (b) Yes: ()-()=8 and ()-()=. (c) No: (-) - (-9) Z =0, so = - Substitute these or =. values into =6.. (, )=(0, 0) or (, )=(, 8): =, so =0 or =. Substitute these values into = (, ) = a, + 89 b and a, - 89 b: - = -, so = Substitute = + into + = 9: ( - ) + = 9 Q = 0. Using the quadratic formula, we find that = ; In the following, E and E refer to the first and second equations, respectivel. 9. (, )=(8, ): E +E leaves =6, so =8.. (, )=(4, ): E +E leaves =44, so =44.. No solution: E +E leaves 0= 7, which is false.. here are infinitel man solutions, an pair a, : - b E +E leaves 0=0, which is alwas true. As long as (, ) satisfies one equation, it will also satisf the other. 7. (, )=(0, ) or (, )=(, ) 9. No solution. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

2 Chapter 7 Sstems and Matrices. One solution. (a) he following is a scatter plot of the data with the quadratic regression equation = superimposed on it. [, ] b [, ]. Infinitel man solutions. [, 0] b [0, 60] (b) he following is a scatter plot of the data with the 4. logistic regression equation = e superimposed on it. [, ] b [, ]. (, ) L (0.69, -0.7) [, 0] b [0, 60] [, ] b [, ] 7. (, ) L (-., -.6) or (0.47,.77) or (.8,.08) (c) Quadratic regression model Graphical solution: Graph the line = 4 with the quadratic regression curve = and find the intersection of the two curves. Since the two graphs do not intersect, the ependitures will never reach 4 billion dollars. [, ] b [, ] 9. (, )=(.,.6) or (, 0) [, ] b [, ] 4. (, ) L (.0,.9) or (.0,.9) [ 4, 4] b [ 4, 4] 4. (, p)=(.7, 4.7): 00-=0+, so 40=0. 4. In this problem, the graphs are representative of the ependitures (in billions of dollars) for public medical research costs for several ears, where is the number of ears past 000. Another graphical solution would be to find where the graph of the difference of the two curves is equal to 0. Since the graph never crosses the -ais, the ependitures will never reach 4 billion dollars. Algebraic solution: 4 = for. Use the quadratic formula to solve the equation = 0. a = -0.6 b =.94 c = = -(.94) ; (.94) - 4(-0.6)(-9.667) (-0.6) = = [0, 0] b [0, 0] -.94 ; ; Since there are no real solutions to the equation, the ependitures will never reach 4 billion dollars. Logistic regression model Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

3 Section 7. Solving Sstems of wo Equations Graphical solution: Graph the line = 4 with the 4. logistic regression curve = and e find the intersection of the two curves. he two intersect at L 4.6. he ependitures will reach 4 billion dollars in about 0. (b) he following is a scatter plot of the Indiana data with the linear regression equation L superimposed on it. [0, 0] b [0, 0000] Another graphical solution would be to find where the graph of the difference of the two curves is equal to Algebraic solution: 4 = for e ( e ) = e = e = = e = = = ln = ln L 4. he ependitures will reach 4 billion dollars in about 0. (d) he long-range implication of using the quadratic regression equation is that the ependitures will eventuall fall to zero. he graph of the function is a parabola with verte at about (9,40) and it opens downward. So, eventuall the curve will cross the -ais and the ependitures will be 0. his will happen at L. (e) he long-range implication of using the logistic regression model is that the ependitures will eventuall level off at about 4. billion dollars. (We notice that as gets larger, e approaches 0. herefore, the denominator of the function approaches and the function itself approaches 4., or about 4..) 47. In this problem, the graphs are representative of the population (in thousands) of the states of Florida and Indiana for several ears, where is the number of ears past 990. (a) he following is a scatter plot of the Florida data with the linear regression equation L superimposed on it. (c) Graphical solution: Graph the two linear equations = and = on the same aes and find the point of intersection. he two curves intersect at L 7.. he population of the two states was the same in the ear 96. [0, 0] b [0, 0000] Another graphical solution would be to find where the graph of the differences of the two curves is equal to 0. Algebraic solution: Solve = for = = = = he population of the two states was the same in the ear =(+) and 00=. hen =00-, so 00=(00-), and therefore = 0 ; 0, and = 0 < 0. Both answers correspond to a rectangle with approimate dimensions.8 m*94.7 m.. If r is Hank s rowing speed (in miles per hour) and c is 4 the speed of the current, (r - c) = and 60 (r + c) =. herefore r = c + (from the first 60 equation); substituting gives, so 60 ac + b = c = 60 mph. Finall, - =, and c = 6 L.06 r = c + mph. = 8 L.6. m +/=.74 and / =m + 0.6, so m+0.6=.74. hen m=$0.79 (79 cents) and /=$0.9 (9 cents).. 4= a+b and 6=a+b, so b=a+4 and 6=a+4. hen a = 4 and b =. [0, 0] b [0, 0000] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

4 ` ` 4 Chapter 7 Sstems and Matrices 7. (a) Let C()=the amount charged b each rental compan, and let =the number of miles driven b Pedro. Compan A: C()= Compan B: C()=+0. Solving these two equations for, =+0. =0.0 00= Pedro can drive 00 miles to be charged the same amount b the two companies. (b) One possible answer: If Pedro is making onl a short trip, Compan B is better because the flat fee is less. However, if Pedro drives the rental van over 00 miles, Compan A s plan is more economical for his needs. 9. False. A sstem of two linear equations in two variables has either 0,, or infinitel man solutions. 6. Using (, )=(, ), ()-( )= +( )= he answer is C. 6. wo parabolas can intersect in 0,,,, or 4 places, or infinitel man places if the parabolas completel coincide. he answer is. 6. (a) = =6 4 = = 4 = 4 -, = (b) [, ] b [ 4, 4] (, ) L (-.9,.9) or (.9, -0.9) (-.9) (c) + (.9) L L and 4 9 (.9)+(.9)=, so the first solution checks. (.9) + (-0.9) L.004 L and 4 9 (.9) + (-0.9) =, so the second solution checks. 67. Subtract the second equation from the first, leaving = 0, or = 0 hen = 4-0 so =., = ;. B 69. he verte of the parabola R=(00-4) =4(-) has first coordinate =. units. Section 7. Matri Algebra Eploration. a = () - () = Set i=j=. a = () - () = Set i=, j=. a = () - () = Set i=, j=. a = () - () = 4 Set i=j=. So, A = c Similar computations show that 4 d.. he additive inverse of A is A and - - -A = c - -4 d. A + (-A) = c - - d + c d = c d = [0] he order of [0] is *.. A-B=c - d - c 4 d 6 = c d - c d = c 8 - d Eploration. det (A)=-a a a + a a a + a a a -a a a - a a a + a a a Each element contains an element from each row and each column due to a definition of a determinant. Regardless of the row or column picked to appl the definition, all other elements of the matri are eventuall factored into the multiplication. a a a a. = a a a (-) a a ` a a a a a a + a a + ` a a a (-) 4 (-) ` ` a a a a = a (a a - a a )-a (a a - a a ) + a (a a - a a ) = a a a - a a a - a a a + a a a + a a a - a a a he two epressions are eactl equal.. Recall that A ij is ( ) i+j M ij where M ij is the determinant of the matri obtained b deleting the row and column containing a ij. Let A=k*k square matri with zeros in the ith row. hen: det(a)= ith row = 0 # A i + 0 # A i + =0 B = c - d. a a a k a a a k : 8 o o o o o o a k a k a kk + 0 # A ik = Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

5 Section 7. Matri Algebra Quick Review 7.. (a) (, ) (b) (, ). (a) (, ) (b) (, ). ( cos, sin ) 7. sin (a + b) = sin a cos b + cos a sin b 9. cos (a + b) = cos a cos b - sin a sin b Section 7. Eercises. *; not square. *; not square. *; not square 7. a = 9. a =4. (a) 0 c - d (b) c 6 9 d 6 9 (c) c - d (d) A-B= c - - c = - d - -4 d c c -9 = c - 0 d -6 - d 4 d. (a) C - 0S (b) C - S -9 (c) C 0 - S (d) A-B=C 0 - S-C - S - -. (a) (b) (c) -6 0 = C 0 - S- C -6 S= C S 4 - C S -4-6 C S (d) A-B=C S- C 0 S= C S-C 0 S =C S - -8 C 6 - S 7. (a) ()() + ()(-) ()(-) + ()(-4) -4-8 c (-)() + ()(-) (-)(-) + ()(-4) d = - -7 d ()() + (-)(-) ()() + (-)() - (b) BA = c d = c (-)() + (-4)(-) (-)() + (-4)() 0-6 d 9. (a) ()() + (0)(-) + ()(0) ()() + (0)() + ()(-) = c ()() + (4)(-) + (-)(0) ()() + (4)() + (-)(-) d - d ()() + ()() ()(0) + ()(4) ()() + ()(-) (b) BA = C (-)() + ()() (-)(0) + ()(4) (-)() + ()(-) S = C S (0)() + (-)() (0)(0) + (-)(4) (0)() + (-)(-) (a) (-)() + (0)(-) + ()(4) (-)() + (0)(0) + ()(-) AB = C (4)() + ()(-) + (-)(4) (4)() + ()(0) + (-)(-) ()() + (0)(-) + ()(4) ()() + (0)(0) + ()(-) = C 7 S ()(-) + ()(4) + (0)() ()(0) + ()() + (0)(0) ()() + ()(-) + (0)() (b) BA = C (-)(-) + (0)(4) + ()() (-)(0) + (0)() + ()(0) (-)() + (0)(-) + ()() S (4)(-) + (-)(4) + (-)() (4)(0) + (-)() + (-)(0) (4)() + (-)(-) + (-)() = C 0 0S -8-0 (-)(0) + (0)() + ()(-) (4)(0) + ()() + (-)(-) S ()(0) + (0)() + ()(-) Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

6 6 Chapter 7 Sstems and Matrices. (a) AB = [()(-) + (-)(4) + ()()] = [-8] (b). (a) AB is not possible. (b) BA=[( )( )+()() ( )()+()(4)]=[8 4] 7. (a) ()(0) + ()(0) + ()() ()(0) + ()() + ()(0) ()() + ()(0) + ()(0) (b) BA = C ()(0) + (0)(0) + ()() ()(0) + (0)() + ()(0) ()() + (0)(0) + ()(0) S (-)(0) + ()(0) + (4)() (-)(0) + ()() + (4)(0) (-)() + ()(0) + (4)(0) In #9, use the fact that two matrices are equal onl if all entries are equal. 9. a=, b=. a=, b=0.. (-)() (-)(-) (-)() BA = C (4)() (4)(-) (4)() S = C ()() ()(-) ()() (0)() + (0)() + ()(-) (0)() + (0)(0) + ()() AB = C (0)() + ()() + (0)(-) (0)() + ()(0) + (0)() ()() + (0)() + (0)(-) ()() + (0)(0) + (0)() + ()(-0.6) ()(-0.) + ()(0.4) AB = c()(0.8) ()(0.8) + (4)(-0.6) ()(-0.) + (4)(0.4) d + (-0.)() (0.8)() + (-0.)(4) BA = c(0.8)() (-0.)() + (0.4)() (0.6)() + (-0.4)(4) d c d - = + (0)(-) 4 - = (-)( - ) (-)(4 + ) = = X=B-A ()() - ()() c - = - c - -. d = c - - d X = B - A = C - S c = ac4 d - c db = - d 4. (a)he entries a ij and a ji are the same because each gives the distance between the same two cities. (b) he entries a ii are all 0 because the distance between a cit and itself is No inverse: he determinant (found with a calculator) is A = ; No inverse, det(a)=0 (found using a calculator) 4. Use row or column since the have the greatest number of zeros. Using column : = ()(-) S 4-6 = c 0 0 d, = c 0 0 d, (0)() + (0)() + ()(4) (0)() + ()() + (0)(4) S ()() + (0)() + (0)(4) so A and B are inverses. - 4 = C 0 S = C 0 S d (a) B A= [$0.80 $0.8 $.00] C 0 70 S (00) 0.80(60) C + 0.8(0) + 0.8(70) S + (00) + (0) = [8 7.0] (b) b j in matri B A represents the income Happ Valle Farms makes at grocer store j, selling all three tpes of eggs. 49. (a) otal revenue=sum of (price charged)(number sold) =AB or BA (b) Profit=otal revenue-otal Cost =AB -CB =(A-C)B + ()(-) -. (a) = [ ] c cos a -sin a, sin a cos a d, =, a = 0 = [ ] = (b) [ ] = [ ] c cos a sin a, -sin a cos a d, =, a = 0 c + = [ ] d L.7 0.7] = c - + d L [0.7.7] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

7 Section 7. Matri Algebra 7. Answers will var. One possible answer is provided for each. (a) c(a+b)=c[a ij + b ij ] = [ca ij + cb ij ] = ca + cb (b) (c+d)a=(c+d)[a ij ]=c[a ij ]+d[a ij ] =ca+da (c) c(da)=c[da ij ]=[cda ij ]=cd[a ij ]=cda (d) # A = # [a ij ] = [a ij ] = A. A # a A - = c a b c d -b c d d ad - bc b -c a d = a ad - bc bca c (since ad - bc = a ad - bc b b d d is a scalar) = a ad - bc bcad - bc cd - cd c d -b -c a d -ab + ba -bc + ad d ad - bc 0 c 0 ad - bc d ad - bc 0 ad - bc = ad - bc 0 ad - bc = c 0 0 d = I 7. If (, ) is reflected across the -ais, then (, ) Q (, ). [ ] = [ ]c d 9. If (, ) is reflected across the line =, then (, ) Q (, ). 0 - [ ] = [ ]c - 0 d 6. If (, ) is horizontall stretched (or shrunk) b a factor of c, then (, ) Q (c, ). [ ] = [ ]c c 0 0 d 6. False. he determinant can be negative. For eample, the determinant of A = c 0 is ( )-(0)=. - d 6. he matri AB has the same number of rows as A and the same number of columns as B. he answer is B. 67. he value in row, column is. he answer is. 69. (a) Let A =[a ij ] be an n*n matri and let B be the same as matri A, ecept that the ith row of B is the ith row of A multiplied b the scalar c. hen: det(b)= a a a n a a a n 6 o 6 ith row ca i ca i ca in o a n a n a nn =ca i (-) i + ƒa i ƒ + ca i (-) i + ƒa i ƒ + Á +ca in (-) in ƒa in ƒ = c(a i (-) i + ƒa i ƒ + a i (-) i + ƒa i ƒ + Á +a in (-) i + n ƒa in ƒ) =c det(a) (b definition of determinant) (b) Use the * case as an eample: det(a)= a 0 =a a -0=a a a a which is the product of the diagonal elements. Now consider the general case where A is an n*n matri. hen: a det(a)= 4 a a o o o a n a n a nn a =a ( ) 4 a a o o o o a n a n a n4 a nn a 0 0 =a (a )( ) 4 a 4 a o o o a n a n4 a nn =a a Á a n - n - 0 a n- n - (-) a n n - a nn =a a Á a n n a n n a nn, which is eactl the product of the diagonal elements (b induction). 7. (a) A # A = ccos a sin a c cos a sin a sin a cos a d sin a cos a d = c cos a + sin a sin a cos a - cos a sin a cos a sin a - sin a cos a sin a + cos d a = c 0 0 d = I (b) From the diagram, we know that: =r cos =r sin =r cos ( -Å) =r sin ( -Å) or cos ( -Å)= sin ( -Å)= r r From algebra, we know that: =r cos ( +Å-Å)=r cos (Å+( -Å)) and =r sin ( +Å-Å)=r sin (Å+( -Å)) Using the trigonometric properties and substitution, we have: =r(cos Å cos ( -Å)-sin Å sin ( -Å)) =r cos Å cos ( -Å)-r sin Å sin ( -Å) =(r cos Å) a -(r sin Å) a r b r b = cos Å- sin Å =r (sin Å cos ( -Å)+cos Å sin ( -Å)) =r sin Å cos ( -Å)+r cos Å sin ( -Å) =(r sin Å) a +(r cos Å) a r b r b = sin Å+ cos Å. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

8 8 Chapter 7 Sstems and Matrices (c) [ ] = [ ] c cos a sin a -sin a cos a d.7. X = A - B = C 7. S which is [ ] A -, the inverse of A (a) det(i -A)= =(-a )( ) =(-a )((-a )(-a )-a a ) +a (( a )(-a )-a a ) -a (a a +(a )(-a )) =(-a )( -a -a +a a -a a ) +a ( a +a a -a a ) -a (a a +a -a a ) = -a -a +a a -a a -a +a a +a a -a a a +a a a -a a +a a a -a a a -a a a -a a +a a a = +( a -a -a ) +(a a -a a +a a +a a -a a -a a )+( a a a +a a a +a a a -a a a -a a a +a a a ) (b) he constant term equals det(a). (c) he coefficient of is the opposite of the sum of the elements of the main diagonal in A. (d) f(a)=det (AI-A)=det (A-A) =det([0])=0 Section 7. Multivariate Linear Sstems and Row Operations Eploration. ++z must equal the total number of liters in the miture, namel 60 L z must equal total amount of acid in the miture; since the miture must be 40% acid and have 60 L of solution, the total amount of acid must be 0.40(60)=4 L.. he number of liters of % solution,, must equal twice the number of liters of % solution, z. Hence =z. 4. C A = C ( a )( ) -a -a -a - a +( a )( ) 4 -a - a -a -a 0 - SC - a -a -a -a - a -a -a -a - a z 60 S = C 4 S 0 S, X = C - a -a -a - a z 60 S, B = C 4 S L of % acid, 7. L of % acid, and 8.7 L of % acid are required to make 60 L of a 40% acid solution. Quick Review 7.. (40)(0.)=.8 liters. (0)(-0.4)=8 liters. (, 6) 7. =w-z c - - d Section 7. Eercises. - + z = 0 + z = z = - () () () Use z= in equation (). + (-) = = 7 = 7 Use z=, =7/ in equation (). - a 7 b + (-) = 0 = So the solution is (/, 7/, ) z = 0 - z = z = z = z = z = z = z = z = z = z = - z = - + z = 0 - z = z = = c d - () = ; = - + = 0; = he solution is (,, ) z = = z = z = = = 7 -z= ( -+z= ) -+z=0 -+z= ( +z= ) +z= - (- + z = -) (z = ) (++z= ) ++z=4 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

9 Section 7. Multivariate Linear Sstems and Row Operations z = = - 0 = he sstem has no solution z = 4 + w = -4 - = + z + w = - z = + w = -4 - = + z + w = - z = + w = -4 - = + z + w = 0 -z - w = + w = -4 - = + z + w = 0 -z - w = + w = -4 - = z = 4 4-= 4+=7 +-z=4 (-=) (-=) +z+w= -z= (+w= 4) (+w= 4) +z+w=0 w + (4) = - ; w = - + a - b = -4; = 7-7 = ; = 9 So the solution is a 9., 7, 4, - b C - S C - S - -. R. ( )R +R - = + w = -4 w + z = - z = 4 - (-z - w = ) For #7 0, answers will var depending on the eact sequence of row operations used. One possible sequence of row operations (not necessaril the shortest) is given. he answers shown are not necessaril the ones that might be produced b a grapher or other technolog. In some cases, the are not the ones given in the tet answers C S In # 4, reduced row echelon format is essentiall unique, though the sequence of steps ma var from those shown.. - C 4S C S 6 0 C 4 7S 4 (-)R + R ()R + R (-)R + R (-)R + R ()R + R (-)R + R - C 0-6S C S C 0-4S 0 - (9/)R + R (-/)R (/)R (/0)R (-/)R - C 0 -. S C 0 - S 0 - (/44)R -4 C S (-)R + R - C 0 -. S 0 ()R + R 0 C 0 - S c d - C S C 0-4S ()R + R c 0 - d (-)R + R c d In #9, the variable names (,, etc.) are arbitrar. 9. += 4+=. +z= += -z= Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

10 60 Chapter 7 Sstems and Matrices. - 8 (-)R + R C - -9 S ()R + R - -+z=8 -z= z=4-4= ; = -( )+4=8; = So the solution is (,, 4).. (,, z)=(,, ): 7. No solution: 9. (,, z)=(-z, +z, z) the final matri translates to +z= and -z=. (-)R c 0 + R c 0 d 0 - d (-)R + R (/)R 4. No solution: C 4 S C S C / S (-4)R + R (,, z)=(z+w+, z-w-, z, w) the final matri translates to -z-w= and -z+w=. - 0 C S c - dc d = c - d = +4= 49. (, )=(, ): - c - d = c 4 d c - - d C 4 0 S 4 (-)R + R (-)R + R = 4 c -4 dc- - d -. (,, z)=(,, 7); C S = C - S z C S C 7 - S =. 4 c -8 4 d = c - d - (-)R + R ()R + R (-)R + R (-)R + R C C 9 S = C - S (,, z, w)=(,,, ); = = z w (,, z)=(0, 0, ): Solving up from the bottom gives z=; then -=, so = 0; then +0=0, so =0. -=0 -=0 -=0 -z= E -E -z= -z= +z= 9 +z= 9 E -E z= 7. (,, z, w)=(,,, 0): Solving up from the bottom gives w=0; then z+0=, so z= ; then +4=, so =; then +6-4=, so =. ++z+w= ++z+w= ++z= E -E -z-w= ++z+w= E -E -z-w= E -E +z+w= E 4 -E -z= 4 E 4 -E ++z+w= ++z+w= -z-w= -z-w= z+w= z+w= z+4w= E 4 +E w=0 - ()R + R (/)R - 8 C S C 0 0 S 0 0 C 0 - S 0 S (-)R + R R (-)R + R ()R + R (-)R + R C 0 0 S 0 0 C 0 - S C S Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

11 Section 7. Multivariate Linear Sstems and Row Operations 6 9. (,, z)= : z can be anthing; once z is chosen, we have +z= 8, so = - ; then z, - z - 4, zb z a -, so =- z z - 4b + z = 6 -+z=6 -+z=6 ++z= E -E +z= 8 6. (,, z, w)=( -w, w+, w, w): w can be anthing; once w is chosen, we have z-w=0, so z= w; then -w=, so =w+; then +(w+)+( w)+w=0, so = -w. ++z+4w= E -E -z= E +E z- w=0 ++z+w= E -E -w= -w= ++z+w=0 ++z+w=0 ++z+w=0 6. (,, z, w)=( w-, 0.-z, z, w): z and w can be anthing; once the are chosen, we have -z= 0., so =0.-z; then since +z=0. we have +0.+w=., so = w-. ++z+w=. E -E -z= 0. ++z+w=. ++z+w=. 6. No solution: E +E gives +-z+w=, which contradicts E f()= --: We have f( )=a( ) +b( )+c=a-b+c=, f()=a+b+c=, and f()=4a+b+c=0. Solving this sstem gives (a, b, c)=(,, ). a-b+c= a-b+c= a-b+c= a+b+c= E -E b= 6 b= 6 4a+b+c=0 E -4E 6b-c= E -E c=6 69. f()=( c-) ++c, for an c or f()=a ++( a-), for an a: We have f( )=a-b+c= 4 and f()=a+b+c=. Solving this sstem gives (a, b, c)=( c-,, c)=(a,, a-). Note that when c= (or a=0), this is simpl the line through (, 4) and (, ). a-b+c= 4 a-b+c= 4 a+b+c= E -E b= 7. In this problem, the graphs are representative of the population (in thousands) of the cities of Richardson, X and Garland, X for several ears, where is the number of ears past 980. (a) he following is a scatter plot of the Richardson data with the linear regression equation L superimposed on it. (c) Graphical solution: Graph the two linear equations = ( ) - ( ) and = 9 on the same aes and find the point of intersection. he two curves intersect at L 40. he population of Garland will be 9,000 more than Richardson in the ear 00. [ 0, 0] b [ 0, 00] (b) he following is a scatter plot of the Garland data with the linear regression equation L superimposed on it. [ 0, 0] b [ 0, 00] [ 0, 0] b [ 0, 00] Algebraic solution: Solve ( ) - ( ) = 9000 for. ( ) - ( ) = = = 8.8 = L 40 he population of Garland will be 9,000 more than Richardson in the ear (,, z)=(8, 40, 6), where is the number of children, is the number of adults, and z is the number of senior citizens. ++z=400 ++z= z=74,000 E -7E 0+=,000 --z=0 E +E =60 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

12 6 Chapter 7 Sstems and Matrices 7. (,, z)=(4,00, 00, 60,000) (all amounts in dollars), where is the amount invested in Cs, is the amount in bonds, and z is the amount in the growth fund. ++z=80,000 ++z=80, z=0,84 000E -67E 6+89z=,48,000 +-z=0 E -E 4z= 40, (,, z) (0, 8,98.0,,06.9): If z dollars are invested in the growth fund, then = (,0,000-88z) 9 7,0.898-z dollars must be invested in bonds, and z-,0.898 dollars are invested in Cs. Since Ú 0, we see that z Ú 06.9 (approimatel); the minimum value of z requires that =0 (this is logical, since if we wish to minimize z,we should put the rest of our mone in bonds, since bonds have a better return than Cs). hen 7,0.898-z=8, z=0,000 ++z=0, z=000 0,000E -7E 9+88z=,0, nickels, dimes, and 7 quarters: 74 C 0 88S - 4 (-)R + R (-)R + R 8. (, p)= : c - a 6 p d = c -0 d c 00, 0 0 d = c - 0 dc00 0 d 74 C 0 0 S C 0 0 S Adding one row to another is the same as multipling that first row b and then adding it to the other, so that it falls into the categor of the second tpe of elementar row operations. Also, it corresponds to adding one equation to another in the original sstem. 8. False. For a nonzero square matri to have an inverse, the determinant of the matri must not be equal to zero. 87. () - (-)() = 8. he answer is. 89. wice the first row was added to the second row. he answer is. 9. (a) he planes can intersect at eactl one point. (b) At least two planes are parallel, or else the line of each pair of intersecting planes is parallel to the third plane. (c) wo or more planes can coincide, or else all three planes can intersect along a single line. 9. (a) C()=(-)(-)-( )( ) = -8+. (b) [, 8.4] b [.,.] (c) C()=0 when =4 appro..7 and.7. (d) det A=, and the -intercept is (0, ). his is the case because C(0)=()()-()()=det A. (e) a +a =+=8. he eigenvalues add to (4- )+(4+ )=8, also. (-)R (/)R (/0)R 74 C 0 0 S C 0 0 S (-)R + R (-)R + R (-)R + R ) Section 7.4 Partial Fractions = c d = Eploration c 6 0 d. (a) -=A (-)+A (+) 4=8 A =A (b) -=A ( -)+A ( +) 6= 8 A =A 0 0 C 0 0 S (a) 4+4+4=A (-) + A (-)+ A 4= A =A (b) 0+0+4=A (0-) +0 A (0-)+0 A 4=4 A =A (c) Suppose = 9+6+4= # (-) + # (-)+A =+6+A 6=A =A Quick Review 7.4 ( - ) + ( - ). = ( - )( - ) - = ( - )( - ) ( + ) + ( + ) +. ( + ) = ( + ) = f() d() = # # Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

13 Section 7.4 Partial Fractions 6 7. Possible real rational zeros: ;, ;, ;, ;4, ;6, ; ; From a graph, = and = seem reasonable: = ( + )( - )( + 4) In #9 0, equate coefficients. 9. A=, B=, C= Section 7.4 Eercises - 7. ( - 4) = A + A - + A ( - ) ( + 9) A = + A + A + A A ( - ) + B + C : +=A(-)+B(+4) =(A+B)+( A+4B) - A + B = ca c d = c - B d = c -A + 4B = Q - 4 d 4 d 7. : ( + ) =(A+B)( +)+(C+) =A +B +(A+C)+(B+) A = 0 B = A + C = B + = = A B Q C 0 0 = - 9. : ( - ) + ( - ) =A(-) +B(-)+C =A +( 4A+B)+(4A-B+C) A = 0 0-4A + B = - Q C S 4A - B + C = 4 - Using a grapher, we find that the reduced row echelon form of the augmented matri is: 0 0 A C 0 0 S Q C B S = C S 0 0 C. : + - ( + ) ( + ) =A(+)( +) +B( +) +(C+)(+) ( +)+(E+F)(+) = (A+C) +(A+B+6C+) 4 +(4A+C+6+E) +(A+4B+C ++6E+F) +(4A+8C++9E +6F)+(A+4B+8+9F) A + C = A + B + 6C + = 4A + C E = 6 A + 4B + C + + 6E + F = Q 4A + 8C + + 9E + 6F = 7 A + 4B F = G W Using a grapher, we find that the reduced row echelon form of the augmented matri is: G W ,so ( - )( - ) = A - + A - A B C G W E F =A (-)+A (-). With =, we see that =A, so A =; with = we have = A,so A = : ,so - = A - + A + 4=A (+)+A (-). With =, we see that 4=A, so A =; with = we have 4= A,so A = : ,so=A (+)+A. + = A + A + With =0, we see that =A, so A = ; with Q =, we have = A,soA = - : = - G W - /. - / + = - / + Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

14 64 Chapter 7 Sstems and Matrices ,so = A - + A + 4 =A (+4)+A (-). With =, we see that 7=7A, so A =; with = 4 we have 4= 7A, so A = : ,so = A + + A - =A (-)+A (+). With =, we see that 4= 7A, so A = ; with = we have A, = 7 - so A =: ,so + ( + ) = B + C + B + C + ( + ) =(B +C )( +)+B +C. Epanding the right side leaves +=B +C +(B +B )+C +C ; equating coefficients reveals that B =0, C =, B +B =0, and C +C =. his means that B =0 and C =:. + + ( + ). he denominator factors into (-),so - +. hen - + = A + A - + A ( - ) -+=A (-) +A (-)+A. With =0, we have =A ; with =, we have =A ; with =, we have 4=A +A +A =+A +4, so A = : ( - ) hen - = A + A + A - -4+=A (-)+A (-)+A. With =0, we have = A, so A = ; with =, we have 8=9A, so A =; with =, we have = A -A +A = A ++,so A =: hen ( + ) = B + C + B + C + ( + ) +4-=(B +C )( +)+B +C. Epanding the right side and equating coefficients reveals that B =, C =0, B +B =4, and C +C =. his means than B =0 and C = :. + - ( + ). he denominator factors into (-)( ++),so + + A. hen = B + C =A( ++)+(B+C)(-). With =, we have 6=A, so A=; with =0, =A-C, so C=0. Finall, with =, we have =7A+B, so B= : In # 6, find the quotient and remainder via long division or other methods (note in particular that if the degree of the numerator and denominator are the same, the quotient is the ratio of the leading coefficients). Use the usual methods to find the partial fraction decomposition A =, so +=A (+) - + A + +A (-). With = and = (respectivel), we find that A = and A = : Graph of : - [ 4.7, 4.7] b [ 0, 0] Graph of = : [ 4.7, 4.7] b [ 0, 0] Graph of : - [ 4.7, 4.7] b [ 0, 0] Graph of - : + = ; r() h() = + - [ 4.7, 4.7] b [ 0, 0] -. + = ; r() h() = - + A =, so -=A +A (+). With + + A = and =0 (respectivel), we find that A = and A = :. + - Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

15 Section 7. Sstems of Inequalities in wo Variables 6 Graph of = - : + [ 4.7, 4.7] b [ 0, ] Graph of = - : [ 4.7, 4.7] b [ 0, ] Graph of = : + [ 4.7, 4.7] b [ 0, ] Graph of = - : [ 4.7, 4.7] b [ 0, ] 7. (c) 9. (d) 4. (a) 4. - : a + a( - a) =A(-a)+B=(A+B)-aA A + B = 0 -aa = Since A= -, - a + B = 0, B = a a c A - B d = a a - 4. : (b - a)( - a) + (b - a)( - b) =A(-b)+B(-a)=(A+B)+( ba-ab) A + B = 0 -ba - ab = Q c 0 -b -a d Q c 0 0 b - a d Q 0 C S Q 0 b - a b - a Q c A B d = b - a b - a b - a 47. rue. he behavior of f() near = is the same as the behavior of =, and lim. - : - - = -q A 49. he denominator factor calls for the terms and A in the partial fraction decomposition, and the denominator B factor + C + calls for the term. + he answer is E.. he -intercept is, and because the denominators are both of degree, the epression changes sign at each asmptote. he answer is B.. (a) =: +4+=A(+)+(B+C)(0) 6=A A= (b) =i: +4i+=( +)+(Bi+C)(i-) 4i=(Bi+C)(i-) 4i= B-Bi+Ci-C -B - C = 0 Q B + C = 0 -Bi + Ci = 4i B - C = -4 Q c d Q c d Q c 0 0 d Q c 0-0 d Q cb C d = c - d = i -4i+=A( -)+( Bi+C)( i-) 4i= B+Bi-Ci-C, which is the =same as above. B=, C= b. = has a greater effect on f() at =. ( - ) Section 7. Sstems of Inequalities in wo Variables Quick Review 7.. -intercept: (, 0); -intercept: (0, ) Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

16 66 Chapter 7 Sstems and Matrices. -intercept: (0, 0); -intercept: (0, 0) For # 9, a variet of methods could be used. One is shown c d Q C = c 0 60 d 4 80 c d Q C S Q c d Q c d = c 0 40 d c d Q c 6 0 d Q c 0 0 d Q c d = c d Section 7. Eercises. Graph (c); boundar included. Graph (b); boundar included. Graph (e); boundar included S Q c d Q c d boundar curve = + ecluded. boundar circle + =9 ecluded. boundar curve = ecluded boundar line =4 included 9. Corner at (, ). Left boundar is ecluded, the other is included boundar line +=7 included 0 Corners at about (.4, 0.0) and (.4, 9.90). Boundaries included. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

17 Section 7. Sstems of Inequalities in wo Variables Corners at about (.,.6) Boundaries included.. 90 Corners at (0, 40), (6.7, 6.7), (0, 0), and (40, 0). Boundaries included Corner points: (0, 0) (0, 80), the -intercept of +=80 a 60 the intersection of +=80, 80 b, and -=0 (, ) (0, 0) (0, 80) a 60, 80 b f 0 40 f min = 0 [at (0, 0)]; L 9. f ma L 9. cat a 60, 80 bd Corners at (0, ), (0, 6), (.8, 4.), (4, 0), and (, 0). Boundaries included Ú - + For #9 0, first we must find the equations of the lines then the inequalities. 9. line : line : m = ( - ) = (0-4) = -4 = -, = - + m = (0 - ) = (6-4) = -, (-0) = - line : =0 line 4: = Ú 0 Ú 0 9 ( - 6), = Corner points: (0, 60) -intercept of +=60 (6, 0) intersection of +=60 and 4+6=04 (48, ) intersection of 4+6=04 and +6=60 (60, 0) -intercept of +6=60 (, ) (0, 60) (6, 0) (48, ) (60, 0) f ` 40 ` 6 ` 44 ` 40 f min =6 [at (6, 0)]; f ma =none (region is unbounded) Corner points: (0, ) -intercept of += (, 6) intersection of += and 4+=0 (6, ) intersection of 4+=0 and +=0 (0, 0) -intercept of +=0 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

18 68 Chapter 7 Sstems and Matrices (, ) (0, ) (, 6) (6, ) (0, 0) ` ` ` ` f f min = 4 [at (0, )]; f ma = none (region is unbounded) For #7 40, first set up the equations; then solve. 7. Let =number of tons of ore R =number of tons of ore S C=total cost=0+60, the objective function Ú 4000 At least 4000 lb of mineral A 60+0 Ú 00 At least 00 lb of mineral B Ú 0, Ú 0 he region of feasible points is the intersection of Ú 4000 and 60+0 Ú 00 in the first quadrant. he region has three corner points: (0, 64), (.48, 0.87), and (0, 0). C min =$,96.0 when.48 tons of ore R and 0.87 tons of ore S are processed (, 4) fails to satisf +. he answer is. 47. (a) One possible answer: wo lines are parallel if the have eactl the same slope. Let l be +8=a and l be +8=b. hen l becomes = a and l becomes =. Since M l = 8 + b 8 8 =M l, the lines are parallel. (b) One possible answer: If two lines are parallel, then a line l going through the point (0, 0) will be further awa from the origin than a line l going through the point (0, ). In this case f could be epressed as m+ and f could be epressed as m+0. hus, l is moving further awa from the origin as f increases. (c) One possible answer: he region is bounded and includes all its boundar points =6 9 = 6-4 = Let =number of operations performed b Refiner =number of operations performed b Refiner C=total cost=00+600, the objective function + Ú 00 At least 00 units of grade A +4 Ú 0 At least 0 units of grade B +4 Ú 00 At least 00 units of grade C Ú 0, Ú 0 he region of feasible points is the intersection of + Ú 00, +4 Ú 0, and +4 Ú 00 in the first quadrant. he corners are (0, 00), (40, 60), (0, 0), and (00, 0). C min =$48,000, which can be obtained b using Refiner to perform 40 operations and Refiner to perform 60 operations, or using Refiner to perform 0 operations and Refiner to perform 0 operations, or an other combination of and such that +4=0 with &6-4 B 6-4 Ú - = = B = = B [ 4, 4] b [, ] Ú - B B - 9 B False. he graph is a half-plane. 4. he graph of +4 Ú is Regions I and II plus the boundar. he graph of - 4 is Regions I and IV plus the boundar. And the intersection of the regions is the graph of the sstem. he answer is A. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

19 Chapter 7 Review. (a) c (b) c - 4 (c) c -6 (d) 8 d 0 - d -8 0 d Chapter 7 Review AB= c- d = c BA is not possible d;. AB=[( )()+(4)() ( )( )+(4)()]=[ 7]; BA is not possible. 7. AB= (0)() + ()() + (0)(-) (0)(-) + ()() + (0)() C ()() + (0)() + (0)(-) ()(-) + (0)() + (0)() (0)() + (0)() + ()(-) (0)(-) + (0)() + ()() (0)(4) + ()(-) + (0)(-) ()(4) + (0)(-) + (0)(-) S (0)(4) + (0)(-) + ()(-) -7 c 4-6 d ()(0) + (-)() + (4)(0) ()() + (-)(0) + (4)(0) ()(0) + (-)(0) + (4)() BA= C ()(0) + ()() + (-)(0) ()() + ()(0) + (-)(0) ()(0) + ()(0) + (-)() S (-)(0) + ()() + (-)(0) (-)() + ()(0) + (-)(0)(-)(0) + ()(0) + (-)() - = C - 4S = C - S Carr out the multiplication of AB and BA and confirm that both products equal I 4.. Using a calculator: = det= =( )( ) ()( ) = (-8)+(4-( 6)) =0+0 =0-4 For # 8, one possible sequence of row operations is shown. 0 0 (-)R ()R. + R + R C S C 0 - S (-)R + R C - S 4 6 (-)R + R (-)R + R C S 0 0 ()R + R ()R + R 0 C 0 - S C 0-0 S 0 0 (-)R ()R + R C S 0 0 For #9, use an of the methods of this chapter. Solving for (or ) and substituting is probabl easiest for these sstems. 9. (, )=(, ): From E, =-; substituting in E gives +(-)=. hen 7=7, so =. Finall, =.. No solution: From E, =-; substituting in E gives 4-4= (-), or 4-4=4- which is impossible.. (,, z, w)=(-z-w, w+, z, w): Note that the last equation in the triangular sstem is not useful. z and w can be anthing, then =w+ and =-z-w. +z+w= +z+w= +z+w= ++z= E -E -w= -w= ++z+w=8 E -E -w= E -E 0=0. No solution: E and E are inconsistent. +-z= +-z= -+z=4 -+z=4 -+4z=6 E +E 0=0 7. (,, z, w)=(-z+w, +z-w, z, w): Note that the last two equations in the triangular sstem give no additional information. z and w can be anthing, then =+z-w and =-6(+z-w)+4z-w=-z+w. -6+4z-w= -6+4z-w= -6+4z-w= ++z-w=4 E +E +z-w= - E -z+w= ++z=6 E +E 0+0z-0w= 0 - E -z+w= 0 -+z-w= 7 E 4 -E -z+w=6 E 4 -z+w= Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

20 70 Chapter 7 Sstems and Matrices 9. (,, z)= a 9 : 4, - 4, b C z. here is no inverse, since the coefficient matri, shown on the right, has determinant 0 (found with a calculator). Note that this does not necessaril mean there is no solution there ma be infinitel man solutions. However, b other means one can determine that there is no solution in this case.. (,, z, w)=(-w, z+, z, w) z and w can be anthing: c d S = C - - S - (-)R + R - C S = - 7 C c d - SC S. ()R + R (-)R c d. (,, z, w)=(,,, ): (-)R + R R (-)R 4 + R (-)R (-)R + R (-4)R + R (-)R + R 4 (-)R + R ()R 4 + R ()R 4 + R (-)R (, p) (7.7, 4.7): Solve 00- =0+ to give 7.7 (the other solution, 0.7, makes no sense in this problem). hen p= (,) (0.4,.9)] [, ] b [, ] 4. (, )=(, ) or (, )=(, ) [, ] b [, ] 4. (, ) (.7,.) [, ] b [, ] 4. (a, b, c, d)= a 7 840, - 80, , 86 b =(0.00 Á, 0.7 Á,.9 Á,.08 ). In matri form, the sstem is as shown below. Use a calculator to find the inverse matri and multipl. 8 4 a b = c d 4-47., so = A + + A - 4 =A (-4)+A (+). With =, we see that = A, so A =; with =4 we have 0=A, so A =: he denominator factors into (+)(+),so = A + + A + + A ( + ) hen +=A (+) +A (+)(+) +A (+). With =, we have =A ; with =, we have =A ; with =0, we have =A +A +A = +A +4, so that A =: ( + ).. he denominator factors into (+)( +),so - -. hen = A + + B + C + --=A( +)+(B+C)(+). With =, we have 4=A, so A=; with =0, =A+C, so C= 4. Finall, with = we have =A+(B+C)()=4+B-8, so that B=: Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

21 Chapter 7 Review 7. (c). (b) Corner points: (0, 90), (90, 0), a 60. Boundaries, 60 b included Corner points: (4, 40), (0, ), and (70, 0). (, ) (4, 40) (0, ) (70, 0) ` ` ` f f min =0 [at (0, )]; f ma =9 [at (4, 40)] Corner points: appro. (0.9,.) and (.4,.80). Boundaries ecluded cos 4 -sin (a) [ ] c sin 4 cos 4 d L c. 0.7 d cos 4 sin 4 (b) [ ] c -sin 4 cos 4 d L c d 7. In this problem, the graphs are representative of the population (in thousands) of the states of Hawaii and Idaho for several ears, where is the number of ears past 980. (a) he following is a scatter plot of the Hawaii data with the linear regression equation = superimposed on it. 6. Corner points: appro. (.,.6) and (.,.6). Boundaries included. [, 0] b [80, 60] (b) he following is a scatter plot of the Idaho data with the linear regression equation = superimposed on it. 6. Corner points: (0, 0), (, 0), and (0, 6). (, ) (0, 0) (0, 6) (, 0) ` ` ` f f min =06 [at (0, 6)]; f ma =none (unbounded). [, 0] b [80, 60] (c) Graphical solution: Graph the two linear equations = and = on the same ais and find their point of intersection. he two curves intersect at L 9.6. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

22 7 Chapter 7 Sstems and Matrices he population of the two states will be the same sometime in the ear 989 and 7 months. [, 0] b [80, 60] Another graphical solution would be to find where the graph of the difference of the two curves is equal to 0. Algebraic solution: Solve = for = = 94.9 = L 9.6 he population of the two states will be the same sometime in the ear 989 and 7 months. 7. (a) N= [ ] (b) P= [$80 $0 $00 $00] $80 (c) NP =[ $0 0] = $9,000 $00 $00 7. Let be the number of vans, 8++zz be the number of small trucks, and z be the number +0+0z 8 of large trucks needed. he + 6+ z requirements of the problem are summarized above (along with the requirements that each of,, and z must be a non-negative integer). he methods of this chapter do not allow complete solution of this problem. Solving this sstem of inequalities as if it were a sstem of equations gives (,, z) =(.77,.0,.4), which suggests the answer (,, z)=(, 4, ); one can easil check that (,, z) =(, 4, ) actuall works, as does (,, ). he first of these solutions requires eight vehicles, while the second requires onl seven. here are a number of other sevenvehicle answers (these can be found b trial and error): Use no vans, anwhere from zero to five small trucks, and the rest should be large trucks that is, (,, z) should be one of (0, 0, 7), (0,, 6), (0,, ), (0,, 4), (0, 4, ), or (0,, ). 77. (,, z)=(60000, 70000, 0000), where is the amount borrowed at 4%, is the amount borrowed at 6.%, and z is the amount borrowed at 9%. Solve the sstem below z=60, z= 46,0 0- z= 0 One method to solve the sstem is to solve using Gaussian elimination: Multipl equation b 0.06 and add the result to equation, replacing equation : ++ z=60, z= z=0 ivide equation b 0.0 to simplif: ++z=60,000 +z=60,000 -z=0 Now add equation to equation, replacing equation : ++z=60,000 +z=60,000 =60,000 Substitute =60,000 into equation to solve for z: z=0,000. Substitute these values into equation to solve for : =70, Pipe A: hours. Pipe B:.4 hours (about hours 7. minutes). Pipe C: hours. If is the portion of the pool that A can fill in one hour, is the portion that B fills in one hour, and z is the portion that C fills in one hour, then solving the sstem above gives (,, z) = a, 60, b One method to solve the sstem is to use elimination: Subtract equation from equation : + + z = / z = / + z = 4/ (convert /.7 to simpler form) Subtract equation from equation : + + z = / z = / = /60 ++z=/ + =/4 +z=/.7 Substitute the values for and z into equation to solve for : =/. 8. n=p the number of columns in A is the same as the number of rows in B. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

23 Chapter 7 Project 7 Chapter 7 Project. he graphs are representative of the male and female population in the United States from 990 to 006, where is the number of ears after [ 0, 0] b [0, 0] [, ] b [0, 60] he linear regression equation for the male population is L he linear regression equation for the female population is L he slope is the rate of change of people (in millions) per ear. he -intercept is the number of people (either males or females) in Yes, the male population is predicted to eventuall surpass the female population, because the males regression line has a greater slope. Since 990, the female population has alwas been greater. Since a span of onl 6 ears is represented, the data are most likel not enough to create a model for 00 or more ears Males: L e Females: L e he curves intersect at approimatel (4, 64); this represents the time when the female population became greater than the male population. he curves intersect at approimatel (9, ); this represents the time when the male population will again become greater the female population Approimatel L 0.49 = 49.% male and 0.9% 8. female. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

24 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.