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# ALGEBRA MATH 102

Texas A&M

GPA 3.92

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This 13 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 102 at Texas A&M University taught by Lidia Smith in Fall. Since its upload, it has received 25 views. For similar materials see /class/225996/math-102-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15

68 The determinant of a square nwirix The determinant of a square matrix A is a number that associated With the matrix and is denoted by A or det A If A is a square matrix of order 1 that is A can define det A an A111 012 021 022 the determinant of A is defined by detA an 122 7 121 112 De nition Let A calj be a square matrix of order n gt 1 The minor M of the element aij is the determinant of the matrix of order n 7 1 obtained by deleting row 239 and column j Example Let A be 011 012 013 A 021 022 023 031 032 033 Find the minors M11 M23 M12 De nition Let A calj be a square matrix The cofactor Aij of an element aij is Au UWMir De nition The determinant of a 3 gtlt 3 matrix A calj is defined detA a11A11 a12A12 113A13 The determinant can also be written detA a11M11 7 a12M12 113MB The computation of Mijs leads to det A 011022033 012023031 021032013 031022013 012021033 123032011 Example Compute the determinant of 71 3 1 A 2 5 0 3 1 2 Theorem Rain of Zerox I f every element in a row or column of a square matrix A zero tlten det A 0 Theorem I f A a square matriaz tlten A invertible if and only if detA 7 0 Cramer s rule Given a linear system in two variables 7 011 1129 b1 021 1229 52 if we let 011 012 b1 012 011 b1 D 7 D 7 D 7 J 021 022 J m J 52 022 J y J 52 J the solution to the system is given by Example Solve the system 39adqs mg pug pm g gg pm 3 147 q nmq any am 30 qdm am mags 39Vg aldmexg 39paugap 301 s 7 30 adqs mg mg sgxzz Ii am 03 maxed s 7 31 z Hg afitmqa W 4 Zx ug a tmqa 4 W 4 Zi 4 w sg 7 30 w adqs aql 397 no suxgod pupsgp aq GHQ pm Zz Iz1d mi pmz sgxzz Ii am 03 maxed 301 S 3219 any p aq 331 gun I 30 adols 911 99 uogqgugaq I 4 zg 30 qdm am 141 39uommba xmug p sg amp 38m paqamgs am qdm asoqm 4 zg uopmxba aql 39z39g aldmexg 393ng p s3tmsaxdax uogwnba xmug p 30 qdm aql 39samngtxa am ug 338 mm am u Immm aq um mantbe xmug p 31219 sumo umpmmba mqm aw amql M 9 a q 29 mm 9 q am mm 319 30 uogmnba 1w 8 FL pm at u uogmnba xmug V 3939g uogqgugaq 2 De nition 35 A horizontal line is a line parallel to the p axis It has equation y b and slope 0 A vertical line is a line parallel to the y axis It has equation x a and its slope is not defined Example 36 Find the equations of the horizontal and vertical line through C7374 De nition 37 The pointslope form for the equation of a line The equation of a line through the point zhyl With slope m is y iyl mz 7 1 Example 38 Find the equation of the line through the points A7l7 73 and B23 De nition 39 The slopeintercept form for the equation of a line The equation of a line With slope m and y intert ept b is y mp b Example 310 Write the equation of the line through the points A7l7 73 and B23 in slope intercept form Theorem 311 Two non ye39rtleal lines are parallel if and only if they have the same slope Two lines with slopes m1 and 7712 are perpendicular if and only if 77117712 62 Systems of linear equations IN TWO VARIABLES Recall that a linear equation in z and y is a equation that can be written in form ax by c where 17 b7 0 are real numbers with a and b not both zero Similarly the equation ax by 02 d is a linear equation in three variables Ly and z A system of linear equations consists of two or more linear equations A solution of the system has to satisfy all the equations sirmiltaneously Systei ns of equations are equivalent systems provided they have the same solutions Theorem Given a system of equations an equivalent system results if 1 two equations are interchanged 2 an equation is i nultiplied or divided by a nonzero constant 3 a constant multiple of one equation is added to another equation The method of elimination This method consists of eliminating one of the variables from one of the equations We can do that by i nultiplying one equation by a number or i nultiplying both equations each by different numbers such that when we add the resulting equations one of the variables cancels Example Solve the system z By 71 2x 7 y 5 Solution We will use transformations to obtain equivalent systems z 3 71 y multiply second equation by 3 2x 7 y 5 x 3 71 i y add the first equation to the second 6x 7 3y 15 z By 71 7x 14 From the second equation we get m 2 and the corresponding y value can be obtained by substituting z 2 in the first equation We obtain y 71 Thus the solution of the system is 2 71 39ammzd aw suogwnba am 3tmsaxdax 3mg sang om am Snugqupxg 39uogmxos ou seq 1933953 319 anl quammms arm 9 ppmquot mth 3219 9 26 xgzzd on s mam mugs uogmxos ou seq 1 398 FL 0 26 0 U633ng aq um uogmnba 38m aql 8 0 9 Ii 968 saAg 391 mp mm 3g 8IX 33RX3QHS pm 2 Sq uopmxba pumas mp 8mmqu 39uozgnyog 0a z 969 9 Ii 968 mmsxs mg was 39suonnxos ou mm suogmxba many 30 HMSKS V aldmexg 39umszxk am 03 summing aw 9 zg any am no smgod up up Sum Idxf 11338th am 03 uoptqos aw 9 FL avg uopmxba am 30 mamas am up anl 9 Ii 968 9 Ii 968 mng 2 Sq uopmxba pumas mp 8mmqu 39uozgnyog a1 z 969 9 Ii 968 mmsxs mg was 39SIX 3HS 30 mqmnu amxgug w mm Suonmxba xmug 30 mms s V aldmexg Z 43 Zeros of Polynomials The zeros ofa polynomial f are the solutions of the equation f 0 Each real zero is an zintercept of the graph if f Theorem Fundamental Theorem of Algebra Ifa polynomial f has positive degree and complex coef cients then f has at least one complex zero Theorem Complete Factorization Theorem If f is a polynomial if degree n gt 0 then there exist 71 complex numbers 017 027 7cn such that f az 7 cl x 7 on7 where a is the leading coef cient of f Each number ck is a zero of f Example Find a polynomial in factored form that has degree 3 has zeros 27 71 and 3 and satis es f1 4 De nition If a factor x 7 0 appears m times in the factorization of fz then C is called a zero of multiplicity m of the polynomial f z or a root of multiplicity m of the equation f 0 De nition 61 An inequality is a statement that two quantities or expressions are not equal An inequality can be expressed using one of the relations lt S gt 2 relation reads as inequality example reads as lt less than a lt b a is le s tl1an b 3 less than or equal to a S b a is le s than or equal to b gt greater than a gt b a is greater than b 2 greater than or equal to a 2 b a is lgreater than or equal to b De nition 62 Solving an inequality that involves a variable x means nding all of its solutions A solution for a inequality in a variable x is a number b that yields a true statement when substituted for z Equivalent inequalities are inequalities that have exactly the same solutions Example 63 Consider the inequality 2x 3 gt 11 We have that the number 5 is a solution for the given inequality but 1 is not since 2z3gtn 1 false statement 13 gt 11 true statement Most inequalities have an in nite number of solutions For example the solutions of the inequality 2 lt z lt 5 consist of every real number between 2 and 5 We call this set of numbers an open interval and denote it by 25 Intervals Methods for solving inequalities The methods for solving inequalities in z are similar to those used for solving equations with one exception When multiplying or dividing an inequality by a negative number we have to reverse the inequality Properties of inequalities 0 we can add or subtract a real number to both sides of an inequality and we obtain an equivalent inequality 0 we can multiply or divide both sides of an inequality by a positive real number and we obtain an equivalent inequality 1Ifaltbtl1en acltbc and aicltbic b 2 If a lt b and c gt 0 then a 0 lt b 0 and lt E b 3 If a lt b and c lt 0 then a 0 gt b 0 and gt 2 Remark 64 We cannot multiply both sides by an expression cotaining the variable x if we cannot say that the expression is either always positive or always negative Example 65 Solve the inequality 5x 7 7 S 3 Solution Given the properties of inequality we obtain the following equivalent inequali ties 5x 7 7 S 3 given 5x777 37 add7 5x 3 10 simplify 5 10 7 S 7 divide by positive 5 5 x g 2 simplify So the solutions to the given inequality are all numbers such that x S 2 This is the interval 7007 2 39 Example 66 Solve the inequality 73z 7 7 lt 13 Solution Given the properties of inequality we obtain the following equivalent inequali ties 73z 7 7 lt 13 given 73 7 7 7 7 lt 13 7 7 subtract 7 73z lt 6 simplify 73 6 J gt 7 divide by 73 negative 73 73 z gt 72 simplify So the solutions to the given inequality are all numbers such that z gt 72 This is the interval 727 00 D Remark 67 We can avoid multiplying by the negative number 73 if we start by adding 3x both sides Thus we obtain 73z 7 7 lt 13 given 7 lt 3x 13 add 3x 7 7 13 lt 3x subtract 13 76 lt 3x divide by 3 positive 72 lt z D Example 68 Solving a continued inequality Solve the inequality and express the solutions in terms of intervals 73 3 2x 7 5 lt 7 Solution A number x is a solution of the given inequality if and only if 7332x75 and 2x75lt7 We can either work each inequality separately or solve both simultaneously as follows 7 z 7 5 lt 7 given 73 2z75lt7 add5 73 5 3 2x lt 7 5 subtract 13 2 3 2x lt 12 divide by 2 positive 1 S x lt 6 So the solutions to the inequality are all numbers in the halfopen interval 17 6 Properties of absolute value The equation lt 3 is equivalent to 73 lt z lt 3 thus its solutions consist of the interval 73 In general the equation zl lt a Where a gt 0 is equivalent to 7a lt z lt a thus its solutions consist of the interval 7a7 a and the equation gt a Where a gt 0 is equivalent to z lt 7a or z gt a thus its solutions consist of the union of intervals 7007 a U a7 00 Example 69 Solving an inequality containing an absolute value Solve the inequality and express the solutions in terms of intervals lz 7 3 lt 1 lz 7 31 lt 1 given 71 lt z 7 3 lt 1 property of absolute value Solution 71 3 lt z lt 1 3 isolate x by adding 3 2 lt z lt 4 divide by 2 positive Thus the solutions are the real numbers in the open interval 24 Example 610 Solving an inequality containing an absolute value Solve the inequality and express the solutions in terms of intervals 12 3 gt 9 12 31 gt 9 given 2x 3 lt 79 or 2x 3 gt 9 property of absolute value Solution 2x lt 712 or 2x gt 6 subtract 3 z lt 76 or z gt 3 divide by 2 positive Thus the solutions are the real numbers in the union of intervals 7007 76 U 37 00 Example 611 Solving a rational 39 l quot u more complicated of these using a sign diagram in section Solve the inequality and express the solutions in terms of intervals gt 0 7 3 Solution Since the numerator equals 5 Which is a positive number for the fraction to be positive we need that the denominator z 7 3 is also positive Thus z 7 3 gt 0 or equivalently z gt 3 and the solutions are all numbers in the interval 37 00 41 Polynoi nial Functions amp Their Graphs De nition A function f is a polynoi nial function if f is a polynoi nial that is if fI anxquot an71n71 a196 107 where the coef cients are real numbers and the exponents are non negative integers If an 7 0 we say that f has degree 71 Example The polynomial p 3 has degree 0 The polynoi nial qx 4x 7 7 has degree 1 The polynoi nial x 2 x has degree 2 and the polynoi nial 5x 2x3 7 6x2 7 10 has degree 3 Graphs of Polynomials o The graphs of polynomials of degree 0 or 1 are lines o The graphs of polynomials of degree 2 are parabolas o If n is even the polynomial f x has the same general graph y 2 As the degree 71 gets larger the graphs become flatter around the origin and steeper elsewhere If n is odd the polynoi nial f x has the same general graph y 3 As the degree 71 gets larger the graphs become flatter around the origin and steeper elsewhere The graph of a polynomial function is always a smooth curve it has no breaks or corners Next are given the graphs of y x3 and y 4 Theorem Intermediate value theorem If f is a polynomial function and f a 31 f b for a lt b then f takes every value between f a and f b in the interval 11 De nition If f is a polynoi nial function then c is called a zero of f if f c 0 Example Using the interi nediate value theorei n Show that f 3 2x2 7 6 has a zero between 1 and 2 f1127673 f2 887610 signs there exists at least one real number 0 between 1 and 2 such that f c 0 1 Solution We have Since f 1 and f 2 have opposite Exercise Sketch the graph of the function by transforming the graph of an appro priate function of the form y x Indicate all z7 and y7intercepts on each graph f 3 7 27 b f 7x 714 16 Using Zeros and Sign Charts to Graph Polynomials If f is a poiynoi niai function then c is called a zero of f if f c 0 In other words the zeros of f are the solutions of the poiynoi niai equation f 0 Note that if f c 0 then the graph of f has an z7intercept at z c so the z7intercepts of the graph are the zeros of the function Exercise Sketch the graph of the polynomial function a f96961962 b 996962962961 Exercise Factor the poiynoi niai and use the factored form to find the zeros Then sketch the graph a pz 3 7 2 7 6x 1 la 4 7 as 3W2

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