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# College Algebra MATH 143

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This 75 page Class Notes was uploaded by Breanne Schaden PhD on Saturday October 3, 2015. The Class Notes belongs to MATH 143 at Boise State University taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/218010/math-143-boise-state-university in Mathematics (M) at Boise State University.

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SECTION 93 Systems of Linear Equations in Several Variables 51 1 56 Let J be the length oftirne John drives and y be the length oftime Mary drives Then y z a 025 so in y 0525 and multiplying by 40 we get 4105 40y 10 Comparing the distances we get 60139 40y 35 or 6035 40y 35 4100 403 10 7 This gives the system Adding we get 2033 45 ltgt 1 225 so y 225 025 25 f 60139 a 401 35 Thus John drives for 2 hours and Mary drives for hours 11 2 7 57 Let m be the tens digit and y be the ones digit of the number Adding 9 times the rst 10yurz2710y equation to the second gives 1827 36 gt an 2 so 2 y 7 4 y 5 Thus the number is 25 58 First let us nd the intersection point of the two lines The ycoordinate of the intersection point is the height of the triangle 39 20 I 4 We have y Adding 2 times the rst equation to the second gives 3y 12 so the triangle has height 4 y 415 20 Furthermore y 2x I 4 intersects the caxis at J 2 and y z 44 3920 intersects the zaxis at I z 5 Thus the base has length 5 2 3 Therefore the area of the triangle is A bh 3 4 6 59 715sozzzlm 112357218 ZLiyk 35669 29 21 magi 13 2 5 3 6 5 6 7 9 124 and 22 z 12 22 32 52 72 z 88 Thus we get the system 18a 55 29 880 18b 124 Subtracting 18 times the rst equation from 5 times the second we get116a98 gt om0845 Then 0 1 a 3 4 5 e 7 39 b i 18 0845 29 m 2758 So the regression line is y 084595 2758 5 Systems of Linear Equations in Several Variables 1 The equation 61 lz 0 is linear 2 The equation m2 y2 22 4 is not linear since it contains squares ofvariables vy 31 2 i 3 The system 7 g y2 52 is not a linear system since the rst equation contains a product of variables In fact 2m yz 3 both the second and the third equation are not linear x 2y3210 4 The system 217 5y 392 is linear y 2z 2 Al 6 7 2y 4z 3 5 y 2 7 Substituting 2 z 2 into the second equation gives y 2 2 7 ltgt y z 3 Substituting 2 2 Z 392 and y 3 into the rst equation gives 22 r 2 3 4 2 3 4quot a 2 1 Thus the solution is 1 3 2 CHAPTER 9 Systems of Equations and Inequalities 512 at y 7 32 8 6 y 7 32 5 Substituting 2 1 into the second equation gives y 7 3 1 5 ltgt y 2 Substituting 2 1 z 71 and y 2 into the rst equation gives m 2 7 3 71 2 8 ltgt I 3 Thus the solution is 3 2 71 m 2y z 7 7 4y 33 9 Solving we get 25 6 42gt z 3 Substituting 2 3 into the second equation gives 2g 6 7y 3 3 9 42gt y Or Substituting 2 3 and y 0 into the rst equation gives 27 2 0 3 7 ltgt z 4 Thus the solution is 4 0 z 2g 32 10 3 2y 7 z 2 Solving we get 32 12 lt2 2 4 Substituting 2 4 into the second equation gives 2 2 12 2y 7 4 2 gt y 3 Substituting 2 4 and y 3 into the rst equation gives as 7 2 3 4 10 lt a 4 Thus the solution is 4 37 4 27 y 62 2 5 9 34 42 0 Solving we get 22 1 lt2 2 Substituting 2 7 into the second equation gives 722 y 4 0 tgt y 2 Substituting 2 7 and y 2 into the rst equation gives 2 7 2 6 5 ltgt 1 5 Thus the solution is 5 27r Solving we get z 4 42gt z 8 Substituting 1 8 into the second equation gives ltgt y 1 Substituting 2 8 into the rst equation gives 4m 3 8 10 lt Thus the solution is 71 8 723172 72317 2 4 11 x y 3H Subtract the rst equation from the second equation 3 4z 4 2x y 20 2r y z 0 w 2y7 2 4 Or subtract times the third equation from the second equation fig i 9 g K21 y z 0 35 y73z3 1 y73z3 12 72139 3y 2 Add 2 times the rst equation to the second equation 5y 4 5 m y2z0 7 y2z2 m y 7 3z 3 Or add 2 times the third equation to the second equation y 5z 2 L y220 SECTION 93 Systems of Linear Equations in Several Variables 513 2139 y3z 2 21397 y32 2 13 z 2y 7 z 4 Add 2 times the rst equation to the third equation x 2y 7 z 4 74325y 210 3y7z14 21 y 3N 2 Or add 4 times the second equation to the third equation 1 2y z 4 13y 7 3 26 m 7 4y z 3 14 y 7 32 10 Add times the rst equation to the third equation 3y 7 82 24 I74y z 3 OI subtract 3 times the second equation from the third equation y 7 32 10 z 6 z 11 2 4 1 9 24 1 y 24 15 t3y3z10 gt 2y22 71XEq 1Eq2 42gt y3z5 1343 23 117 z 3 y3z 2XEqllXEq3 2y26 Eq y 4 gt y32 5 gt z1andy318ltgty2tThen12 igtl4 ltgt 74274 72 X Eq 2 Eq 3 139 1 So the solution is 1 2 1 1 y 10 xyz0 my z 0 7w2y523 4i y2z21 xEq1 xEq2 gt y22 1 3w y 6 741734 73gtltEqlEq3 5110 4xEq2Eq3 SozZandyt221ltgty73Thenm 7320 lt gt 11Sothesoiutionis1732i 11 4z1 3 4z1 z 7 4z 1 2757 y 6z4 ltgt 7 y2z2 72ltEqlEq2 lt2quot 7y 22 2 233 3y 7228 3y6 2 XEq1Eq3 1212 3XEqZEq3 Soz1and7y212ltgty0Thenz7411 ltgt x5Sothesolutionis501 m erZZ 2 z7y2z 2 w erZZ 2 3xy5 8 ltgt y76z 1l 3gtltEqi5q3 ltgt y762z 11 2x y 212 7 4y 2 2 2gtltEqlt Eq2 232 46 4thq2EQ3 SozQandy762711 ltgty1 Thenml22 lt3 71Sothesolutionis 112 2m4y39 z 2 962y73274 w2y7 32 4 12y 374 ltgt 24y z 2 4 77y102 13 Soz2and Sac y z 1 317 14 2 1 5 10 77y 10 2 13 42gt y 1 Thenx2 1 7 3 2 2 74 ltgt z 0 Sothe solution is 0 12 514 CHAPTER 9 Systems of Equations and Inequalities 2xy7 278 my z 3 727 yz 3 7xy z 20 7xy z 3 42gt 2zy7 z78 ltgt y 276 117 z 723 42 18 7226 43 18 3yz72 3942 So 2 4 and y 4 76 lt y 72 Then 7z 72 4 3 gt x 71 So the solution is 71724i 3722 0 7172y z71 772y 271 7x72y 271 21 23y 2 ltgt y722 O ltgt y Qz 0 ltgt 1722 7x72y 271 213y 2 7y22 0 0 O The system is dependent so when 2 t we solve for y to get y 7 275 lt2 y 2t Then 7m 7 2 216 t 71 gt 7x 7 3t 71 ltgt cc 3t 1 So the solutions are 73t 12tt where t is any real number 214 z 3 x By 72 m7 3g 72 731 22 5m4y32 1 2 2y 2 3 gt 2y z 3 lt2 2y 1737 72 5x4y3271 19y32 9 713 Soz3and2y33 yOTlietim73D 2 ltgt 3372Sothesolutionis7203 m2y7 z 1 z2y7 z 1 23 213y74z 3 lt2 y22 5 Since01isfalse this system is inconsistent 3r6y73z 4 0 1 x2y 554 z 7 220 1 7230 7 7220 24 7 7 220 gt 7m2y 5 ltgt 2y3z4 lt3 2y334 4m 2y7llz2 4172y7114 72y73 06 Since 0 6 is false this system is inconsisteuL 2x3y7z1 12y 3 x2y 3 z2y 3 25 952y 3 lt2quot 23y7z1 lt2 y7z75 4 7y7z75 3yz x3y 7z4 yz 1 74 Since 0 74 is false this system is inconsistent x72y735 r 2y73z 5 72y732 5 26 273 y7 15 lt9 5y52 75 ltLgt 5y5z 753 417374 7225 5y5z715 0710 Since 0 710 is false this system is inconsistent rc y7 50 26 747 30 m y7 0 27 12y 3z73 ltgt y72273 lt25 y 273 23y74z73 y7 273 00 Soztandy72t73 ltgt y2t73iThenx2t737t0 gt x t3Sothesolutionsare 7t 3 2t 7 8 t where t is any real number m72y 23 172y z 3 x72y 2 3 28 2175314763 gt 7y4z 1 gt 7y4z 1 2x73y7 z y 271 0 0 Then we have 7y 475 1 5 y 4t 1 Substituting into the rst equation we have 2 2 4t 7 1 1 3 gt 5139 77 1 So the solutions are 7t 1 4t 115 where t is any real numbeit 536 CHAPTER 9 Systems of Equations and lnequalities 12031000 12 0 3 1 000 20310 0 21 01110100 R3424 01 1 1 0 100 11101 0 01010010 R4414 0071 0 0 110 44 0100171 12020001 00 07171 001 000110 0 1 10 211 2 0 0 1001 1072 0 RIJRPRI 0 1 0 1 0 0 1 0 RIHRPRI 0 1 0 0 71 0 1 1 whim Rerkarh 00 100 171 0 249 0010 01 1 0 00 011 0 0 1 0001 10 0 1 1000 00 2 1 00s21 0100710 1 1 1011 Therefore the Inverse matrix is 0010 011 0 01a1 0 0001 10 071 10 071 10101000 010 1000 1010 1 000 22 01010100 RFRHM 101 0100 R4412 010 1 0 100 1 0 11100010 RS RIARS 0100 1010 RaRWRS 00071 1 110 11110001 010171001 000 0717101 Therefore there is no inverse matrix 526 311 4 1 5 3 z 4 I 23 IS equivalent to the matrix equation Usmg 1J1e Inverse from Exercrse 3 Sz 21 0 3 2 0 z 2 73 4 8 Therefore an 8 and y 12 y 73 5 0 12 3x 41 10 3 4 r 10 24 J is equivalent to the matrix equation ltgt 7x 9y 20 7 9 y 20 m 79 4 10 10 2 Therefore7 an 710 and y 2 10 y 7 A3 20 10 2x quot 2 2 5 0 2 25 T W is equivalent to the matrix equa don Using the inverse from 75m 7 137 20 75 13 y 20 m 13 5 2 126 Exeruse 5 Therefore I 126 and y 750 y 5 2 2O 50 en 41 0 quot 4 0 26 J is equivalent 10 the matrix equation I I lt3 8 7 50 100 8 5 y 100 5 4 4 0 36 ii 5 0 T 400 700 Therefore m fiend y i y 7 7 we 3 3 SECTION 96 lnverses of Matrices and Matrix Equations 537 2x 4y z 7 2 4 1 x 27 y 7 5 0 is equivalent to the matrix equation 71 1 1 y O Usingthe inverse from at 4y 2 72 1 4 0 z 72 m 4 4 5 7 38 Exercise 115 y 1 1 71 0 g Therefore 2 3938 y 9 and z 47 z 5 4 6 2 47 5x7y4z1 5 74 139 1 23 30 y 32 2 1 is equivalent to the matrix equation 3 71 3 y 1 Q 635 7 5g 6 7 5 z 1 1 26 7 25 1 8 y 3 71 3 1 1 Therefore x 8 y 71 and z 78 z 727 77 26 1 8 2y22 12 0 2 2 x 12 29 3g y 32 72 is equivalent to the matrix equation 3 1 3 y 72 Using the inverse from a 7 2y 32 8 1 2 3 z 8 z 7 71 4 12 720 Exercise 15 y 3 1 3 2 10 Therefore cc 2 720 y 10 and z 16 z 539 1 3 8 16 z 2y 3w 2 0 1 2 0 3 z 0 1 U 1 1 1 1 30 y Z w is equivalent to the matrix equation y cgt y w z 2 0 1 0 1 z 2 x 2y 271 3 1 2 0 2 w 3 r 0 0 72 1 0 71 y 71 0 1 1 1 5 Therefore er 71 y 71 and w 31 e 2 U 1 1 O 2 71 w 1 0 0 71 3 73 31 Using a calculator we get the result 3 2 1 32 Using a calculator we get the result 71 21 3 33 Using a calculator we get the result 3 72 2 34 Using a calculator we get the result 6 12 24 35 Using a calculator we get the result 8 1 0 3 36 Using a calculator we get the result 8 4 2 1 37 This has the form MIX 2 C so A171 MX VI 10 and llI391llX llI IM X X 71 3 72 1 3 2 3 2 1 Now IlI A Since X Ill 0 we get 74 3 9 e 8 4 3 4 3 at y z 7 3 2 1 O 1 7 2 3 u 11 w 4 3 2 1 3 10 3 5 538 CHAPTER 9 Systems of Equations and Inequalities 7 1 4 3 6 7 3 9 39 38 Usingthe inverse matrixfromExercise 19We seethat 3 1 23 6 12 15 30 g 1 3 0 0 33 1 M 73279 739 Hence y 0 15 30 z w 33 71 390711 1i 00 aui 11 a 1 12 742 0 a 2 12 a 0 2a 71 1 00001000 ARI 10001000 0 40 00000100 12 010001000 3900c00010 Rs 00100 01c0 8R4 O O 0 d 0 0 0 1 D 0 0 1 0 U 0 111 a O 0 O 10 0 0 0 0 b 0 0 0 1b 0 0 Thus the mamx has mverse 0 0 c 0 0 0 1 0 0 0 0 d 0 0 0 ld 71 2 1 1 62 7x 1 x2 var 1 1139 41 2 7 The inverse does not exxst when ac x2 2m T ix 2 I 39x 2 7195 Zz2 z z 0 71 42 ex 52r 1 Par 62 v 1 e e l t f 11 62 31 e4re i 76h em 5 fin e 3 1T1emverse emsts oxa z it z z1 171 z x 75 m x I The inverse exists for E 3 0 1 A1 45 cos 6 sin 6 1 cos 0 7 sin 9 cos 9 v sin 9 sin 9 cos 0 C052 9 Sing 9 sin 9 cos 0 sin 19 cos 6 SECTiON 9 6 inverses of Matrices and Matrix Equations 539 1 51306 tam k 1 sec tan9 I 5669 tana tam sec Secgg tanag itanO sec itan0 seeO 313100 424 010 1117110 47a424010 313100 313100 324001 011 101 011 101 1 1171 10 R1421 101 1 0 123111 2 040430 1 0104 go LEE 0 1 1 1 0 1 0 0 1 1 3 1 1 O 0 0 1 1 0 1 1 0 1 O 2 g 0 iThereforeJhe inverse ofthe matrix is 2 0 0 0 1 1 gt3 1 1 e 1 A 0 1 1 10 1 b B 72 g 0 14 1 C39 1 7 1 13 2 Therefore he should feed the rats 1 oz of food A 1 02 offood B and 2 02 of food C A 0 1 1 9 2 c B 2 g 0 12 0 0 1 10 1 Therefore he should feed the rats 2 oz of food A no food B and 1 oz of food C A 0 1 1 2 7 d B 72 g 0 4 7 2 C 73 1 11 7 Since A lt 0 there is no combination of foods giving the required supply 314100 314100 112110 48426010 112410 3 314100 W 3 1 R3 325001 0117101 0114101 2 71 1 0 1 1 2 1 1 0 272 4 30 02 24730 0 1 1 1 0 1 0 0 0 1e 1 Since the inverse matrix does not exist it would not be possible to use matrix inversion in the solutions of parts 13 c and d 540 CHAPTER 9 Systems of Equations and Inequalities J y2z675 1 1 2 5 675 49 a 22 34 2600 b 2 1 1 y I 600 w2y 2625 1 2 1 z 625 112100 112100 10 17110 c211010 07173 210 0 1 37210 121001 01 17101 00 4 311 10 1 1 10 100 37 R2 R R 8R 4 75139s 0 1 3 2 l 0 R217 33 0 1 0 3 Therefore the inverse ofthe 001ei i Delgeivi g 1 4 7g 675 125 matrixis 7 i g and y 71 7 g 600 150 Thus he eams 125 nna g z 3 7 J 625 200 standard set 150 on a deluxe set and 200 on a leatherbound set 0 1 0 2 50 No Consider the following counterexample A 0 0 and B 2 O Then AB 0 but neither A 0 nor 0 9 0 1 B 0 There are in nitely many matrices for which A 0 One example is A Then A2 0 but A 7 O 0 0 Determinants and Cramer s Rule 2 0 O 71 1 The matrix 0 3 has determinant 2 The matrix has determinant 2 0 ml 2 3 e 0 0 Z 5 lDl 0 0 i 1 2 2 Al 5 I 2 1 3 The matrix 0 1 has determinant 4 The matrix 3 2 has determinant lDl 4 1 i 5 0 4 IDi 2 2 e 1 3 Z 1 3 5 The matrix I 2 5 does not have a determinant because 6 The matrix 0 does not have a determinant because the the matrix is not square I matrix is not square I 1 The matrix 2 1 NIH nip gt 22 7 4 has determinant 8 The matrix has determinant 05 pl 2 5 1E 37 lDl221107015 14 2207 29 Chapter 9 Systems of Equations Operations for Transformation of a Linear System of Equations to Obtain a Solution AKA Row Operations R a R 1 Interchange any two equations kRi c R 2 Multiply or divide any equation of the system by a nonzero real number kRi R c R 3 Replace any equation of the system by the sum of that equation and a multiple of another equation in the system Solving Systems with the Substitution Method Substitution Method Use one equation to find an expression for one variable in terms of the other Substitute that expression into the other equation of the system Solve for the variable Substitute that value into the expression and solve for the next variable Example 4x y 7 l 3x 2y 30 ll Solve for y interms ofx using equation I y 4x 7 Sustitute 4x 7 for y in equation ll solve for x 3x24x 7 30 3x 8x 14 30 11x 14 30 11x 44 x 4 Substitute 4 for x in the equation y 4x 7 to find that y 9 So the solution is x 4 and y 9 Solving Systems with the Elimination Method Elimination Method Use multiplication and addition to eliminate a variable from an equation by getting the coefficients of a variable to be additive inverses Combine the two equations to eliminate the variable Solve for the variable Substitute that value into the equation and solve for the next variable Example x 2y z 6 l x y z 6 ll 2x y z 7 III Add equation I to equation II to eliminate 2 Add equation II to equation III to eliminate 2 2x 3y 12 IV 3x 2y 13 V Multiply equation IV by 3 Multiply equation V by 2 6x 9y 36 lVb 6x 4y 26 Vb Add the two equations to eliminate x 5y 10 y 2 Substitute 2 for y in the equation 3x2y 13 to find that x 3 Substitute 2 for y and 3 for x in the equation x y z 6 to find that z1 So the solution is x 3 and y 2 and 21 or 321 Solving Systems with the Graphing Method Graphing Method Solve each equation for y in terms of x Put each equation in the graphing calculator Graph the equations with the calculator or by hand Solve for the intersection points Example 4x y 7 l 3x 2y 30 ll Solve for y interms ofx uto get the equations ready for input to the calculator y 4x 7 l 3 ix15 ll y 2 Solving Systems with the Gaussian Elimination Method Gaussian Elimination Method Use the Row Operations to transform the augmented matrix a row echelon form Example x2y z6 l x yz6 ll R1R2R2 2x y z7 lll R2R3R3 2x 3y12 ll 3x 2y13 lll 3R22R3R3 1 5y 10 lll g R3R3 y2 lll Substitute 2 in equation II for y and solve for x x3 Substitute 3 in equation I forx and 2 in equation I for y then solve for z z 1 or use the calculator ref button on the augmented matrix Matrices The dimension of a matrix is the number of rows by the number of columns nx m Matrix Operations Matrix Addition Subtraction a11 a1quot b11 b1n a11ib11 a1nib1n 3 V 3 i E E E I bm1 V V V bmn am1 i b1m amn i bmn Dimensions must be the same to add or subtract a m1 a mu Example 1 2 3 2 4 6 12 23 36 3 5 9 4 5 6 8 10 12 48 510 612 12 15 18 7 8 9 14 16 18 714 816 918 21 24 27 Multiplication by a Scalar constant a11 a12 a13 ka11 k312 k313 k a21 a22 a23 kaz1 kazz kaza a31 332 333 kaa1 kaaz kaaa Example 1 2 3 51 52 53 5 10 15 5 4 5 6 54 55 56 20 25 30 7 8 9 57 58 59 35 40 45 Matrix Multiplication a11 a1quot a11 a1p a11 quot a1p am1 amn an1 anp am1 amp In order to multiply matrices the number of columns of the first matrix MUST the number of rows of the second and the resulting matrix will have the number of rows of the first matrix and the number of columns of the second A mx n matrix multiplied by a ngtlt p matrix yields a mx p matrix Further matrix multiplication IS NOT COMMUNTATIVE AB a BA Multiplication of matrices is accomplished by multiplying the row of the first matrix times the column of the second Example 1 2 3 2 4 6 1228314 14210316 16212318 60 72 84 4 5 6 8 10 12 4258614 44510616 46512618 132 162 192 7 8 9 14 16 18 7288914 74810916 76812918 204 252 300 Examples of specific types of matrices n xm matrix diagonal matrix a11 0 0 0 a11 a1quot 0 a o o 22 am amquot 0 0 333 0 0 0 0 a44 upper triangular matrix lower triangular matrix a11 a12 a13 a14 a11 o o o 0 a22 a23 a24 a21 a22 o o o o 333 a34 a31 a32 333 o o o 0 a44 a41 a42 343 a44 identity matrix augmented matrix 1 0 0 0 a11 a12 an b1 0 1 o 0 a21 a22 a23 b2 0 o 1 0 a31 a32 333 b3 0 o o 1 a41 a42 343 b4 matrix in row echelon form matrix in reduced row echelon form 1 a12 a13 a14 b1 1 0 0 0 b1 0 1 an a24 b2 0 1 0 0 b2 0 0 1 a34 b3 0 0 1 0 b3 0 0 0 1 b4 0 0 0 1 b4 Solving Systems with the Gaussian Elimination by calculator For any system of linear equations a Enter the augmented matrix into the calulator b use the function ref to transform it into row echelon form ref c read the solution from the resulting matrix Gaussian Eliminaton Method Example x 2y z 6 l x y z 6 ll 2x y z 7 III The augmented matrix representing this system 1 2 1 6 1 1 1 6 2 1 1 7 Enter the augmented matrix into the calculator a Click for the matrix menu b Highlight to create a new matrix or to change an existing matrix c Highlight the name for the matrix A E and click d Enter the dimensions rows by columns and click e Enter each of the matrix elements click after each Note If you make a mistake entering an element you can use the arrow keys to make corrections You can also add or delete a row or column f When you are finished entering elements click to leave the matrix editor To view the matrix a Click MATRIX b Highlight the name of the matrix you wish to view c Click ENTER twice d you should see your matrix on the viewing screen Transform the matrix into row echelon form a Click click b Select click c To select the matrix you wish to transform by clicking d Highlight the name of the matrix you wish to transform click e A matrix in row echelon form should now be on the viewing screen 1 1 1 6 0 1 1 6 you can read off the solution 2 3 0 0 1 7 f The values for x and y can be found with two steps of backsubstitution To save or store this matrix a Click b Click MATRIX c Highlight the name of the matrix you wish to call this matrix d Click ENTER twice Solving Systems with the GaussJordan Elimination by calculator For any system of linear equations a Enter the augmented matrix into the calulator b use the function rref to transform it into reduced row echelon form rref c read the solution from the resulting matrix GaussJordan Method Example x2y z6 l x yz6 ll 2x y z7 Ill The augmented matrix representing this system 1 2 1 6 1 1 1 6 2 1 1 7 Enter the augmented matrix into the calculator a Click for the matrix menu b Highlight to create a new matrix or to change an existing matrix c Highlight the name for the matrix A E and click d Enter the dimensions rows by columns and click e Enter each of the matrix elements click after each Note If you make a mistake entering an element you can use the arrow keys to make corrections You can also add or delete a row or column f When you are finished entering elements click QUIT to leave the matrix editor To view the matrix a Click MATRIX b Highlight the name of the matrix you wish to view c Click ENTER twice d you should see your matrix on the viewing screen Transform the matrix into reduced row echelon form a Click click b Select click c Select the matrix you wish to transform by clicking d Highlight the name of the matrix you wish to transform click e A matrix in reduced row echelon form should now be on the viewing screen 1 0 0 6 0 1 0 6 you can read off the solution x 1 y 2 and z 3 0 0 1 7 To save or store this matrix a Click b Click MATRIX c Highlight the name of the matrix you wish to call this matrix d Click ENTER twice Inverse Matrices If A is a SQUARE matrix then A has an inverse if and only if detA at 0 Inverse of a 2 x2 Matrix a b 1 d b lfA then A 1 7 c d ad bc c a Find the Inverse of a nx n Matrix with your calculator a Click b Highlight matrix A click c Click d Click e The inverse matrix ofA A4 is now in your viewing screen Inverse of a nx n Matrix by hand Augment the matrix with its corresponding identity matrix all 312 313 1 0 0 all 312 313 1 0 0 A a21 a22 a23 I 0 1 0 Augmented matrix a21 a22 a23 0 1 0 331 332 333 0 0 1 331 332 ass 0 0 1 Use Row Operations to change the left side of the augmented matrix to the identity matrix row echelon form The inverse matrix of A is now the right half of the augmented matrix a11 a12 a13 1 0 0 1 0 0 b11 b12 b13 Augmented matrix a21 a22 a23 0 1 0 2 Row Operations 0 1 0 b21 b22 b23 a a a 0 0 1 0 0 1 b31 b321 b33 31 32 33 b Aquot1 b21 b22 b23 or use the calculator use the button bill Solving Systems with the Inverse Matrix Method with your calculator Matrix Equations Form a matrix equation that represents the system of linear equations Find the matrix inverse and use it to solve the system AX B 2 A391AX A391B IX A391B X A391B Example x 2 y z 6 l x y z 6 ll 2x y z 7 III Form the matrix equation representing the system 2 1 x 6 1 X y B 6 Enter A and B into the calculator z 7 1 A1 2 1 1 1 Find the inverse matrix of A a Click b Highlight matrix A click c Click d Click e The inverse matrix ofA A4 is now in your viewing screen To save or store this matrix a Click b Click MATRIX c Highlight the name of the matrix you wish to call this matrix say E d Click ENTER twice Find A391B a Click b Highlight E click c Click E d Click e Highlight B click twice f The matrix product A B should be in your viewing screen g You should be able to read off the solution set 15 15 35 25 35 13 A 15 J 1 35 25 5 15 15 35 25 15 x3 y2 z1 321 X YZ Solving Systems with the Inverse Matrix Method by hand Matrix Equations Form a matrix equation that represents the system of linear equations Find the matrix inverse and use it to solve the system AX B gt AquotAX A391B IX A391B X A391B Example x 2y 2 6 X y z 6 2x y z 7 III Form the matrix equation representing the system x 6 A111 Xy 56 2 7 Find the inverse matrix ofA and then find A391B 12 1100 12 1100 1 1 1 010 gtR2 R1gtR2R3 2R1gtR3gt o 1 2 1 10 gt 21 1oo1 0 31 201 103 120 1R2gtR22R2R1gtR1 3R2R3gtR3gt 01 2 1 1 o gt 00 51 31 213 1 555 1 2 3 312 Rgt Rgtgtgto10777 53R35R32R25R3R4R4 555 oo1 1 1 55 5 31 i 31 3 z616 7j 335 555 555 6 5 5 5 555 3 so A 1 1E A 1B 1E6 616E7 E E2 55 5 55 5 7 5 5 5 555 1 131 131 1 3 1 6187 quot r as 76 7 47 7 55 5 55 5 5 5 5 555 x3 y2 21 Determinants Determinant of a 2x 2 Matrix a b a b lfA then detA W ad bc c d d C Calculating a Determinant of a nx n Matrix with your calculator a Click b Highlight MATH in the menu c Highlight 1 det d Click e Click f Select matrix D click g Click again to see the determinant of matrix D in the viewing screen Cramer39s Rule If a system ofn linear equations in the n variables x1x2x3xn is equivalent to the matrix equation DX B and if D 0 then its solutions are Dxl sz st Dxn X1 iDi39Xz lol39 lol lo where DX39 is the matrix obtained by replacing the ith column of D with the ngtlt1 matrix B X3 Solving Systems with Cramer s Rule by calculator Cramer39s Rule Determinant Method Form the matrix equation for the system of linear equations DX B Verify that lDlae 0 Then form and calculate the solutions x y g z Example x 2y z 6 X Y z 6 II 2X y z 7 III Form the matrix equation representing the system 1 2 1 x 3 D 1 1 1 X y B 6 2 1 1 z 7 Form the by substituting the column matrix B for each of the columns of D 62 1 16 1 126 DX611 Dy161 Dz116 71 1 27 1 217 The determinants of D DX Dy Dz can be calculated with your calculator Using determinants the solution set may be calculated D xi 3 Fig2 02 D 5 D 5 D 5 Calculating a Determinant of a ngtlt n Matrix by hand minor MD of the element a is the determinant obtained by deleting the ith row and thejth column cofactor A of the element ans 1ij My examples an an an am a21 a22 a23 aZn A 331 632 333 a3n am anZ ans am an 312 an aln 321 322 323 aZn detA A z a31 332 333 aan a11AI1a12A12a13A13quot alnAln am anZ ans arm A 12 th lAl 12 1432 46 2 34 e 34 1 4 7 1 4 7 A 2 5 8 detAlAl 2 5 8 159 68 429 38726 35 3 6 9 145 48 418 24712 15 3 35 2111 Singular matrix If a matrix A has a determinant 0 W 0 then the matrix is singular and has no inverse PRECALCULUS Chapter 1 Fundamentals 11 Real Numbers Rational numbers Irrational numbers r rquot 3quot 1 46 017 06 0317 o3 V15 lz 712 7 77 Integers Natural numbers 32 l0 l23 I C Complex Numbers xlx a bi where a be Rz39 R U Imaginary Numbers Number Sets Imaginary Numbers xl x bi where b e R 139 R Real numbers Q U Irrational Numbers Irrational Numbers xl x cannot be written where a be Z and b at 0 Q Rational Numbers xl x can be written where a be Z and b at 0 Z Integers 4 3 2 101234 Whole Numbers 01 2 3 4 N Natural Numbers 1 23 4 Language of Sets Set Operations U Union quotorquot combines the elements of two sets Intersection quotandquot yields the elements that two sets have in common 39 or Complement a set containing all elements not in a given set Set Language Symbols proper subset always quotpoints at the quotsmallerquot set 0 g improper subset may be the same set or a quotsmallerquot set Q empty set this is NOT an empty set Q e is an element ofas in l e 1 2 3 4 quotequalquot sets have exactly the same elements sets can be de ned either by listing the elements or de ning how to create the elements U the Universal set includes all elements Set Builder Notation xx where ab e Z and b at 0 is read as the set of x s that can be written where a b are elements of the set of Integers and b 0 Venn Diagram a good way to visualize unions and intersections of sets PROPERTIES OF REAL NUMBERS Property 39 4 Examine quot 39 bescriptton Commuta ve39Properties j y r y y W quot a 39 l a 7 3 z 34 7 Whenfweadd tWol39numbers order doesn trmarleri hail rbu WhenV39we rnnI39 39 yirw numbers order doesn t Associativer Propertiesquot abc aibc 355393 lt7s39 2H 4 L7 2 2 4 7 matter When wearirl three numbers it doesn t matter which two We add first then wemultipty thr ee numbers it doesn t mat abc abc V 39 39 ter which twp we multiply first r Distributive Property I 39 39 r r 39 I ab c 4117 ac 39 2 5 2 13 2 5 When We multiplya39number by 39a sum of 39 b ca 2 ab ac 3 5 r 2 2 3 2 5 I two numbers we get the sameresultasmultiply ingf the nuxnber by 39each of the terms and then 39 adding the results The Real numbers are closed under the operations of I What does closed mean Are the integers closed under the four operations Why subtraction for any two real numbers x y x y division additive identity 0 additive inverse x 1 for any two nonzero real numbers x y x for any real number x0 0x x for any real number x x x x 0 multiplicative identity 1 for any real number x 1 1 x x 1 1 multipllcatlve 1nverse for any real number x x 1 x PROPERTIES OF NEGATIVES Property Example 1 x PROPERTIES OF FRACTIONS Property M Example H 3 L i quot 39 Description To multiply fractions multiply numerators and39denominators 39 r 4T6 divide ffatti ons invert the divisor and multiply 39 j To add fractions with the same denominaton dd lekhiim eratofs 39 tions with different denominators r find a common denominatdr Then add the numeratofs a C an39cci numbers that are common factors in numbraior and denominator a I 6 Itb rthennd be gzso2936 Crossmultiplyi The real number line is dense between any two real numbers there is another real number Are the rationals dense Are the integers dense Why A 1 l 7 2 44 49999 49 47 41725 4 lo a 4 i F 3 4 w 2463 I 42 Mg VHI 17 w 4 FIGURE 3 5 74 3 2 71 0 if 1 2 3 24 455 Thei eal line 4i85 0 Translation Matrix important Notation i Set description Graph ab xnltxltb 4 1119 xla xsb gz E 11b xiaxltbi 42 73 11111 ixialtxl7i 43 3 6 a4 3 ix i a lt xi 42 gt as mi X i a X gt WJI xixltbj gt 4 xixsbl 00 Dlt3 R set of all real numbers gt wc unvc uAc luuuvvlus mu mu DEFINITION OF ABSOLUTE VALUE PROPERTIES OF ABSOLUTE VALUE I I I I I I I I A 39 I Description V The absdlute value0f a hum V 531 iS39falwhys positive plzero 39 A number and its negative haVe the same absolute Value The absolute value of a prod 39S thepmduct of the I IG VEIIIljSg 39 nabs l te valhenf a quo entis the q otientjof rhe39 abso i te values 39 ON THE REAL LINE DISTANCE BETWEEN POINTS I istajlc 39betwsen the points a and b in Think about distance on the number line 12 Exponents and Radicals EXPONENTIAL NOTATION nd39n i positive rmteger then the nthrpowe r of 39a 39lffa39yisfianyiryejalfnumii a CL61quot a nfact39ors The39rjumb39e rdjs calledth as mug is called the exponent ZERO AND NEGATIVE EXPONENTS If 0 is ny ta I M y39 n39is a positive integer then he number add ihejt nmb rgsubtraict the L rm ittpiy the ch factor to the em to apo wgr 1 misc bomnmneratoiiand39 I oyt eip39 r a I LAWS OF EXPONENTS V V reuse a ftaehon to a neganve power invert the 1 r g ftac onandchangethesign of the exponent quotTo 319 a number raised to a power from numeratgr r 39 i V i 39 l chang the sign f the cxpon nt Scientific notation H x 4 a39X 10quot where 1 51 lt10 and n is an integer db means bza and 1120 DEFINITION OF nth ROOT V V I If r i39svaLny positive integerajha the principal nth root of a 13 de ned as 39 follows means b 2 a Property IveJaw 3 82vxafg a 23 6 a I 3916 16 g 2 quotI7ZW 45 1 3 3 Wrfl 3 4 750 ifnisodd 3 5 3 z 5 6532 5 WIaI ifniseven 3quotI 3 3 illquot32 DEFINITION OF RATIONAI EXPONENTS 39 Edi any rational expo gpt 171quot in lowest terms where m and n are integers quot and n gt 0 We define amn Na 39 or equalen y mr39 m a Z1 39 If is even Eben we reqpirefth xt a B 0 Rationalizing the denominator removing radicals from the denominator 15 354 V3 324 Rationalizing the denominator multiply by 1 name it carefully multiply by what is necessary to make number under root sign a perfect quotnt quot power 15 3324Y 15WM1533VE3VE5V M 3 3242 3 3243 324 2 Note this isl 13 Algebraic Expressions umacudbcbd W T T T T F O I L The acron cm FOIL h7lps us remember that the pr ig o two binomials is the sum of the prod ets of the First terms the Outer tenns the Inner mums and the Last terms B I I A5 3AZBlt 3A327 T9 3 A3 3AZB 53A33 513 n 1 u 4 duo amnch nr nnv other formula in algebra 13 the FAcrommlt9 x24x 2x2 y 39 EXPANDING FORMULAS Formula 39 FACTORING 1 Ale 3 e 4 B 14 Fractional Expressions I Simplifying Fractional Expressions To simplify fractional expressians we factor both numerator and denominator and use the following property of radians 115 says mar w um cancel common factors from numcmwr and denominator Multiplying and Dividing Fractional Expressions To multiply l39raclhmnl expressiuns we use the following property of 39uClinnS This says that to muiuply we fractions we mullipr than nunmraiurs and multiply thuir ii lllllllll lallll39h Tn divide fractional expre ans we use the following property of fractions This buy ha an aura 31 fraclion by another fraction we invcrl Lhc divisor and multiply I Adding and Subtracting Fractional Expressions To add Il39 Suhlmcl fractional expressions we rst nd a Gammon clenumilmmr and then use the ullowing ropcrty ril39 l muliuns I Compound Fractions A compound fractinn is u fraction in which Lhc numcr39umr the dcnuminamr or both are Kl lenweh rs fractiunal mprussions I Ra onaliling he Denominamr at he uumeraxov II n Incmu hm 1 ummmm r M In mm A I J M nu nummlm m llumzrmlulur by mnluplymg mmmmr mm L39nnmlmlur by m mnjugan mdin L N ix dk ulm Mcnuw by Pmdun anuulk 1 In Scmvn LN prud at or he ummmmnum u wlungnr mam um nut ummm H mm A Nam 7 m A m y w I Avoiding nmmon Errars um x mm m rman I mum mum m mulm llmuuu to m aparwun my mum Mmyufllkuumm u nmd humruhcd n W m w e was mm pnm e urmummmmn um vlluumhv m rm m npplmlg quotmu m mum m M a p n u w 41 20 15 Equations PROPERTIES OF EQUALIT 39 Fibpa w39 1 1 A39B A 5053 C 39 Miami 391 4xde the s nie qtian lyio both sides of gn equa on gives an equivql ilenltequa on39 5 u 2quot1 2 CA CB I Mulliplyingvborh siiie ofan equation a I v 39 39 hy zhe same no nzem quamity giyes quot m f v neiqulVaientequanon Conklin s Rule for Solving Equations Nothing can go wrong if you ALWAYS do the SAME thing to BOTH sides of the equation I Linear Equatians The 39simpiusi 13p nl39 equation 15 1 linear EQUEIUUEL ur 1m dugruc ithiiElL llL which I is an miuzitinn in which such term is either 1 mnsimn or a nonzero multiple of the variable This means that ii is equivalent to an equation of the form at b 0 Here a and F rupmscnl ruzii nunth with r7 71 0 and x is lit unknnwn variable than we are suiting fur ThL equation in the following L Mhlllplt 15 linear QUADRATIC EQUATIONS I 7 I r H r 3 A 39quadra c equali on39is39an equation of Lhc form I I i ax2 bzt390 where Mr and are real numbersWith a 0 ZEROPRODUCT PROPERTY 39 quot1530 ifandmiyif AO 39o39r 3o SOLVING AHSVIMPLE QUADRATIC EQUATION 1 quotlieSim I COMPlE39IING TN SQUARE Turns 2 peri cx squm 2m hnqum nflnlflhcmzmz m a e or Thiv givcs me per zct square r ms qunm mnc romuu Thnmtsybtymrqnmh39mic q39 mu 39hxd c u whch a r WA III c m Hhegcnez mmmcmrhr c 0 Hum me DISCRIMINANY Th 1 Mn uc m1 sululinns 4 2 WI 339 up lt 01111quot w gquaum m up 511 mug o 4uc 1 1sz gt 31le we maneurhn nvu um me equn en has many mama snluum Mm m nhc m equalmn wt my and n or mum exlrunmus ml amr m emamm mhmun E mncn Nummm quotmy be um we M an uqnnuou he use m quotpcmlmn uulum mm d we we Fur ewmnl 1 x hm on may be mm m mun Mn 0139 the valuable E Hun me u ma equauau Tm n why you mm WW mch ynur Jnmcn n mm W um mm mm m m mum mu um Aneqmmxmmlhulonnuw HmM r whucwhun wmwcxpxessmn My nqumun ul39quadmlir lvpc We salve uqummm m mum ype by sumw mung in me A gehmlc cxmcwum u wu m um um lwo examples up mm o lions m n pmmnu mlnnun um do um MIN me Ungng cun is H me In J 7 1 15 16 Modeling with Equations GUIDELINES FOR MODELING WITH EQUATlONS l IDENTJFY THE VARIABLE Identify the quanlily thal the pmhlcrn asks yuu In nd This quanlily can usually bu dclcrrninad by a careful reading if hr ques Lion posed at the end of the problem Introduce notation for he vuxlahle call i1 x or some other letter 2 EXPRESS ALL UNKNOWN QUANTITIES IN TERMS OF THE VARIABLE Read each sentence in the problem again and express all the quantities memlonecl in the problem in terms of rhr variable ynu de nml in Slap I To organize this information il is smuelimcs helpful LU draw a diagram or make a table sec Examplcs 2 3 4 and 7 in this SOCUUIII 3 SET UP THE MODEL Find the crucial fact in the problem that gives a rela tionship between he expressinns you listed in Slap 2 Set up an equalinn nr model that cxpresscs this relationship 4 SOLVE THE EQUATION AND CHECK YOUR ANSWER Solve the equation check your answer and express it as a wntence that answers the question pusle in he pmhlern 17 Inequalities RULES FOR INEQUALITIES V39Dgsqipio g V h Sidc of39ng inequalitygiy es an pquitf Ell I in a h of n inequality gives tin naily by thezsame positi e q mi yl 39 ainnah 39 1 u my 1qu a quotcan jitteqimtity involvingpqa39it ive 39 thevineQuality 39 v v Solution Equation 4x 739 l9 x 3 IncqualiLy 4x 7 L 19 x S 3 I Nonlinear Inequalities To solve inequalities involving squares and other puwms of the variable we use factoring together mm the following principle cumulst rou smvms NONLINEAR INEQUALITIES MOVE ALL rzmsm ONE smz Ifnkccssary rewme the mowth u link 1 mum mus appear m m m a me ingquuhry Hg mm uonzem Sid mm Inequality mom quomms mug mum to a common denommmor 2 mum a mum m a lhe mcquamy E US m Mm Minn m nd an aokuuum of he quaunn mumde m m g a inoqun ty Thus numbcn will mm m m3 lme mm murals Lid he intervals duelminw by mm numbers 4 MAKEA mm ox m E AGRAM Us m wlum to make a nbl a diagram om gm ofmch factor on bash mmml In me um mw of me mm deter mm 11 mgn om pmduul 0r qualicnt uf mes factors 5 souE Devannmc m4 501mm m m inzqmmy lmm m m mw DI me I signtable 5 mm m chock whethenhemaqualny 15 aus ed by 50mm oral of m ndpoims arm imennk mix may happen if m maqnamy involves l r3 I Absu ute Value Inequalities w us me mHleng propcmc u Maw Vquthhu um mum mom vnlnc or AHSDLuIE VALUE INEQUAL r is lnequaliw qu nlva em form Gvaph 1 qlt 1 z m a 1mm a h 18 Coordinate Geometry I The Coordinate Plane ur Jun l uh pullllh on a line can bi identified mm real numbers 0 fonn the mnrdiiiatc Eiuc points in Ll plant can be idcnli cii wiih ordered pairs of numhcrs 10 form the coordinate plane or Cartesian plane 1390 do this we draw two perpendicular real lines that imem cl ill I on each line Usually one line is horizontal with pus 39u Jii39cuimn m ihc right 39 39 1 he raxis Liic other line is vertical with pa iv direction upward and is called Lhc y axis i39lie point of intL ion of the xuxis an the 1ixis is the nrigin 0 und the 1ch axes divide the plane iniu i39uur quadrants labeled 1 ll ill and 1V in Figure l m pagu 90 The points on he coordinate axes at nm assigned to any quadrant S L39 57 1 7 h 1 I e i ax nnu139 IGUREl FIGURE 2 1mm rm 42 hr VIx M1 39 Graphs of Equa nns Suppme W hm m mumon ang m vmahlm and v mch a yuan m pm A w Ea L um cquunm n39 11 eqmdon s we when m coordxmle n as of u pom are subm nuled into the squadsquot For example the pom 3 4V smis c m m eqummu mu 11 1 41 2 25V hm um punu 7 7 dncx mm mm 11 3 13 25 AN EQUAYION 4 Mn I fundamemal Prmdnle n39 mink aeomeuy Apmm 4x y has an be graph inn mm and July irm momma many m Squaw mrn f wkrm hrmlvlyrk r hwuumnn u Tm m m mum aqunmu I S y m In any 5ng m xhaws um graph 1139 y r Nam um um pun m the gwn to me 1m 1 ms 15 m mmrlmn c if me an m m ngh arm vau m reason is if Lhc pain x 3 Is on m graph X y and mm pmnh m m nclym nfensh cum aboul 4 w my m g aph 5 symmnvrlc w r pm to LhcyJu39is Sx um v w a a graph symmetric win 139 9mm xaxls if whenmx lhe point my is on me gmph Lhcu 0 b u gm in with respch n he nrigin 1 whunever A v is an m mph 5 y Dreamm my d rgnqnx Jun mm admin In m gap 1mm n 19 Solving Equations and Inequalities Graphically SOLVING AN EQUATION 39 Mgehrglc wlequ 39 quot i the uvnlma39wn xmi one sideof the equation 39 Em leizx39 6 4 x7 39 gt 33 6 Adxu39 39 quot 72 Div39iaahy 391 The39sqiulicm is x 2 L Im he ml s 39of algebra39t SbLhm 39 equal to y Sketch the graph to find 1116 value af 3 where y l Example 22 639 I Graphical39Methpd Move ail 10 me side andvse39 39 06 3ru Sc y 6 3x and Ef ph From 1119 graph the Solution is x 393 L 110 Lines SLOPE or A LINE 39s39Io39pe ma an enical kin Railings 15 thrdugh iihe minis Aifo OFORM OF HE EQUATION OF A LINE se ih39r iigh magiaux hasjlqpe V SLOPElNTERCEPT FORM OF THE EQUATION OF A LINE 39 VAn39 cQ miQh 3901quot mc39xi c39It nEMsfsibpg m and yint fc pt his GENERAL EDUAIION or A LINE I ParaHel and Perpendicular Lines Smut up quotmm mm steepness of n 1m u m 51mm hm me same slope In fast w m WM um Wumu Hmlpumllul hm ma lint mm J opew 4 m an Wpudi umi It and om39ymwa g mauvemmcm mam ammo x u Chapter 1 Fundamentals Terms to know definitions of Midpoint Distance Tangent line a line that intersects a graph in one point Secant line a line that intersects a graph in two points Chapter 2 Functions Terms to know definitions of Function A set of ordered pairs in which no two ordered pairs have the same first element A rule or process that assigns each element x in a set A exactly one element fx in a set Independent variable Dependent variable Domain Range Graph of a function Linear function Family of functions Step function Greatest integer function Power function Vertical line test Root function Reciprocal function Absolute value Absolute value function Increasing function Decreasing function Direct variation Inverse variation Joint variation Average rate of change Symmetry the property of remaining invariant under transformation Transformation the operation of changing as by rotation reflection mapping etc one configuration or expression into another in accordance with a mathematical rule Shifting a function Reflecting a function Stretching a function Shrinking a function Even function Odd function Quadratic function Extreme point Local maximum Local minimum Composite function Sum function Difference function Quotient function Product function Identity function Onetoone function Inverse function Horizontal line test Jtrac 4 then divide by 3 tare then add 2 hine diagram for f as 392 Functions What is a Function 2433 4 f I W5 5 6 Divide by 3 than subtract 4 8 Add 2 then take the square root 10 Machine diagram for f I 2 712 12gt2w3 I 39 i r Hr m gm 1 21 1 128 3 12g3313 5quot 2 12 z 2 2 2 2 3 1 2391 1393o o 2033 324132 1 213g5 5139 33719s 3 23319 i 143f212fa2a12a 1 3 2 ab12a2b1 7 33 23939615f3 3 f 1 I22 IZZmflt 41 2 I f 391 21 3 9 v z 7 1 a11a12 a g1 25 171 7171 2h22 l 7377g z 37130 74 ND 7 22232gt74 864 7 2748767472f v 23 74 7 i J 7 2 71273I174 7 2x24z23z374 21377r7 7 gt s 7 7 00f1137412217473f113 4Lv1 1 3 J E 7 lt gtBi4lt gtgt iwwgt2W7 5 7 6 f0 7 210711 21 2 f 272J21212fm1gt2ltzn7172 A7 z 2 7 z2122122sincem21gt0 I 4 r 7 7 1L7i171fm isnotde nedalx0f5 l 7 39 7 722 4 Since 71 lt 0we havef71 712 1 SinceD 2 v3 ha ef1112 SinceZZOwehave f2213 I quot 7 inceDg2wehavef05Since 2wehavef25Smce L gtgt2 we have f 525737 7 7 7 4 4 274 16 7 8 8 Since 7 g 71 we have 7 7 77 73 Since 71g71wehavef71 1227117271 s 5gt1vehavef2571 47 15Since0S g2wehavef0011 Since0 132 x havef2213SinceSgt2wehavef55722E 5 7 74m5fxf2m21221x2141r3 I 7 39 39 39 quot 39 423a3h2 L7 3af3h2 3a 23ih3l h h a 7 7 739 39i71 a22ahh21 meLw SECTION 21 What is a Func cn 7 127 32fltaa1rahgtrl 1 1 21 a39h1 fah fa ah1m a1ah1 r11ah1 h h h h 7 a1uh1 1 7 h aluh1 a ah 33f a1 f h h l ah a aha1 aah1 fah fa ahla10h1 11ah1a1 h h h aha1 aah1 ahla1 u2aahhia2aha h hah71a1 1 ah71a1 s4fagta2f1fah 2ah 2a 2aha 1 2aah71 fah fa ah71 a 1ah71a 1 nh71a1 h h h 2aha 1 2uah 1 ah 1a 1 2azah72ah 2 9 I1 hah71a li 7 ah h 7 u1 hah 1a 1 ah71a 1 35fa35a4a2 fah3 5ah4ah235a75h4a22ahh2 3 5a 5h4a28ahtlhg fah fa 35u 5h4a28ah4h2 73 5a4n h h 35a 5h4a28ah4h2735a 4a2 5h S W 58a4h 35 f a 13 f a h a h3 z 13 3a2h 311 l13 f V h39 f a f a3 Bazh I mL2 13 a3 7 gag 35 h3 h 7 7 h h 1132393 I2 a w 3a 3ahh2 Since there is no resirictions ihc domain is the set of real numbers fix r 1 Since There is no restrictions the domain is all real numbers x 3 The domain is restricted by the exercise to 15 1 The domain is restricted by the exercise to 0 5 1 i i j Since the denominator cannot equal Owe have I 8 7i 0 39 2 V ion the domain is oo3 U 3 cc 128 42 A ll m f1x 5 Werequirex7520 gt e r 61 9 Since even roots are only de ned for nonnegative numbers we must have x 9 gt 0 CHAPTER 2 Functions f as 3m 6 Since the denominator cannot equal 0 we have 31 7 6 0 gt 3x E 6 a I ye 2 In in notation the domain is 700 2 U 2 oo f Since the denominator cannot equal 0 we have 12 7 1 75 0 ltgt m2 74 1 gt domain is r 75 i1 In interval notation the domain is 00 71 U 71 1 U 1 oo 4 f Since the denominator cannot equal O 2 I 7 6 75 0 ltgt x 3 ac 7 2 quot O D I 73 orz 2 In interval notation the domain is 00 73 U 73 2 L 2 00 z 2 5 Thus the domain is I x Z 5 The domain can also E7 expressed in interval notation as 57 oo 2 77 the domain is 79 00 t 9H 7 1 Since the odd root is de ned for all real numbers the domain is the set of real numbers 73c x s xT 7 32 For the square root to be de ned we must have 7 7 3m 2 0 42gt 7 2 3x tgt g 2 7 5 Since the square root is de ned as a real number only for nonnegative numbers we requi x 2 So the domain is z 2 In interval notation the domain 1 a 225 cgt 35229 e lz23 gt mZBOrxg Zl 27 7 139 2 0 and the denominator cannot equal 0 Now 2 as 2 0 gt E be domain is x i z 2 72 and z 397quot 3 which can be expressed in interval r have 1 2 0 for the numerator and 2902 x 7 1 75 0 for the denominator S Z il IIl7 0 gt 21719 00rx10 gt m orrli m to an even root must be nonnegative we have 12 7 61 2 0 ltgt r 7 7 61 2 Cu 8 B 39aver2 2a 8 Z 0 ltgt m 4m2 20 Wemakeatable 700 72 72 4 4 oo 7 7 7 7 UI to a 512 47T22 a SECTION 21 W Since the input to an even root must be nonnegative and the denominator cannot 3 374 I r 4 gt 0 lt2 7 gt 4 Thus the domain is 400 a I Since the m ut to an even met must be nonneaative and the denominator cannot 71 i 2e M6 p v m 6 z gt 0 lt9 6 gt 3 Thus the domain is 00 6 x 11 r Lug 1 f L 2 1 Since the Input to an even root must be nonnegmwe and the denon nnator car39 7 39 quot 39 m 211gt0 lt 2gtThusthedomainis f x Since the input to an even root must be nonnegative and the denominator cam1 sr 39 39i r a 7 1 932 gt 0 11gt 37031 gt0 Wemakeatable Interval of3 139 of3 of 1 Thus the domain is 33 a C10 1500 3 10 002102 00001 103 1500 302 01 15321 0 100 1500 3 100 002 1002 000011003 1500 300 200 T 100 b C 10 represents the cost of producing 10 yards of fabric and C 100 represents the fabric c c 0 1500 3 0 002 02 00001 03 1500 101T m 5027 S 3 47 32 367r w 11310 b 5 I 2 represents the surface area of a sphere of radius 2 and S 3 represents the SLZI D 01 23060 010127020 ms miles 7 1137 5230 miles 0215 miles D 0215 V52 3900 7 72 554 89 2356 miles 41 2501 2 50andV20 50 1 2 0 J 50 represents the volume ofthe full tank at time t O and L 0 represents the volume of the empty tank twenty minutes 130 I CHAPTER 2 Functions 3 v V001 18500 025 012 4440 r 18500 025 f 049 1665 C s that the blood ows much faster about 275 times faster I r Ihe center than 01 cm from the edge b m 148 H100M 10mm x zc including shipping Ii SECTION 22 Graphs c1quot Z39E 131 71 T 72 TA 60 0 I 50 0 5 m 73 PA 900 800 z H 1988 1992 1996 1000 I 74 Answers will vary 75 Answers will vmy y 1 Graphs of Fu nctions 1 v 2 139 f1392 7 f11 9 2 4 S3 6 2 1 2 is 73 2 b W r b T J 5 D 2 I 2 3 3 7 4 3 393 2 6 3 37 4 3 1 fl l39LL 1 392 3 1 I v U 3 u 1 3 g 2 2 I 4 3 73 3 l3 113111111111 11391 11251 139 a1011m 17 1 9112 1111111155 111 Em i 11131515 9111 11151125 111111 91111 51111 uo 1111011 X1112 31111 311125 91111113521112 RZ 7L41TZ 171 f1 5110111211113 359111 231113 lt39 h 139 I SE 11 4 1395 3 m 171 lt 1 141 1 gt 1 I 511101111105 9111 511111 3917 3 lt Eng 3914 a gt grg 11 g 1394 fi 19 am 15111 9111 111011 1101113an 13110305 o111 111131111q113 A 9 139 9 m5 h 1 5 511101111105 9111 51 1111 1 Ii 1 15 Ii xg m S gt g 12 g 03 x 4 111 xg 19 em 3111ppv 0 175 131113 114312 51101111105 21115110111211139 1110111951195 51111011 1110q 931113 3915 91 4 91 17 f p 31 za pun 4 17 w a 0512 517 0114310Hz134 1m 3 0 4 a4 uo1111105 1 51 11112111111311 01 511011111111 11011 111 1101111105 9111158011 110129 1139113 17 1745 81 3M 111915115 911 01 51101111 05 am 17 quotg pus 1 34 1211 5190111117 11 1111213 911 11 1101133115111 Kg 1 l l I J Z A L a I H 5 p112 9112 51101111105 3111 5n111139z lt5 Z 11 Z5 113111 391EJI Z1 JO 4x lt2 0111505 1zgr1rz1 z ltgt z 1 4 13 137 am 395u011enb9 9111 111ppv 1111011 121110 3111 101 serum 9131110312 133 01 5 z 119111 lt2 0 vgzgg 0 4xlt5471 1111311111 51 11111 111315 3111 01 1101111105 2 s1 g 5 12111 511321111113 11 11111313 311110 1101109115111 Kg 1 393 Z 391 21 39174 17 11111 H 3934 211 51101111105 9111 5110111211b9 1110111 A 81111011 111011 231115 17 oz 4 9139 1719 4 C1 171m 17 1 1 1119 05115171 01 v z 9 g z1 111112 8 H 9 11 a 9 1101111105 12 s1 11 12111 1911311 01 5110111211113 1110111 111 1101111105 2111155011 113139 439113 3 111 4 LE fi 391u91ss 91110151101111105 3112 1a 17 p112 171 3 113111 5112011111 11 11111318 911111 1101139d5111 51 A Z 393 391395 139013111 51 11011n1os 9111131 9 09911119 LEW ONES 0 Z 4 E UK 4 E PUB A E V 1 a j 1101111105 2111151 11111111191121 0151101 0 g 4 11011 u1 5111 139113 9111 39111215K5 3111 01 11011n1os 9111 51 1 3 111111 51129dd13 11 1111116 311110 101139dsu1 Kg szigw 1111911133 6 131119113 sa1111anbau1pue 51101151111310 511191513 5 a It 12 13 CHAPTER 9 Review 571 is false Hence there is no soiutini The lines are parallel 2m 7 y z 1 v 139 3t 1t39l Solving the rst equation for y we get y 2 721 1 3139 39r 4 15 Substituting into the second equation gives 1 3 721 t U i 75139 7 gt z39 7 So 3 7 7 t 2 gatedng the point m l 4 ml I l l ml 11 H SJY I y 395 9 n5 a 6 m 9 3 9 r N 1 D E D 5 4 Q o E i Y4 G I E 5 9 Fl Adding the rst equation to twice the second equation gives y 1 Substituting back into the second equation we get 71 31 1 4 1 11 Checking paint 41 in the third equation gives 7 4 7 2 1 26 14 Thus there is no solution and the lines do not intersect at one point Substituting For 1 gives 739 1 32 27 ltgt 10 a 7 6 z 0 Factoring we have r 2 it 3 0 Thus 4139 393 or 73 ltd 2 than y z 3 and iftr 73 then y 3 Thus the solutions are t73 3 andt lS 1392 32 8 t a buhstituting for y in the rst equation gives r 1 2 8 ltgt 1 139 r 72 4 0 lt3 1 r T 2139 7 0 Usmg the quadratic tonnula we have 1 7 1 717 then I t v 2 1 7 M5 and il r 7t V3 them 71 2 2 l x 3 Thus the solutions ate717 t 7 and 71 M11 l g Adding twice the rst equation to the second gives 71 16 1 7 U My 5 28y 121 16 ltgt y 7131 Thus the solution is my mm x 5 2 19 21 CHAPTER 9 Systems of Equations and Inequalities 139 212 10 12 2341 Ty O Subtracting the rst equation from the second gives 1f 71y 10 lt y277y100 1172y509112y5ll y2thenm2410 4 263 and ify 5 then r2 25 10 lt 12 5 which leads to no real solution Thus the solutions are V x 0327 04331 o y 72 12y 341 The solution is approxitrlatsly 2141l71593 r yzz 10 1ir 10 12y12 y22r 12 a 2 x 3 13 Yes this matrix is in rowechelon form c this matrix is not in reduced rowechelon form since the leading 1 in the second row dues not have a 0 above it 1 2y J3 d 9 3 u a 3 x 4 b Yes this matrix is in rowechelon form c Yes this matrix is in reduced rowechelon fonni ir 3 y 660 713m 303111 20000 39 y iggjx r 15 The solution is approximately 6104 10573 39 5 1 18 y r The solutions are approximately y 1 5 145 7135 16and 15141293 20 a 2 X 3 b Yes this matrix is in row echelon form D Yes this matrix is in reduced rowechelon form d 1 6 y 0 22 a 3 1 4 b No his matrix is not in rowechelon form since leading 1 in the second row is not to the left ol39tl CHAPTER 9 Revxew 573 23 a 1 X 4 24 a 4 x 4 M a 25 b No this matrix is not in row echelon form The leading l in the second row is not to the left ofthe one above it C this matrix is not in reduced row echelon form b No this matrix is not in rowechelon Form The leading in the fourth row is not to the le ol Ihe on above it C No this matrix is not in reduced rowechelon form Lr8y6z 4 d utl y 1731 5 d 12yT 2 22 1 Jy 2 0 c y z 1 U23 l y23 6 392r 5212 b 2517 539 39 9 1y 36 gt y1undr1226 m 2y321 zit3yA 30 lt4 392 76 6 I m 213U1 Gy7lt139 l4 4 Therefore 3 6 42gt 3 lt2 1 1 Hence the solution is 11 2 m72y 321 1 4 Thus 031 1 and 2820 3 and so the solution is 110 n 2y 321 r 2y 33 r Qy3z 1 27 21 y z c 3 4 3y 7 s z 1 3y 51 2 1 which i impossible 2L39 T39y112 Gy 10l 0 1 Therefore the sysrem has no solution w l y 2 12 m y 2 1 2 4 y 11 2 2139 3 5 2y5s21u71 2y 5 u 1 28 c I Zy 1uI9 3y 3w7 13712i39r11 r l y223u395 2w 3 ins 3 1 g z w 2 2g 5 211 391 1 7 Therefore 14m 28 gt w 2 13 122 11 7 3 12w 11 11111 28 211512271 lt2 21 1 1 ltgt 1 10 12 1320 1404 1 10 1 1 gt 291 10 1 1925 0327 v R3 DR1R3 213 2137 0339 3y3y1undr71 1 4 1 U and 50 ha 3 1 391 1 2 1 v1 1 2 1 71 1 2 30 1 1 3 6 Elif 0 2 2 1 1 o 2 2 1 trims 131 U 2 3 5 1 2 3 5 U 0 1 1 gt y 1 and E 139 1 Z 9 139 392mdsolhesolutinnis211 5Equaiion5 and Inequauhes 7 1 7 11 2 3 2 7112 13756 rquot 11 21 11 71 5 5 R3 R3 113 1 2 3 v2 1 3 5 G 0 O O 1 112 112 1412 7171127 136 1 221 0221 1 0112 1310 L 24 00004 JODOJ 1 0 2 4 0112 Lelf111eny t392 s y2tandrr2t 134 2tandso1 U U 0 solutions a1 e4 7 21 2 115 Where 1 is 011 131 Hummer 1 1 1 1 0 1 1 1 1 0 1 1 1 1 U 1 1 4 71 1 13 11 72 75 2 71 R3R2 R3 0 1 4 is 6 1 Z 0 4 T R4 9 1R4 11 73 71 3 77 U 3 1 3 7 2 2 3 4 3 D L 1 2 73 U U l 2 3 1 1 1 1 0 F1 1 1 1 0 1 1 1 1 0 EEK 0 1 4 is 6 r1 1 v1 5 6 WRWRJ 0 1 4 5 6 0 U 713 12 11 U 0 l 2 3 O 0 1 392 3 O 0 1 2 73 O U 13 712 11 U 0 O 14 28 Therefore 14w 28 Q 111 72 2172 73 9 2 11741 52 G 4 y 05211 J U1 A U 1 11 80111250101100 is 10172 1 O 3 U 1 1 O 3 0 1 1 U 3 0 1 010 4 5 Rrgm ql 01 0 4 5 01 0 1 5 Ra 0 2 1 1 0 O 2 1 1 El 0 0 1 9 10 L2 1 5 4 4 0 l 1 4 6 0 0 71 0 l 1 U 3 0 l 1 I 3 0 1 1 U 3 0 A1 0 1 0 74 5 RHSRPM 0 1 0 4 5 7R 0 1 0 4 00710 1 39 00 101 1R1 1101071 0 0 1 9 10 0 O I U 79 1 0 U 1 71 1 0 0 0 392 U 1 0 0 Tl1ereibre1l1e solution is 21 1 1 71 0 1 1 1 0 71 0 0 0 1 1 1735117 1 L3y 4 W I I gt T111131 the sys1em has 111111111er many SDILIUOI IS gwen by z 1 41 Therefare the solutions are CHAPTER 9 Review 575 3921 L 7 n l 7 0 z 721 Since this last equation is impossmle 3 inconsistent and has no solution r4y 8 8 r4y 37 2iu Gy r D lt77 3yl332 7 2 2y37 31767174z71l3 5114 95221 00 Thus he system has in nitely many solutions Letting t we nd 2g 3t 7 e y 57 gt and u4 7 t 8 cgt 1 675t Therefore the solutions are 6 5t tt wheret is any real number 1 w 2 r 7 2 w2 1 7 w 3 33 2139 y 721 2 t a y2z7 411 23 c 3y 7i 3 w 4 3y z w 4 5 713w x 20 17 1117 10 y 7 w8 2 4w 20 1 7 2 w 2 L Therefore 33m 7120 lt2 5 7 13w 20 33w 2 120 8 lt2 y ganclw777 1 7 1 71 3 9 0 3 7 7 0 3 75 2 1 Cl l R2 Am 1 l 7 The system is dependent so let t y 7 r 7 5i 0 1 l 1 Since the system is dependent let t Titet1 39 1 n 0 0 0 n 0 n n 1 7 171andrL 2 5 v 7172 So the solution is L 393 7 1 t where t is an rca 1 1 1710 17 0 2 4 22 A J 172 11 1 1 u MJRWRE 11 1 1 o b 1 LAv Since the system is dependent Let 2 z s and w L Then y 7 25 7 t l A y 42 1 s 1 So the solution is N 1 2s 7 t 1 s f where 5 and t are any real numbers 578 42 44 46 47 48 CHAPTER 9 Systems orquot Equations and 139 l1 391 1 t 3 2 1 6 1134 it I 1 Since the second l OWCOrl eSpUndsm 1 72 s 6 n l1 1 10 1711111 17110 393 2 1 6 o 4 11 0 1 6 Thelastrowoflhis 13 Rt Rg l1 4 33 5 13 0 0 03 matrix corresponds to the equation 0 3 which is aluays false Hence there is no solution 1 392 3 392 1 2 3 2 1 393 3 2 2 1 5 1 0 75 11 3 Wis 11 1 11 3 Since me third row 4 316 075 1172 t D 01 corresponds to the equation 0 1 which is always false this system has no solution 1 1 A1 2 1 O O 1 1 1 U 0 1 1 171 1710 1 A1 11 u g n 71 1 9 71 392 0 l 2 2 1 1 1 l 2 1 litml 4 U 1 71 2 1 24 4726 24 4720 04v v44 1 0 0 1 1 1 0 U 1 l 1 O 0 ll 1 0 1 1 2 1 lm1 U L 1 72 1 O 1 10 71 0 0 1 v4 1 77 o t o o 0 m me 0 o 0 1 0 0 0 0 712 0 O U 0 1 0 U U 0 U 0 This system is dependent Let 1150 y 1 1 r y z t 11 1 gt 1 1 So the solution is 1 t 1 t O wheret is any real number 1 1 230 1 17230 1 1 230 o 1 1 1 1 U 1711 1 0 1 1 1 1 Sincethelhird 37 7102 017112 00001 row corresponds to the equation 0 l which is always false this system has no solution Let 239 he the amount in the EDi account and y the amount in the 7 7 account The system is y 2m 00611 007 500 Substituting gives 006m 0OT2L 600 cgt 0321 600 lt2 1 3000 so 1 2 3000 6000 Hence the man has 3000 invested at 5 and 86000 invested at 7 Let n he the number of nickels d the number ofclimes and q the number ot quarter in the piggy bank We get the Following 17 d 50 system 571 10d 560 Since 10d 211 we have I 371 so substituting into the rst equation we 10d 5 5n get 71 gin q 2 at t g1 q 90 42gt 50 7 11 Now substituting this into the second equation we have 571 10 in 25 50 7 9 500 r 2511 1250 1250 7 392 n 560 e 11550 7112 ritendg hank contains 12 nickels 30 dimesi and 8 quarts 39s 3 CHAPTERS Renew 577 49 Li 1 4quot r39 11 in Bank A y the amount invested in Bank B and 3 the amount invested in Bank C 1 y 60000 J y 60000 We get the 101100111 5131ern 0021 003925y 003 z 1575 4 21 25y 3 157500 1 230 2y 2117 y2z 0 1 1 1 60000 1 l 1 60000 which has matrix representation 2 25 3 157500 0 05 1 37500 1 33 392 1 2 O O 3 0 7120000 1 1 1 60000 1 0 1 20000 1 0 1 2500 0 1 0 40000 13125ng 0 1 0 40000 0 1 0 40000 Thus she invests 0 05 1 37500 0 0 1 17500 0 0 1 17500 2500 in Bank A 110000 in Bank B and 17500 in Bank C 50 Let I be the amount ofhaddack y the amount ofsea bass and z the amount ofred snapper in pounds Our system 1 1 1 560 1 1 1 560 hasthe matrix representation 125 075 200 575 243 0 075 0 255 3quot3925 R3 125 0 200 320 125 0 200 320 1 1 1 560 1 1 1 560 1 1 1 560 0 0750 255 0 1 0 340 010 340 0 l3925 075 380 0 125 01r 380 0 0 075 45 111560 110500 100160 0 1 0 310 0 1 0 340 X39R lnm 0 1 0 340 11mshe caught 100 lb of 00160 00160 00160 haddock 310 lb 01 553 bass and 60 lb 01quot red snapper In Solutions 51 62 the mairices A B C D E F and G are de ned as follows i 3 1 2 4 3 A3920 1 B 6 7 2 392 1 0 2 1 1 lt1 1 0 392 2 1 00 1 E 11 F 7110 G5 2 0 9 7 5 0 51 141 B is 1101 de ned because the matrix dimensions 1 x 3 and 2 X 3 are not compatible g3 1 4 1 52 0 D 2 U 1 2 1 2 0 m1 1 3 1 4 1 0 3 12 11s 53203D2 2 3 0 1 43 0 73 4 0 21 2 0 712 6 0 2 2 54 5B 2C is not de ned because the matrix dimensions 2 x 3 and 3 X 2 are not compatible 55 GA 5 i 2 0 110 0 Chapter 9 Systems of Equations Terms to know definitions of System of Equations Substitution Method Elimination Method Graphical Method independent system one solution inconsistent system no solution dependent system infinite number of solutions triangular form linear equation nonlinear equation Matrix Matrices dimension augmented matrix elementary row operations Gaussian Elimination rowechelon form ref GaussJordan Elimination reduced rowechelon form rref equality of matrices sum of matrices difference of matrices scalar product matrix multiplication NOT commutative inner product Matrix Equation Inverse of a Matrix Identity Matrix Determinant of a Matrix minor cofactor singular matrix determinant 0 Cramer s Rule Chapter 4 Exponential and Logarithmic Polynomials Terms to Know Definitions Of Exponential function base Natural Exponential function e compound interest principal interest interest rate compounding period compounded continuously compounded semiannually compounded quarterly compounded monthly compounded weekly compounded daily present value accumulated value Logarithmic function Natural logarithmic function exponential form logarithmic form log In properties of logs laws of logs change of base logarithmic equation exponential equation exponential growth logistic growth radioactive decay Newton s Law of Cooling logarithmic scales pH scale Richter scale decibel scale SECTION 98 Systems of Linear Equations in Several Vanables 515 z3y 2z 39 r3y 2O 29 Zr ltgt 6y l 834 431 611 00 4 SozLand 6ysc 4 42gt 76y78t4ltzgtyz Z Tlienv35tiv2t0 a 212 Q l 2 31 t where t is any real number So the solutions are 1 a 2x4y7 23 12y 22 L392y 20 m l 2y72520 30 I2y4z6 2x4y 9 323 ltgt 323 2y 30 w2y4z6 626 00 So 2 1 and substituting into the rst equation we have x 2t 7 2 1 0 I 2 721 2 Thus the solutions are 2t 1t 1 where t is any real number 6 x r z 2w 6 31 3 2 b y 7 2 73 lt2 2 7 w 1 0 32 6w 79 7321 6 1 1 2 2 6 4 w z 1 Thus the solutionis 17112 2 y z w0 75 11 5 10 5 34 22 2 0 2 w0 2 32 IT 3 w e e 1 2x2y3z4wl 22w1 23y4z5w2 y23w2 95 y 2 w0 y2z3w2 zl 1110 11121 and w 1 1 1 0 lt1 I 71 So the solution is 1 1 1 1 Soui2andz72 1 5 11Theny 213 42gt y1and 3 1110 z3w2 U 9 3 w0 z2w1 Sow1anclz10 ltgt z 1Theny271312 gt y1 33 Let 2 be the amount invested at 4 y the amount invested at 5 and z the amount invested at 6 We set up Total money 5 y z 100000 a model and get the following equations Annual income 004 0051 0062 0051 100000 gt Equal amounts 111 y 2 100000 7 100000 5 y z 100000 41 5y 6239 2 510000 5 110000 gt y 22 110000 So an 7 y 0 72y z 100000 33 120000 3 40000 and y 2 40000 110000 ltgt y 30000 Since 0 y I 30000 She must invest 30000 in shortterm bonds 30000 in intermediatetenn bonds and 40000 in longterm bonds 516 CHAPTER 9 Systems of Equations and Inequalities Total money I y z 100000 004m 006y 008z 6700 lt 9 7 100000 the following equations 43 6y 82 670000 Annual income Equal amounts y z y z 0 1 7 z 100000 J y Z 100000 lt3 23 42 270000 r 2y 4z 270000 SO Z 45000 and y Z 2 45000 Since y z z 0 62 270000 as 45000 45000 100000 Q 1 10000 She must invest 10000 in shortterm bonds 45000 in intermediateterm bonds and 45000 in longterm bonds 35 Leta l7 and c be the number of ounces of Type A Type B and Type C pellets used The requirements for the different 2a3b c 9 2a3b c 9 vitamins gives the following system 3a bSe14 ltgt 77174 30 1 8a5b7c32 7b30274 Equations 2 and 3 are inconsistent so there is no solution 1112 130 11I2 130 1112 130 36 1611 812 4 ltgt 42412 1613 4 812 413 5 812 413 5 812 413 5 2813 19 Sorg eo68and8124 5 e 12z029Thenh7 0 es n039 37 Let x y and 2 be the number of acres of land planted with corn wheat and soybeans We set up a model and I Total acres 2 y z 1200 get the following equations Market demand 23 y Substituting 20 for y we get Total cost 451 60y 50 63750 I2ft 2 1200 31 z 1200 31 z 1200 q 21 y 21 y 0 ltgt 2x 7 y 0 So 4531 60 2x 502 63750 1652 50 2 63750 1527 3750 15 3750 egt y 250500z 1200 gt of soybeans 2 50 and y 2 250 500 Substituting into the original equation we have 2 450 Thus the Farmer should plant 250 acres ofcom 500 acres ofwheat and 450 acres 38 Let a b and c be the number of shares of Stock A Stock l3 and Stock C in the investor s portfolio Since the total value remains unchanged we get the following system 100 2517 296 74000 10o 251 29C 74000 10a 251 29c 74000 12a l 2017 32c 74000 ltgt 5017 14c 74000 ltgt 50b 14c 74000 160 l 1517 32c 74000 125b 72c 222000 74c 74000 So 5 1000 Back substituting we have 50b 14 1000 74000 ltgt 50b 2 60000 gt b 1200 And nally 10a 25 1200 29 1000 74000 100 30000 29000 74000 ltgt 10a 15000 cgt o 1500 Thus the portfolio consists 011500 shares ofStock A 1200 shares of Stock B and 1000 shares of Stock C 34 Let m be the amount invested at 4 y the amount invested at 6 and z the amount invested at 8 We set up a model and gm g 2 L i i l i l i l I i SECTlON 94 Systems of Linear Equations Matrices 517 39 a we begin by substituting m0 1 7 and into the lefthand side of the rst equation 2 a1 bx 11110 blyu mm 1111 biyt 61zt d1d1dl Thus the given ordered triple satis es the rst equation We can show that it satis es the second and the third in exactly the same way Thus it is a solution ofthe system b We have shown in part a that if the system has two different solutions we can nd a third one by averaging the two solutions But then we can nd a fourth and a h solution by averaging the new one with each of the previous two Then we can nd Four more by repeating this process with these new solutions and so on Clearly this process can continue inde nitely so there are in nitely many solutions 13x2 22gtlt L 32x1 7 a Yes this matrix is in rowechelon Form 12 Yes this matrix is in reduced rowechelon form 9 a Yes this matrix is in rowechelon form b No this matrix is not in reduced rowechelon form since the leading 1 in the second row does not have a zero above it 1 2y 82 O c y 3 2 O 0 11 a No this matrix is not in rowechelon form since the row of zeros is not at the bottom b No this matrix is not in reducer rowechelon form 139 O c 0 o y 5 13 a Yes this matrix is in rowechelon form b Yes this matrix is in reduced roweehelon form 13y w0 z2w2 0 01 00 Notice that this system has no solution 794 Systems of Linear Equations Matrices 43x1 51X3 62x2 8 a Yes this matrix is in rowechelon form b No this matrix not in reduced rowechelon form The entry above the leading 1 in the second row is not 0 at 31 33 C y 5 10 a Yes the matrix is in rowechelon form b Yes the matrix is in reduced rowechelon form 1 7 0 C 12 a Yes the matrix is in rowechelon form In Yes the matrix is in reduced row echelon form a 1 c y 2 z 3 14 a No this matrix is not in rowechelon form since the fourth column has the leading 1 oftwo rows 3 No this matrix is not in reduced rowechelon form z3y w 0 y 410 0 c wu2 w 0 38 39 SECTlON 94 Systems of Linear Equations Mamces 521 1 2 3 75 73 5 1 Z 73 75 2 74 75 10 RZHRIM Z 0 0 12 0 rm 0 1 7 2 T11eref0re712z0 Rg SRl Rg V 3 7 42 13 U 1 7 2 O 0 12 O ltgt Z 011 T 7 U 2 ltgt y 2 andu 2 e 3 75 lt27 m 791 Hence the solution is 9 2 0 3 1271 1 321 1 321 fix Ra R274R1gtR2 RzHgRa 4 2 1 77 7R1 4 2 1 77 R M u 10 7 711 1 3724 3 1271 0 874 A4 1 3 2 1 1 3 2 1 0 2 1 1 W3 0 2 1 v1 Thus 21 76 gt z 3 O 10 7 711 D O 2 76 2y 3 71 a 2y 2 ltgt y Lands 3123 1 ltgt x 2 Hence the solution is 213 7121733 1721373 12A1373 RniiiR all 3 74 1 1 9 3 74 1 1 9 ganga 0 2 4 78 18 112 71 71 1 1 0 71 71 1 1 0 Rrwmh 0 3 0 4 A3 a 2 1 4 2 3 2 1 4 2 3 0 5 6 78 9 1 72 71 3 3 1 2 1 3 73 1 72 1 3 73 0 1 2 74 9 1131321 0 1 2 74 9 Emlyn 0 1 2 74 9 0 73 0 4 3 R4492 0 0 6 8 24 0 0 6 s 24 0 5 6 78 9 0 U 4 12 736 0 O 0 20 760 y73andx6 973 Therefore 20w 60 gt u 73 62 24 24 ltgt z 0 Then y 12 9 gt ltgt m 0 Hence the solution is 0 73 0 73 11 141 6 117171 6 11 171 6 2 0 1 43 8 gaj ljki2 0 2 3 71 4 Egi jj j 0 0 5 1 72 1 1 0 4 710 beam Mu 0 72 1 5 16 512 0 0 3 7 14 3 5 71 71 20 0 2 2 2 2 0 1 1 1 1 1141 1 6 11 171 6w 0 1 1 1 1 0 1 1 1 M W Thus32w 64 gt 4572 Wm 0 0 5 1 72 0 0 5 1 0 0 3 7 714 0 o o 32 64 52 272 cgt 520 4 z0y0721 y3andm3 Oi26 gtm1 Hence the 501111101115 13072

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