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P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 CHAPTER 1 Matrices Objectives To be able to identify when two matrices areequal To be able to add and subtract matrices of thesame dimensions To be able to perform multiplication of a matrix and ascalar To be able to identify when the multiplication of two given matrices is possible To be able to performmultiplicationon two suitable matrices To be able to find theinverseof a2×2matrix To be able to find thedeterminantof a matrix To be able to solvelinear simultaneous equationsin two unknowns using an inverse matrix 1.1 Introduction to matrices Amatrixis a rectangular array of numbers. The numbers in the array are called the entries in the matrix. The following are examples of matrices: ⎡ ⎢ ⎣−12 −34 56⎤ ⎥ ⎦[ 2156]⎡ ⎢ ⎣√ 23 00 1 √ 20⎤ ⎥ ⎦[ 5] Matrices vary in size. The size, ordimension, of the matrix is described by specifying the number of rows (horizontal lines) and columns (vertical lines) that occur in the matrix. The dimensions of the above matrices are, in order: 3×2,1×4,3×3,1×1. The first number represents the number of rows and the second, the number of columns. 1

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 2 Essential Advanced General Mathematics Example 1 Write down the dimensions of the following matrices. a 112 210 b⎡ ⎢ ⎢ ⎢ ⎣1 2 3 4⎤ ⎥ ⎥ ⎥ ⎦c 223 Solution a2×3b4×1c1×3 The use of matrices to store information is demonstrated by the following two examples. Four exportersA,B,CandDsell televisions (t), CD players (c), refrigerators (r) and washing machines (w). The sales in a particular month can be represented by a 4×4arrayof numbers. This array of numbers is called a matrix. rcwt A B C D⎡ ⎢ ⎢ ⎢ ⎣120 95 370 250 430 380 1000 900 60 50 150 100 200 100 470 50⎤ ⎥ ⎥ ⎥ ⎦row 1 row 2 row 3 row 4 column 1 column 2 column 3 column 4 From the matrix it can be seen: ExporterAsold 120 refrigerators, 95 CD players, 370 washing machines and 250 televisions. ExporterBsold 430 refrigerators, 380 CD players, 1000 washing machines and 900 televisions. The entries for the sales of refrigerators are made in column 1. The entries for the sales of exporterAare made in row 1. The diagram on the right represents a section of a road map. The number of direct connecting roads between towns can be represented in matrix form. ABC D A B C D⎡ ⎢ ⎢ ⎢ ⎣0211 2010 1100 1000⎤ ⎥ ⎥ ⎥ ⎦ B A C D IfAis a matrix,a ijwill be used to denote the entry that occurs in rowiand columnjofA. Thus a 3×4 matrix may be written A=⎡ ⎢ ⎣a 11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 ⎤ ⎥ ⎦

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 3 ForB,anm×nmatrix B=⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣b 11 b12 .....b 1n b21 b22 .....b 2n .. .. .. .. .. b m1 bm2 .....b mn ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Matrices provide a format for the storage of data. In this form the data is easily operated on. Some graphics calculators have a built-in facility to operate on matrices and there are computer packages which allow the manipulation of data in matrix form. A car dealer sells three models of a certain make and his business operates through two showrooms. Each month he summarises the number of each model sold by a sales matrixS: S= s 11 s12 s13 s21 s22 s23 , wheres ijis the number of cars of modeljsold by showroomi. So, for example,s 12 is the number of sales made by showroom 1, of model 2. If in January, showroom 1 sold three, six and two cars of models 1, 2 and 3 respectively, and showroom 2 sold four, two and one car(s) of models 1, 2 and 3 (in that order), the sales matrix for January would be: S= 362 421 A matrix is, then, a way of recording a set of numbers, arranged in a particular way. As in Cartesian coordinates, the order of the numbers is significant, so that although the matrices  12 34 , 34 12 have the same numbers and the same number of elements, they are different matrices (just as (2, 1), (1, 2) are coordinates of different points). Two matricesA,B, areequal, and can be written asA=Bwhen each has the same number of rows and the same number of columns they have the same number or element at corresponding positions. e.g.⎡ ⎣21−1 01 3⎤ ⎦ =⎡ ⎣1+11−1 1−116 2⎤ ⎦

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 4 Essential Advanced General Mathematics Example 2 If matricesAandBare equal, find the values ofxandy. A= 21 x4 B= 21 −3y Solution x=−3 andy=4 Although a matrix is made from a set of numbers, it is important to think of a matrix as a single entity, somewhat like a ‘super number’. Example 3 There are four rows of seats of three seats each in a minibus. If 0 is used to indicate a seat is vacant and 1 is used to indicate a seat is occupied, write down a matrix that represents athe 1st and 3rd rows are occupied but the 2nd and 4th rows are vacant bonly the seat on the front left corner of the bus is occupied. Solution a⎡ ⎢ ⎢ ⎢ ⎣111 000 111 000⎤ ⎥ ⎥ ⎥ ⎦b⎡ ⎢ ⎢ ⎢ ⎣100 000 000 000⎤ ⎥ ⎥ ⎥ ⎦ Example 4 There are four clubs in a local football league. Team A has 2 senior teams and 3 junior teams Team B has 2 senior teams and 4 junior teams Team C has 1 senior team and 2 junior teams Team D has 3 senior teams and 3 junior teams Represent this information in a matrix. Solution ⎡ ⎢ ⎢ ⎢ ⎣23 24 12 33⎤ ⎥ ⎥ ⎥ ⎦ Note: rows represent teams A, B, C, D and columns represent the number of senior and junior teams respectively.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 5 Exercise1A 1Write down the dimensions of the following matrices. Example 1 a 12 34 b 21−1 01 3 c[ abcd] d⎡ ⎢ ⎢ ⎢ ⎣p q r s⎤ ⎥ ⎥ ⎥ ⎦ 2There are 25 seats arranged in five rows and five columns. If 0, 1 respectively are used to Example 3 indicate whether a seat is vacant or occupied, write down a matrix which represents the situation when aonly seats on the two diagonals are occupied ball seats are occupied. 3If seating arrangements (as in2) are represented by matrices, consider the matrix in which thei,jelement is 1 ifi=j,but0ifi =j. What seating arrangement does this matrix represent? 4At a certain school there are 200 girls and 110 boys in Year 7, 180 girls and 117 boys in Example 4 Year 8, 135 and 98 respectively in Year 9, 110 and 89 in Year 10, 56 and 53 in Year 11 and 28 and 33 in Year 12. Summarise this information in matrix form. 5From the following, select those pairs of matrices which could be equal, and write down Example 2 the values ofx,ywhich would make them equal. a 3 2 , 0 x ,[ 0x],[ 04] b 47 1−2 , 1−2 4x , x7 1−2 ,[ 4x1−2] c 2x4 −1103 , y04 −1103 , 204 −1103 6In each of the following find the values of the pronumerals so that matricesAandBare equal. a A= 21−1 01 3 B= x1−1 01y b A= x 2 B= 3 y cA=[ −3x]B=[ y4] dA= 1y 43 B= 1−2 4x 7A section of a road map connecting townsA,B,C andDis shown. Construct the 4×4 matrix which shows the number of connecting roads between each pair of towns. BD A C

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 6 Essential Advanced General Mathematics 8The statistics for the five members of a basketball team are recorded as follows. Player A: points 21, rebounds 5, assists 5 Player B: points 8, rebounds 2, assists 3 Player C: points 4, rebounds 1, assists 1 Player D: points 14, rebounds 8, assists 60 Player E: points 0, rebounds 1, assists 2 Express this data in a 5×3 matrix. 1.2 Addition, subtraction and multiplication by a scalar Addition will be defined for two matricesonlywhen they have the same number of rows and the same number of columns. In this case the sum of two matrices is found by adding corresponding elements. For example,  10 02 + 0−3 41 = 1−3 43 and⎡ ⎢ ⎣a 11 a12 a21 a22 a31 a32 ⎤ ⎥ ⎦+⎡ ⎢ ⎣b 11 b12 b21 b22 b31 b32 ⎤ ⎥ ⎦=⎡ ⎢ ⎣a 11 +b 11 a12 +b 12 a21 +b 21 a22 +b 22 a31 +b 31 a32 +b 32 ⎤ ⎥ ⎦ Subtraction is defined in a similar way. When the two matrices have the same number of rows and the same number of columns the difference is found by subtracting corresponding elements. Example 5 Find a 10 20 − 2−1 −41 b 23 −14 − 23 −14 Solution a 10 20 − 2−1 −41 = −11 6−1 b 23 −14 − 23 −14 = 00 00 It is useful to definemultiplication of a matrix by a real number.IfAis anm×nmatrix, andkis a real number, thenkAis anm×nmatrix whose elements arektimes the corresponding elements ofA. Thus 3 2−2 01 = 6−6 03 These definitions have the helpful consequence that if a matrix is added to itself, the result is twice the matrix, i.e.A+A=2A. Similarly the sum ofnmatrices each equal toAisnA (wherenis a natural number). Them×nmatrix with all elements equal to zero is called thezero matrix.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 7 Example 6 LetX= 2 4 ,Y= 3 6 ,A= 20 −12 ,B= 50 −24 FindX+Y,2X,4Y+X,X−Y,−3A,−3A+B. Solution X+Y= 2 4 + 3 6 = 5 10 2X=2 2 4 = 4 8 4Y+X=4 3 6 + 2 4 = 12 24 + 2 4 = 14 28 X−Y= 2 4 − 3 6 = −1 −2 −3A=−3 20 −12 = −60 3−6 −3 A+B= −60 3−6 + 50 −24 = −10 1−2 Example 7 IfA= 32 −11 andB= 0−4 −28 ,find matricesXsuch that 2A+X=B. Solution If 2A+X=B,thenX=B−2A ∴ X= 0−4 −28 −2× 32 −11 = 0−2×3−4−2×2 −2−2×−18−2×1 = −6−8 06

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 8 Essential Advanced General Mathematics Using the TI-Nspire 2-by-2 matrices are easiest entered using the 2-by-2 matrix template ( 5 )as shown. Notice that there is also a template for enteringm-by-nmatrices. To enter the matrixA= 36 67 , use the NavPad to move between the entries of the 2-by-2 matrix template andstore( / ) the matrix asa. Define the matrixB= 36 5−6.5 in a similar way. Entering matrices directly To enter matrixAwithout using the template, enter the matrix row by row as [[3,6][6,7]] andstore( / ) the matrix asa. Alternatively, enter the matrix by typing [3,6;6,7]. Semicolon (;) can be obtained by typing / .

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 9 Addition, multiplication and multiplication by a scalar OnceAandBare defined as above,A+B, ABandKAcan easily be determined. Using the Casio ClassPad Matrices are entered using the 2D CALC menu on the k .Tap , enter the numbers required then store this as a variable (using VAR). Calculations can be performed as shown in the screen at the far right. Exercise1B 1LetX= 1 −2 ,Y= 3 0 ,A= 1−1 23 ,B= 40 −12 FindX+Y,2X,4Y+X,X−Y,−3Aand−3A+B. Example 6 2Each showroom of a car dealer sells exactly twice as many cars of each model in February as in January. (See example in section 1.1.) aGiven that the sales matrix for January is 362 421 , write down the sales matrix for February.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 10 Essential Advanced General Mathematics bIf the sales matrices for January and March (with twice as many cars of each model sold in February as January) had been 100 423 and 210 614 respectively, find the sales matrix for the first quarter of the year. cFind a matrix to represent the average monthly sales for the first three months. 3LetA= 1−1 02 Find 2A,−3Aand−6A. 4A,B,Carem×nmatrices. Is it true that aA+B=B+Ab(A+B)+C=A+(B+C)? 5A= 32 −2−2 andB= 0−3 41 Calculate a2Ab3Bc2A +3Bd3B–2A 6P= 10 03 ,Q= −11 20 ,R= 04 11 Calculate aP+QbP+3Qc2P−Q+R 7IfA= 31 −14 andB= 0−10 −217 ,find matricesXandYsuch that Example 7 2A−3X=Band 3A+2Y=2B. 8MatricesXandYshow the production of four modelsa,b,c,dat two automobile factories P,Qin successive weeks. X=P Qab cd  150 90 100 50 1000750 Y=P Qab cd  160 90 120 40 1000500 week 1 week 2 FindX+Yand write what this sum represents.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 11 1.3 Multiplication of matrices Multiplication of a matrix by a real number has been discussed in the previous section. The definition for multiplication of matrices is less natural. The procedure for multiplying two 2×2 matrices is shown first. LetA= 13 42 andB= 51 63 ThenAB= 13 42  51 63 = 1×5+3×61×1+3×3 4×5+2×64×1+2×3 = 23 10 32 10 andBA= 51 63  13 42 = 5×1+1×45×3+1×2 6×1+3×46×3+3×2 = 917 18 24 Note thatAB =BA. IfAis anm×nmatrix andBis ann×rmatrix, then the productABis them×rmatrix whose entries are determined as follows. To find the entry in rowiand columnjofABsingle out rowiin matrixAand columnjin matrixB. Multiply the corresponding entries from the row and column and then add up the resulting products. Note: The productABis defined only if the number of columns ofAis the same as the number of rows ofB. Example 8 ForA= 24 36 andB= 5 3 findAB. Solution Aisa2×2 matrix andBisa2×1 matrix. ThereforeABis defined. The matrixABisa2×1 matrix. AB= 24 36  5 3 = 2×5+4×3 3×5+6×3 = 22 33

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 12 Essential Advanced General Mathematics Example 9 MatrixXshows the number of cars of modelsaandbbought by four dealers,A,B,CandD. MatrixYshows the cost in dollars of modelaand modelb. FindXYand explain what it represents. ab X=A B C D⎡ ⎢ ⎢ ⎢ ⎣31 22 14 11⎤ ⎥ ⎥ ⎥ ⎦Y= 26 000 32 000 a b Solution ab XY=A B C D⎡ ⎢ ⎢ ⎢ ⎣31 22 14 11⎤ ⎥ ⎥ ⎥ ⎦ 26 000 32 000 a b 4×22×1 The matrixXYisa4×1 matrix XY=⎡ ⎢ ⎢ ⎢ ⎣3×26 000+1×32 000 2×26 000+2×32 000 1×26 000+4×32 000 1×26 000+1×32 000⎤ ⎥ ⎥ ⎥ ⎦=⎡ ⎢ ⎢ ⎢ ⎣110 000 116 000 154 000 58 000⎤ ⎥ ⎥ ⎥ ⎦ The matrixXYshows dealerAspent $110 000, dealerBspent $116 000, dealerC spent $154 000 and dealerDspent $58 000. Example 10 ForA= 234 567 andB=⎡ ⎢ ⎣40 12 03⎤ ⎥ ⎦findAB. Solution Aisa2×3 matrix andBisa3×2 matrix. ThereforeABisa2×2 matrix. AB= 234 567 ⎡ ⎢ ⎣40 12 03⎤ ⎥ ⎦ = 2×4+3×1+4×02×0+3×2+4×3 5×4+6×1+7×05×0+6×2+7×3 = 11 18 26 33

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 13 Exercise1C 1IfX= 2 −1 ,Y= 1 3 ,A= 1−2 −13 ,B= 32 11 ,C= 21 11 ,I= 10 01 , Examples 8,10 find the productsAX,BX,AY,IX,AC,CA,(AC)X,C(BX),AI,IB,AB,BA, A 2,B 2,A(CA) andA 2C. 2aAre the following products, of matrices given in1, defined? AY,YA,XY,X 2,CI,XI bIfA= 20 00 andB= 00 −32 ,findAB. 3The matricesAandBare 2×2 matrices, andOis the zero 2×2 matrix. Is the following argument correct? ‘IfAB=O, andA =O, thenB=O’. 4IfL=[2−1],X= 2 −3 , findLXandXL. 5AandBare bothm×nmatrices. AreABandBAdefined and, if so, how many rows and columns do they have? 6Suppose ab cd  d−b −ca = 10 01 . Show thatad−bc=1. What is the product matrix if the order of multiplication on the left-hand side is reversed? 7Using the result of6, write down a pair of matricesA,Bsuch thatAB=BA=Iwhere I= 10 01 . 8Select any three 2×2 matricesA,BandC. CalculateA(B+C),AB+ACand (B+C)A. 9It takes John five minutes to drink a milk shake which costs $2.50, and twelve minutes to Example 9 eat a banana split which costs $3.00. Calculate the product 512 2.50 3.00  1 2 and interpret the result in milk bar economics. Suppose two friends join John. Calculate 512 2.50 3.00  120 211 and interpret the result.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 14 Essential Advanced General Mathematics 10The reading habits of five studentsA,B,C,DandEare shown in the first matrix below where the columnsp,q,r, andsrepresent four weekly magazines. The second matrix shows the cost in dollars of each magazine. Find the product of the two matrices and interpret the result. pq r s A B C D E⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣0011 1011 1000 1111 0101⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦p q r s⎡ ⎢ ⎢ ⎢ ⎣2.00 3.00 2.50 3.50⎤ ⎥ ⎥ ⎥ ⎦ 11LetS= s 11 s12 s13 s21 s22 s23 be the sales matrix for two showrooms selling three models of cars. Heres ijis the number of cars of modeljsold from showroomi. Let the prices of the three models of cars be $c 1,$c 2,$c 3. Call the 3×1 matrix,C=⎡ ⎢ ⎣c 1 c2 c3 ⎤ ⎥ ⎦the price matrix. aFindSC.bWhat is the practical meaning ofSC? cSuppose the car dealer sells both new and used cars and the price of two-year-old used cars for the three models is $u 1,$u 2and $u 3,respectively. Form a new cost matrix C=⎡ ⎢ ⎣c 1 u1 c2 u2 c3 u3 ⎤ ⎥ ⎦ FindSCand state its meaning. dSuppose the car dealer makes 30% profit on his selling of new cars and 25% on used cars. IfV= 0.30 00.25 , what is the meaning ofCV? 1.4 Identities, inverses and determinants for 2×2 matrices Identities A matrix with the same number of rows and columns is called a square matrix. For square matrices of a given dimension, e.g. 2×2, a multiplicative identityIexists. For example, for 2×2 matricesI= 10 01 and for 3×3 matricesI=⎡ ⎢ ⎣100 010 001⎤ ⎥ ⎦

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 15 IfA= 23 14 ,AI=IA=A, and this result holds for any square matrix multiplied by the appropriate multiplicative identity. Inverses Givena2×2 matrixA, is there a matrixBsuch thatAB=BA=I? LetB= xy uv andA= 23 14 ThenAB=Iimplies 23 14  xy uv = 10 01 i.e. 2x+3u2y+3v x+4uy+4v = 10 01 ∴ 2x+3u=1 and 2y+3v=0 x+4u=0y+4v=1 These simultaneous equations can be solved to findx,u,y, andvand henceB. B= 0.8−0.6 −0.20.4 Bis said to be theinverseofAasAB=BA=I. LetAbea2×2 matrix withA= ab cd and letB= xy uv whereBis the inverse ofA. ThenAB=I. In full this is written ax+bu ay+bv cx+ du cy+dv = 10 01 Henceax+bu=1ay+bv=0 cx+du=0cy+dv=1 which form two pairs of simultaneous equations, forx,uandy,vrespectively. Taking thex,upair and eliminatingu,(ad−bc)x=d Similarly, eliminatingx,(bc−ad)u=c These two equations can be solved forxandurespectively providedad−bc =0 i.e.x=d ad−bcandu=c cb−ad=−c ad−bc In a similar way it can be found that y=−b ad−bcandv=−a cb−ad=a ad−bc Therefore the inverse=⎡ ⎢ ⎢ ⎢ ⎣d ad−bc−b ad−bc −c ad−bca ad−bc⎤ ⎥ ⎥ ⎥ ⎦

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 16 Essential Advanced General Mathematics The inverse of a square matrixA, is denoted byA −1. The inverse is unique.ad−bchas a name, thedeterminantofA. This is denoted det(A). i.e.forA= ab cd ,det(A)=ad−bc A2×2 matrix has an inverse only if det(A) =0 A square matrix is said to beregularif its inverse exists. Those square matrices which do not have an inverse are calledsingularmatrices; i.e. for asingularmatrix det(A)=0. Using the TI-Nspire The operation of matrix inverse is obtained by raising the matrix to the power of−1. TheDeterminantcommand is found in theMatrix and Vectormenu ( b 72 ) and used as shown. (ais the matrixA= 36 67 , defined on page 8.) Using the Casio ClassPad The operation of matrix inverse is obtained by entering A ∧−1 in the entry line. The determinant is obtained by entering and highlightingAand tappingInteractive, Matrix-Calc, det.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 17 Example 11 For the matrixA= 52 31 find adet(A)bA −1 Solution adet(A)=5×1−2×3=−1bA −1 =1 −1 1−2 −35 = −12 3−5 Example 12 For the matrixA= 32 16 find adet(A)bA −1 cXifAX= 56 72 dYifYA= 56 72 Solution adet(A)=3×6−2=16 bA −1 =1 16 6−2 −13 cAX= 56 72 Multiply both sides (from the left) byA −1. A −1AX=A −1  56 72 ∴ IX=X=1 16 6−2 −13  56 72 =1 16 16 30 16 0 = 12 10 dYA= 56 72

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 18 Essential Advanced General Mathematics Multiply both sides (from the right) byA −1 YA A −1 =1 16 56 72  6−2 −13 ∴ YI=Y=1 16 24 8 40−8 ∴ Y=⎡ ⎢ ⎢ ⎣3 21 2 5 2−1 2⎤ ⎥ ⎥ ⎦ Exercise1D 1For the matricesA= 21 32 andB= −2−2 32 find Example 11 adet(A)bA −1 cdet(B)dB −1 2Find the inverse of the following regular matrices (is any real number,kis any non-zero real number). a 3−1 4−1 b 31 −24 c 10 0k d cos−sin sincos 3IfA,Bare the regular matricesA= 21 0−1 ,B= 10 31 , findA −1,B −1. Also findABand hence find, if possible, (AB) −1. Also find fromA −1,B −1, the productsA −1B−1 andB −1A−1. What do you notice? 4For the matrixA= 43 21 Example 12 afindA −1 bifAX= 34 16 , findXcifYA= 34 16 , findY. 5IfA= 32 16 ,B= 4−1 22 andC= 34 26 , find aXsuch thatAX+B=CbYsuch thatYA+B=C 6IfAisa2×2 matrix,a 12 =a 21 =0,a 11  =0,a 22  =0, then show thatAis regular and findA −1. 7LetAbe a regular 2×2 matrix,Ba2×2 matrix andAB=0. Show thatB=0. 8Find all 2×2 matrices such thatA −1 =A.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 19 1.5 Solution of simultaneous equations using matrices Inverse matrices can be used to solve certain sets of simultaneous linear equations. Consider the equations 3x−2y=5 5x−3y=9 This can be written as  3−2 5−3  x y = 5 9 IfA= 3−2 5−3 the determinant ofAis 3(−3)−5(−2)=1 which is not zero and soA −1 exists. A −1 = −32 −53 Multiplying the matrix equation 3−2 5−3  x y = 5 9 on the left hand side byA −1 and using the fact thatA −1A=Iyields the following: A −1 A x y =A −1  5 9 ∴ I x y =A −1  5 9 ∴  x y = 3 2 sinceA −1  5 9 = 3 2 This is the solution to the simultaneous equations. Check by substitutingx=3,y=2 in the equations. When dealing with simultaneous linear equations in two variables which represent parallel straight lines, a singular matrix results. For example the system x+2y=3 −2x−4y=6 has associated matrix equation  12 −2−4  x y = 3 6

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 20 Essential Advanced General Mathematics Note that the determinant of 12 −2−4 =1×−4−(−2×2)=0. There is no unique solution to the system of equations. Example 13 IfA= 2−1 12 andK= −1 2 , solve the systemAX=KwhereX= x y . Solution IfAX=K,thenX=A −1K A −1K=1 5 21 −12 × −1 2 = 0 1 ∴ X= 0 1 Example 14 Solve the following simultaneous equations. 3x−2y=6 7x+4y=7 Solution The matrix equation is 3−2 74  x y = 6 7 LetA= 3−2 74 ThenA −1 =1 26 42 −73 and x y =1 26 42 −73  6 7 =1 26 38 −21 Since any linear system ofnequations innunknowns can be written as AX=K,whereAis ann×nmatrix,X=⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣x 1 x2 . . x n ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦andK=⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣k 1 k2 . . k n ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦,

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Chapter 1 — Matrices 21 this method can be applied more generally whenAis regular. In fact, as shown, an expression for the solution can be written at once. MultiplyAX, andK, on the left byA −1, and A −1(AX)=A −1KandA −1(AX)=(A −1A)X=IX=X. HenceX=A −1K, which is a formula for the solution of the system. Of course it depends on the inverseA −1 existing, but onceA −1 is found then equations of the formAX=Kcan be solved for all possiblen×1 matricesK. Again this process can be completed using a calculator as long as matricesAandKhave been entered onto the calculator. Example 15 Consider the system of five equations in five unknowns. 2a+3b−c+d+2e=9 a+b−c=4 a+2d−3e=4 −b+2c−d+e=−6 a−b+d−2e=0 Use a graphics calculator to solve fora,b,c,dande. Solution Enter 5×5 matrixAand 5×1 matrixBinto the graphics calculator. A=⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣23−112 11−100 1002−3 0−12−11 1−101−2⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦B=⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣9 4 4 −6 0⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ThenA −1B=⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣4 9 23 9 −1 7 9 −2 3⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ∴ a=4 9,b=23 9,c=−1,d=7 9ande=−2 3 It should be noted that just as for two equations in two unknowns, there is a geometric interpretation for three equations in three unknowns. There is only a unique solution if the equations represent three planes intersecting at a point.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 22 Essential Advanced General Mathematics Exercise1E 1IfA= 3−1 4−1 , solve the systemAX=KwhereX= x y , and Example 13 aK= −1 2 bK= −2 3 2IfA= 31 −24 , solve the systemAX=Kwhere aK= 0 1 bK= 2 0 3Use matrices to solve the following pairs of simultaneous equations. Example 14 a−2x+4y=6 3x+y=1b−x+2y=−1 −x+4y=2 c2x+5y=−10 y=x+4d1.3x+2.7y=−1.2 4.6y−3.5x=11.4 4Use matrices to find the point of intersection of the lines given by the equations 2x−3y=7 and 3x+y=5. 5Two children spend their pocket money buying some books and some CDs. One child spends $120 and buys four books and four CDs. The other child buys three CDs and five books and spends $114. Set up a system of simultaneous equations and use matrices to find the cost of a single book and a single CD. 6Consider the system 2x−3y=3 4x−6y= 6 aWrite this system in matrix form, asAX=K. bIsAa regular matrix? cCan any solutions be found for this system? dHow many pairs does the solution set contain? 7Consider the system of four equations in four unknowns. Example 15 p+q−r−s=5 r+s=1 2p−q+2r=−2 p−q+s=0 Use a graphics calculator to solve forp,q,rands.

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Review Chapter 1 — Matrices 23 Chapter summary Amatrixis a rectangular array of numbers. Two matricesAandBare equal when: each has the same number of rows and the same number of columns, and they have the same number or element at corresponding positions. The size ordimensionof a matrix is described by specifying the number of rows (m) and the number of columns (n). The dimension is writtenm×n. Addition will be defined for two matrices only when they have the same dimension. The sum is found by adding corresponding elements.  ab cd + ef gh = a+eb+f c+gd+h Subtraction is defined in a similar way. IfAis anm×nmatrix andkis a real number,kAis defined to be anm×nmatrix whose elements arektimes the corresponding element ofA. k ab cd = ka kb kc kd IfAis anm×nmatrix andBis ann×rmatrix, then the productABis them×rmatrix whose entries are determined as follows. To find the entry in rowiand columnjofAB, single out rowiin matrixAand columnjin matrixB. Multiply the corresponding entries from the row and column and then add up the resulting products. The productABis defined only if the number of columns ofAis the same as the number of rows ofB. IfAandBare square matrices of the same dimension andAB=BA=IthenAis said to the inverse ofBandBis said to be the inverse ofA. IfA= ab cd thenA −1 =⎡ ⎢ ⎢ ⎣d ad−bc−b ad−bc −c ad−bca ad−bc⎤ ⎥ ⎥ ⎦ det(A)=ad−bcis thedeterminantof matrixA. A square matrix is said to beregularif its inverse exists. Those square matrices which do not have an inverse are calledsingularmatrices. Simultaneous equations can be solved using inverse matrices, for example ax+by=c dx+ey=f can be written as ab de  x y = c f and x y = ab de −1  c f

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Review 24 Essential Advanced General Mathematics Multiple-choice questions 1The matrixA=⎡ ⎢ ⎢ ⎢ ⎣10 2−1 −23 30⎤ ⎥ ⎥ ⎥ ⎦has dimension A8B4×2C2×4D1×4E3×4 2IfA= 20 −13 andB= 1−34 −1−3−1 thenA+B= A 3−3 −20 B 34 −22 C −12 23 D 21 1−3 ECannot be determined 3IfC= 2 −31 10−2 andD= 1−31 23−1 thenD−C= A 100 −1−3−1 B 2−64 −20−4 C −100 131 D 1−60 131 ECannot be determined 4IfM= −40 −2−6 then−M= A −40 −2−6 B 0−4 −6−2 C 40 −2−6 D  04 62 E 40 26 5IfM= 02 −31 andN= 04 30 then 2M−2N= A 00 −92 B 0−2 −61 C 0−4 −12 2 D 04 12−2 E 02 6−1 6IfAandBare bothm×nmatrices, wherem =n, thenA+Bis an Am×nmatrixBm×mmatrixCn×nmatrix D 2m×2nmatrixECannot be determined 7IfPis anm×nmatrix, andQis an×pmatrix, the dimension of matrixQPis An×nBm×pCn×pDm×nECannot be determined 8The determinant of matrixA= 22 −11 is A4B0C−4D1E2 9The inverse of matrixA= 1−1 1−2 is A−1B 21 −1−1 C 11 −1−2 D 11 −12 E 2−1 1−1

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Review Chapter 1 — Matrices 25 10IfM= 0−2 −31 andN= 02 31 thenNM= A 0−4 −91 B −4−2 2−8 C 04 91 D −62 −3−5 E 6−2 −3−5 Short-answer questions (technology-free) 1IfA= 10 23 andB= −10 01 , find a(A+B)(A−B)bA 2−B 2 2Find all possible matricesAwhich satisfy the equation 34 68 A= 8 16 . 3LetA= 12 3−1 ,B=[ 3−12],C= 6 1 ,D= 24 andE=⎡ ⎢ ⎣5 0 2⎤ ⎥ ⎦. aState whether or not each of the following products exist:AB,AC,CD,BE bEvaluateDAandA −1. 4IfA= 1−21 −512 ,B=⎡ ⎢ ⎣1−4 1−6 3−8⎤ ⎥ ⎦andC= 12 34 ,evaluateABandC −1. 5Find the 2×2 matrixAsuch thatA 12 34 = 56 12 14 6IfA=⎡ ⎢ ⎣200 002 020⎤ ⎥ ⎦,findA 2and henceA −1. 7If 12 4x is a singular matrix, find the value ofx. 8aIfM= 2−1 13 ,find the value of iMM=M 2 iiMMM=M 3 iiiM −1 bFindxandygiven thatM x y = 3 5

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Review 26 Essential Advanced General Mathematics Extended-response questions 1A= 31 1−4 ,B= 2−1 52 aFind iA+BiiA−Biii2A+3BivCsuch that3A+2C=B bFind iABiiA −1 iiiXsuch thatAX=BivYsuch thatYA=B 2IfA=⎡ ⎢ ⎣1−22 20−1 134⎤ ⎥ ⎦,B=⎡ ⎢ ⎣−20 1 42−2 13 3⎤ ⎥ ⎦andC⎡ ⎢ ⎣20 2 30−1 13 1⎤ ⎥ ⎦, find aABbACcBC dXsuch thatAX=CeYsuch thatYA=B fXsuch thatAXC=CBgYsuch thatCYA=BA 3aConsider the system of equations 2x−3y=3 4x+y =5 iWrite this system in matrix form, asAX=K. iiFind detAandA −1. iiiSolve the system of equations. ivInterpret your solution geometrically. bConsider the system of equations 2x+y=3 4x+2y=8 iWrite this system in matrix form, asAX=K. iiFind detAand explain whyA −1 does not exist. cInterpret your findings in partbgeometrically. 4The final grades for Physics and Chemistry are made up of three components, tests, practical work and exams. Marks out of 100 are awarded in each component each semester. Wendy scored the following marks in each of the three components for Physics. Semester 1: tests 79, practical work 78, exam 80 Semester 2: tests 80, practical work 78, exam 82 aRepresent this information in a 2×3 matrix. To calculate the final grade for each semester the three components are weighted so that tests are worth 20%. Practical work is worth 30% and the exam is worth 50%. (cont’d)

P1: FXS/ABE P2: FXS 9780521740494c01.xml CUAU033-EVANS October 5, 2008 6:21 Review Chapter 1 — Matrices 27 bRepresent this information in a 3×1 matrix. cCalculate Wendy’s final grade for Physics in each semester. Wendy also scored the following marks in each of the three components for Chemistry. Semester 1: tests 86, practical work 82, exam 84 Semester 2: tests 81, practical work 80, exam 70 dCalculate Wendy’s final grade for Chemistry in each semester. Students who gain an aggregate score for Physics and Chemistry of 320 or more over the two semesters are awarded a Certificate of Merit in Science. eWill Wendy be awarded a Certificate of Merit in Science? She asks her teacher to remark her Semester 2 Chemistry Exam hoping that she will gain the necessary marks to be awarded a Certificate of Merit. fHow many extra marks does she need? 5A company runs Computing classes and employs full-time and part-time teaching staff as well as technical support staff, cleaners and catering staff. The number of staff employed depends on demand from term to term. In one year they employed the following teaching staff: Term 1: full-time 10, part-time 2 Term 2: full-time 8, part-time 4 Term 3: full-time 8, part-time 8 Term 4: full-time 6, part-time 10 aRepresent this information in a 4×2 matrix. Full-time teachers are paid $70 per hour and part-time teachers are paid $60 per hour. bRepresent this information in a 2×1 matrix. cCalculate the cost per hour to the company for teaching staff for each term. In the same year they also employed the following support staff Term 1: technical staff 2, catering staff 2, cleaning staff 1. Term 2: technical staff 2, catering staff 2, cleaning staff 1. Term 3: technical staff 3, catering staff 4, cleaning staff 2. Term 4: technical staff 3, catering staff 4, cleaning staff 2. dRepresent this information in a 4×3 matrix. Technical staff are paid $60 per hour, catering staff $55 per hour and cleaners are paid $40 per hour. eRepresent this information in a 3×1 matrix. fCalculate the cost per hour to the company for support staff for each term. gCalculate the total cost per hour to the company for teaching and support staff for each term.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 CHAPTER 2 Algebra I Objectives To express a number instandard form To solvelinear equations To solve problems with linear equations andsimultaneous linear equations To usesubstitution and transpositionwith formulas To add and multiply algebraic fractions To solveliteral equations To solvesimultaneous literal equations 2.1 Indices In this section a review of indices is undertaken. Review of index laws am×a n=a m+n am÷a n=a m−n (am)n=a mn a−n =1 an n√a=a 1n (ab) n=a nbn a0=1 Example 1 Simplify each of the following. ax 2×x 3 bx 4 x2 cx 12÷x 45 d(x 3)12 Solution ax 2×x 3=x 2+3 =x 5 bx 4 x2=x 4−2 =x 2 cx 12÷x 45=x 12−4 5=x −310 d(x 3)12=x 32 28

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 29 Example 2 Evaluate a125 23 b 1000 27 23 Solution a125 23= (125) 132=5 2=25 b 1000 27 23=⎛ ⎝ 1000 27 13 ⎞ ⎠2 = 10 3 2 =100 9 Example 3 Simplify 4 x2y3 x12y23 . Solution 4 x2y3 x12y23 =(x 2y3)14 x12y23 =x 24y34 x12y23 =x 24−1 2y34−2 3 =x 0y112 =y 112 Exercise2A 1Simplify each of the following using the appropriate index laws. ax 3×x 4 Example 1a ba 5×a −3 cx 2×x −1 ×x 2 dy 3 y7 Example 1b ex 8 x−4 fp −5 p2 ga 12÷a 23 Example 1c h(a −2)4 i(y −2)−7 j(x 5)3 k(a −20 )35 l x −12−4 m(n 10)15 n2x 12×4x 3 o(a 2)52×a −4 p1 x−4 q 2n −255÷(4 3n4)rx 3×2x 12×−4x −32 s(ab 3)2×a −2b−4 ×1 a2b−3 t(2 2p−3 ×4 3p5÷(6p −3))0

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 30 Essential Advanced General Mathematics 2Evaluate each of the following. Example 2 a25 12 b64 13 c 16 9 12 d16 −12 e 49 36 −12 f27 13 g144 12 h64 23 i9 32 j 81 16 14 k 23 5 0 l128 37 3Use your calculator to evaluate each of the following, correct to two decimal places. a4.35 2 b2.4 5 c√ 34.6921d(0.02) −3 e 3√0.729f 4√2.3045 g(345.64) −13 h(4.568) 25 i1 (0.064) −13 4Simplify each of the following, giving your answer with positive index. aa 2b3 a−2b−4 b2a 2(2b) 3 (2a) −2b−4 ca −2b−3 a−2b−4 da 2b3 a−2b−4 ×ab a−1b−1 e(2a) 2×8b 3 16a −2b−4 f2a 2b3 8a −2b−4 ÷16ab (2a) −1b−1 5Write2 n×8 n 22n ×16in the form 2 an+b . 6Write 2 −x ×3 −x ×6 2x ×3 2x ×2 2x as a power of 6. 7Simplify each of the following. a2 13×2 16×2 −23 ba 14×a 25×a −110 c2 23×2 56×2 −23 d 2 132× 2 125 e 2 132×2 13×2 −25 8Simplify each of the following. Example 3 a 3√a3b2÷ 3√a2b−1 b√ a3b2×√ a2b−1 c 5√a3b2× 5√a2b−1 d√ a−4b2×√ a3b−1 e√ a3b2c−3 ×√ a2b−1c−5 f 5√a3b2÷ 5√a2b−1 g√ a3b2 a2b−1c−5 ×√ a−4b2 a3b−1 ×√ a3b−1 2.2 Standard form Often when dealing with real world problems, the numbers involved may be very small or very large. For example, the distance from the Earth to the Sun is approximately 150 000 000 kilometres, and the mass of an oxygen atom is approximately 0.000 000 000 000 000 000 000 026 grams. In order to deal with such numbers, a more convenient way to express them can be used. This involves expressing the number as a product of a number between 1 and 10 and a power of ten and is calledstandard formorscientific notation.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 31 These examples written in standard form would read 1.5×10 8kilometres and 2.6×10 −23 grams respectively. Performing multiplication and division with very small or very large numbers can often be simplified by first converting the numbers into standard form. When simplifying algebraic expressions or manipulating numbers in standard form, a sound knowledge of the index laws is essential. Example 4 Write each of the following in standard form. a3 453 000b0.00675 Solution a3 453 000=3.453×10 6 b0.00675=6.75×10 −3 Example 5 Find the value of32 000 000×0.000 004 16 000. Solution 32 000 000×0.000 004 16 000=3.2×10 7×4×10 −6 1.6×10 4 =12.8×10 1 1.6×10 4 =8×10 −3 =0.008 Example 6 Evaluate 5√a b2 ifa=1.34×10 −10 andb=2.7×10 −8. Solution 5√a b2 = 5√1.34×10 −10 (2.7×10 −8)2 =(1.34×10 −10 )15 2.7 2×(10 −8)2 =1.454 43··· ×10 13 =1.45×10 13 to three significant figures. Many calculators have scientific notation. The actual display will vary from calculator to calculator. For example, in standard form 3 245 000=3.245×10 6may appear as 3.245 E6or3.245 06.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 32 Essential Advanced General Mathematics Using the TI-Nspire The TI-Nspire can be set to express answers in standard form by selectingDocument SettingsorSystem Settingsfrom 8: Systems Info. The number 3 245 000 will then appear as 3.245E6. The number of significant figures can also be set through these menus. For example, if two significant places are selected (Float 2), 3 245 000 will appear as 3.2 E6. Using the Casio ClassPad The Classpad calculator can be set to express decimal answers in various forms. To select a fixed number of decimal places, including specifying scientific notation with fixed decimal accuracy, tap and inBasic formattap the arrow to select from the various Number formats available. Exercise2B 1Express each of the following numbers in standard form. Example 4 a47.8b6728c79.23d43 580 e0.0023f0.000 000 56g12.000 34h50 million i23 000 000 000j0.000 000 0013k165 thousandl0.000 014 567 2Express each of the following in scientific notation. aX-rays have a wavelength of 0.000 000 01 cm. bThe mass of a hydrogen atom is 0.000 000 000 000 000 000 000 0166 g. cVisible light has wavelength 0.000 05 cm.dOne nautical mile is 1853.18 m. eA light year is 9 463 000 000 000 km. fThe speed of light is 29 980 000 000 cm/s. 3Express each of the following as an ordinary number. aThe star Sirius is approximately 7.5684×10 13 km from Earth. bA single blood cell contains 2.7×10 8molecules of haemoglobin. cThe radius of an electron is 1.9×10 −13 cm. 4Find the value of Example 5 a324 000×0.000 0007 4000b5 240 000×0.8 42 000 000 5Evaluate the following correct to three significant figures. Example 6 a 3√a b4 ifa=2×10 9andb=3.215b 4√a 4b 4ifa=2×10 12 andb=0.05

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 33 2.3 Solving linear equations and linear simultaneous equations The solution to many problems may be found by translating them into a mathematical equation which may then be solved using algebraic techniques. An equation is solved by finding the value or values of the unknowns that would make the statement true. Linear equations are simple equations that can be written in the formax+b=0. There are a number of standard techniques that can be used for solving linear equations. Example 7 Solvex 5−2=x 3 Solution x 5−2=x 3 x 5×15−2×15=x 3×15 Multiply both sides of the equation by the lowest common multiple of 3 and 5. 3x−30=5x 3x−5x=30 −2x=30 x=30 −2 x=−15 Example 8 Solvex−3 2−2x−4 3=5 Solution x−3 2×6−2x−4 3×6=5×6 3(x−3)−2(2x−4)=5×6 3x−9−4x+8=30 3x−4x=30+9−8 −x=31 x=31 −1 =−31

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 34 Essential Advanced General Mathematics Simultaneous linear equations Finding the intersection of two straight lines can be done graphically, however the accuracy of the solution will depend on the accuracy of the graphs. (1, –2) x + 2y = –3 x 2x – y = 4 0 1 123 –1 –1 –2 –3 –2 –3 –42 3 4 y Alternatively this point of intersection may be found algebraically by solving the pair of simultaneous equations. Three techniques for solving simultaneous equations will be considered. Example 9 Solve the equations 2x−y=4 andx+2y=−3. Solution 1: By substitution 2x−y=4 (1) x+2y=−3 (2) First express one unknown from either equation in terms of the other unknown. From equation (2) we getx=−3−2y. Then substitute this expression into the other equation. Equation (1) then becomes 2(−3−2y)−y=4 reducing it to one equation in one unknown. Solving (1)−6−4y−y=4 −5y=10 y=−2 Substituting the value ofyinto (2)x+2(−2)=−3 x=1 Check in (1) LHS=2(1)−(−2)=4 RHS=4 N.B. This means that the point (1, –2) is the point of intersection of the graphs of the two linear relations.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 35 2: By elimination 2x−y=4 (1) x+2y=−3 (2) If the coefficient of one of the unknowns is the same in both equations, we can eliminate that unknown by subtracting one equation from the other. It may be necessary to multiply one of the equations by a constant to make the coefficients ofx orythe same for the two equations. To eliminatexmultiply equation (2) by 2 and subtract the result from equation (1). Equation (2) becomes 2x+4y=6(2 ) Then 2x−y=4 (1) 2x+4y=−6(2 ) Subtracting (1)−(2 )−5y=10 y=−2 Now substitute foryin (1) to findx, and check as insubstitutionmethod. Using the TI-Nspire The simultaneous equations can be solved in aCalculatorapplication.Solve( )from the Algebramenu ( b 31 ) can be used with either thesimultaneous equations template ( / ) or withandas shown. Theandcan either be typed or found in the catalog ( 1 ). The simultaneous equations can also be solved graphically in aGraphs & Geometry application. The equations are rearranged to makeythe subject. The equations in this form arey=2x−4 andy=−3−x 2. Enter these as shown.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 36 Essential Advanced General Mathematics The intersection point is found by selecting Intersection Point(s)from thePoints and Linesmenu ( b 63 ). Use the NavPad to move the hand to select each of the two graphs as shown. The coordinates of the intersection point will appear on the screen. Press to exit theIntersection Point(s) menu. Using the Casio ClassPad The simultaneous equations can also be solved graphically. First, the equations need to be rearranged to makeythe subject. In this form the equations arey=2x−4 and y=−1 2x−3 2. Enter these in area as shown. Select both equations by ticking the box at the left then press to produce the graph. To find the solution, click into the graph screen to select it and then clickAnalysis, G-Solve,Intersect.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 37 Exercise2C 1Solve the following linear equations. a3x+7=15b8−x 2=−16c42+3x=22 d2x 3−15=27e5(2x+4)=13f−3(4−5x)=24 g3x+5=8−7xh2+3(x−4)=4(2x+5)i2x 5−3 4=5x Example 7 j6x+4=x 3−3 2Solve the following linear equations. ax 2+2x 5=16 Example 8 b3x 4−x 3=8 c3x−2 2+x 4=−18 d5x 4−4 3=2x 5ex−4 2+2x+5 4=6 f3−3x 10−2(x+5) 6=1 20 g3−x 4−2(x+1) 5=−24 h−2(5−x) 8+6 7=4(x−2) 3 3Solve each of the following pairs of simultaneous equations. Example 9 a3x+2y=2 2x−3y=6b5x+2y=4 3x−y=6c2x−y=7 3x−2y=2 dx+2y=12 x−3y=2e7x−3y=−6 x+5y=10f15x+2y=27 3x+7y=45 2.4 Solving problems with linear equations Many problems can be solved by translating them into mathematical language and using an appropriate mathematical technique to find the solution. By representing the unknown quantity in a problem with a symbol (called a pronumeral) and constructing an equation from the information, the value of the unknown can be found by solving the equation. Before constructing the equation, state what the pronumeral is and what it stands for (including the unit). It is essential to remember that all elements of the same type in the equation must be in the same units. Example 10 For each of the following, form the relevant linear equation and solve it forx. aThe length of the side of a square is (x−6) cm. Its perimeter is 52 cm. bThe perimeter of a square is (2x+8) cm. Its area is 100 cm 2.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 38 Essential Advanced General Mathematics Solution a The perimeter=4×length of a side 4(x−6)=52 Thereforex−6=13 andx=19 bThe perimeter of the square is 2x+8 The length of one side=2x+8 4=x+4 2 Therefore x+4 2 2 =100 In this casex+4 2=10 as side length must be a positive number. Thereforex=16 Example 11 An athlete trains for an event by gradually increasing the distance she runs each week over a five-week period. If she runs an extra 5 km each successive week and over the five weeks runs a total of 175 km, how far did she run in the first week? Solution Let the distance run in the first week=xkm. Then the distance run in the second week=x+5 km. The distance run in the third week=x+10 km. So the total distance run=x+x+5+x+10+x+15+x+20 ∴ 5x+50=175 5x=125 x=25 The distance she ran in the first week was 25 km. Example 12 A man bought 14 CDs at a sale. Some cost him $15 each and the remainder cost $12.50 each. In total he spent $190. How many $15 CDs and how many $12.50 CDs did he buy?

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 39 Solution Letnequal the number of CDs costing $15. Then 14 –n=the number of CDs costing $12.50. ∴ 15n+12.5(14−n)=190 15n+175−12.5n=190 2.5n+175=190 2.5n=15 n=6 He bought 6 CDs costing $15 and 8 CDs costing $12.50. Exercise2D 1For each of the cases below, write down a relevant equation involving the variables defined and solve the equation for partsa,bandc. Example 10 aThe length of the side of a square is (x−2) cm.Its perimeter is 60 cm. bThe perimeter of a square is (2x+7) cm.Its area is 49 cm 2. cThe length of a rectangle is (x−5) cm.Its width is (12−x)cm.The rectangle is twice as long as it is wide. dThe length of a rectangle is (2x+1) cm.Its width is (x−3) cm.The perimeter of the rectangle isycm. enpersons each has a meal costing $p. The total cost of the meal is $Q. fSpersons each has a meal costing $p. 10% service charge is added to the cost. The total cost of the meal is $R. gA machine working at a constant rate producesnbolts in 5 minutes. It produces 2400 bolts in 1 hour. hThe radius of a circle is (x+3) cm.A sector subtending an angle of 60 ◦at the centre is cut off. The arc length of the minor sector isacm. 2Bronwyn and Noel have a women’s clothing shop in Summerland. Bronwyn manages the shop and her sales are going up steadily over a particular period of time. They are going up by $500 a week. If over a five-week period her sales total $17 500, how much did she earn in the first week? Example 11 3Bronwyn and Noel have a women’s clothing shop in Summerland and Bronwyn manages the shop. Sally, Adam and baby Lana came into the shop and Sally bought dresses and handbags. The dresses cost $65 each and the handbags cost $26 each. The total number of items was 11 and in total she spent $598. How many dresses and how many handbags did she buy? Example 12 4A rectangular courtyard is three times as long as it is wide. If the perimeter of the courtyard is 67 m, find the dimensions of the courtyard.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 40 Essential Advanced General Mathematics 5A wine merchant buys 50 cases of wine. He pays full price for half of them but gets a 40% discount on the remainder. If he paid a total of $2260, how much was the full price of a single case? 6A real estate agent sells 22 houses in six months. He makes a commission of $11 500 per house on some and $13 000 per house on the remainder. If his total commission over the six months was $272 500, on how many houses did he make a commission of $11 500? 7Three boys compare their marble collections. The first boy has 14 less than the second boy, who has twice as many as the third. If between them they have 71 marbles, how many does each boy have? 8Three girls are playing Scrabble. At the end of the game, the total of their scores adds up to 504. Annie scored 10% more than Belinda, while Cassie scored 60% of the combined scores of the other two. What did each player score? 9A biathlon event involves running and cycling. Kim can cycle 30 km/h faster than she can run. If Kim spends 48 minutes running and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run? 10The mass of a molecule of a certain chemical compound is 2.45×10 −22 g.If each molecule is made up of two carbon atoms and six oxygen atoms and the mass of one oxygen atom is1 3that of a carbon atom, find the mass of an oxygen atom. 2.5 Solving problems using simultaneous linear equations When the relationships between two quantities is linear then the constants which determine the linear relationship can be determined if two sets of information satisfying the relationship are given. Simultaneous linear equations enable this to be done. Another situation in which simultaneous linear equations may be used is where it is required to find the point of the cartesian plane which satisfies two linear relations. Example 13 There are two possible methods for paying gas bills: Method A: A fixed charge of $25 per quarter+50c per unit of gas used Method B: A fixed charge of $50 per quarter+25c per unit of gas used. Determine the number of units that must be used before method B becomes cheaper than method A.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 41 Solution Let C 1=charge in $ using method A C 2=charge in $ using method B x=number of units of gas used NowC 1=25+0.5x C 2=50+0.25x It can be seen from the graph that if the number of units exceeds 100 then method B is cheaper. 100 50 25C 2 = 0.25x + 50 C 1 = 0.5x + 25 25 50 75 100 125 150 0 Units Dollars C x The solution could also be obtained by solving simultaneous linear equations: C 1=C 2 25+0.5x=50+0.25x 0.25x=25 x=100 Example 14 If 3 kg of jam and 2 kg of butter cost $29, and 6 kg of jam and 3 kg of butter cost $54, find the cost of 1 kg of jam and 1 kg of butter. Solution Let the cost of 1 kg of jam=xdollars and the cost of 1 kg of butter=ydollars. Then 3x+2y=29 1 and 6x+3y=54 2 Multiply 1 by 2: 6x+4y=58 1 Subtract 1from 2: −y=−4 y=4 Substituting in 2 gives: 6x+3(4)=54 6x=42 x=7 ∴The jam costs $7 per kilogram and the butter, $4 per kilogram. Exercise2E 1A car hire firm offers the option of paying $108 per day with unlimited kilometres, or $63 per day plus 32 cents per kilometre travelled. How many kilometres would you have to travel in a given day to make the unlimited kilometre option more attractive? Example 13

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 42 Essential Advanced General Mathematics 2Company A will cater for your party at a cost of $450 plus $40 per guest. Company B offers the same service for $300 plus $43.00 per guest. How many guests are needed before Company A’s charge is less than Company B’s? 3A basketball final is held in a stadium which can seat 15 000 people. All the tickets have been sold, some to adults at $45 and the rest for children at $15. If the revenue from the tickets was $525 000, find the number of adults who bought tickets. Example 14 4A contractor employed eight men and three boys for one day and paid them a total of $2240. Another day he employed six men and eighteen boys for $4200. What was the daily rate he paid each man and each boy? 5The sum of two numbers is 212 and their difference is 42. Find the two numbers. 6A chemical manufacturer wishes to obtain 700 litres of a 24% acid solution by mixing a 40% solution with a 15% solution. How many litres of each solution should be used? 7Two children had 220 marbles between them. After one child had lost half her marbles and the other had lost 40 marbles they had an equal number of marbles. How many did each child start with and how many did each child finish with? 8An investor received $31 000 interest per annum from a sum of money, with part of it invested at 10% and the remainder at 7% simple interest. She found that if she interchanged the amounts she had invested she could increase her return by $1000 per annum. Calculate the total amount she had invested. 9Each adult paid $30 and each student paid $20 to attend a concert. A total of 1600 people attended. The total paid was $37 000. How many adults and how many students attended the concert? 2.6 Substitution and transposition of formulas An equation that states a relationship between two or more quantities is called a formula, e.g. the area of a circleA=r 2. The value ofA, the subject of the formula, may be found by substituting a given value ofrand the value of. Example 15 Using the formulaA=r 2, find the value ofAcorrect to two decimal places, ifr=2.3, =3.142 (correct to two decimal places). Solution A=r 2 =3.142(2.3) 2 =16.621 18 A=16.62,correct to two decimal places. The formula can also be transposed to makerthe subject. When transposing formulas a similar procedure to solving linear equations is followed. Whatever has been done to the pronumeral required is ‘undone’.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 43 Example 16 Transpose the formulaA=r 2to makerthe subject and find the value ofr, correct to two decimal places, ifA=24.58,=3.142 (correct to three decimal places). Solution A=r 2 A =r 2 A =r ∴ r= A  = 24.58 3.142 =2.79697 ∴ r=2.80, correct to two decimal places Exercise2F 1Substitute the specified values to evaluate each of the following, giving the answers correct to two decimal places. avifv=u+atandu=15,a=2,t=5 bIifI=PrT 100andP=600,r=5.5,T=10 cVifV=r 2handr=4.25,h=6 Example 15 dSifS=2r(r+h) andr=10.2,h=15.6 eVifV=4 3r 2handr=3.58,h=11.4 fsifs=ut+1 2at 2andu=25.6,t=3.3,a=−1.2 gTifT=2 l gandl=1.45,g=9.8 hfif1 f=1 v+1 uandv=3,u=7 icifc 2=a 2+b 2anda=8.8,b=3.4 jvifv 2=u 2+2asandu=4.8,a=2.5,s=13.6 2Transpose each of the following to make the symbol in brackets the subject. av=u+at(a)bS=n 2(a+l)(l) cA=1 2bh(b) dP=I 2R(I) Example 16 es=ut+1 2at 2 (a) fE=1 2mv 2 (v) gQ=√ 2gh(h)h−xy−z=xy+z(x) iax+by c=x−b(x) jmx+b x−b=c(x)

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 44 Essential Advanced General Mathematics 3The formulaF=9C 5+32 is used to convert temperatures given in degrees Celsius (C) to degrees Fahrenheit (F). aConvert 28 degrees Celsius to degrees Fahrenheit. bTranspose the formula to makeCthe subject and findCifF=135 ◦. 4The sum (S) of the interior angles of a polygon withnsides is given by the formula S=180(n– 2). aFind the sum of the interior angles of an octagon. bTranspose the formula to makenthe subject and hence determine the number of sides on a polygon whose interior angles add up to 1260 ◦. 5The volume (V) of a right cone is given by the formulaV=1 3r 2hwhereris the radius of the base andhis the height of the cone. aFind the volume of a cone with radius 3.5 cm and height 9 cm. bTranspose the formula to makehthe subject and hence find the height of a cone with base radius 4 cm and volume 210 cm 3. cTranspose the formula to makerthe subject and hence find the radius of a cone with height 10 cm and volume 262 cm 3. 6The sum (S) of a particular sequence of numbers is given by the formulaS=n 2(a+l), wherenis the number of terms in the sequence,ais the first term andlis the last term. aFind the sum of the sequence of seven numbers whose first term is –3 and whose last term is 22. bWhat is the first term of a sequence containing thirteen terms, whose last term is 156 and whose sum is 1040? cHow many terms are there in the sequence 25+22+19+···+ −5=110? 2.7 Algebraic fractions The principles involved in addition, subtraction, multiplication and division of algebraic fractions are the same as for simple numerical fractions. To add or subtract, all fractions must be written with a common denominator. When multiplying, first try to simplify the fractions by cancelling down. This process will involve factorisation of either the numerators or denominators or both. Addition and subtraction Example 17 Simplify ax 3+x 4b2 x+3a 4 c5 x+2−4 x−1d4 x+2−7 (x+2) 2

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 45 Solution ax 3+x 4=4x+3x 12 =7x 12b2 x+3a 4=8+3ax 4x c5 x+2−4 x−1=5(x−1)−4(x+2) (x+2)(x−1) =5x−5−4x−8 (x+2)(x−1) =x−13 (x+2)(x−1)d4 x+2−7 (x+2) 2=4(x+2)−7 (x+2) 2 =4x+1 (x+2) 2 Multiplication and division Before multiplying and dividing algebraic fractions, it is best to factorise numerators and denominators where possible so that common factors can be readily identified. Example 18 Simplify a3x 2 10y 2×5y 12xb2x−4 x−1×x 2−1 x−2 cx 2−1 2x−2×4x x2+4x+3dx 2+3x−10 x2−x−2÷x 2+6x+5 3x+3 Solution a3x 2 10y 2×5y 12x=x 8y b2x−4 x−1×x 2−1 x−2=2(x−2) x−1×(x−1)(x+1) x−2 =2(x+1) cx 2−1 2x−2×4x x2+4x+3=(x−1)(x+1) 2(x−1)×4x (x+1)(x+3) =2x x+3 dx 2+3x−10 x2−x−2÷x 2+6x+5 3x+3=(x+5)(x−2) (x−2)(x+1)×3(x+1) (x+1)(x+5) =3 x+1 Example 19 Express3x 3 √4−x+3x 2√4−xas a single fraction.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 46 Essential Advanced General Mathematics Solution 3x 3 √4−x+3x 2√4−x=3x 3+3x 2√4−x√ 4−x √4−x =3x 3+3x 2(4−x) √4−x =12x 2 √4−x Example 20 Express( x−4) 15−(x−4) −45as a single fraction. Solution ( x−4) 15−( x−4) −45=( x−4) 15−1 ( x−4) 45 =(x−4) 15(x−4) 45−1 (x−4) 45 =x−5 (x−4) 45 Exercise2G 1Simplify each of the following. Example 17 a2x 3+3x 2b3a 2−a 4 c3h 4+5h 8−3h 2d3x 4−y 6−x 3 e3 x+2 yf5 x−1+2 x g3 x−2+2 x+1h2x x+3−4x x−3−3 2 i4 x+1+3 (x+1) 2 ja−2 a+a 4+3a 8 k2x−6x 2−4 5xl2 x+4−3 x2+8x+16 m3 (x−1)+2 (x−1)(x+4)n3 x−2−2 x+2+4 x2−4 o5 x−2−3 x2+5x+6+2 x+3px−y−1 x−y q3 x−1−4x 1−xr3 x−2+2x 2−x

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 47 2Simplify each of the following. Example 18 ax 2 2y×4y 3 xb3x 2 4y×y 2 6xc4x 3 3×12 8x 4 dx 2 2y÷3xy 6e4−x 3a×a 2 4−xf2x+5 4x 2+10x g(x−1) 2 x2+3x−4hx 2−x−6 x−3ix 2−5x+4 x2−4x j5a 2 12b 2÷10a 6bkx−2 x÷x 2−4 2x 2 lx+2 x(x−3)÷4x+8 x2−4x+3 m2x (x−1)÷4x 2 x2−1nx 2−9 x+2×3x+6 x−3÷9 x o3x 9x−6÷6x 2 x−2×2 x+5 3Express each of the following as a single fraction. a1 x−3+2 x−3b2 x−4+2 x−3c3 x+4+2 x−3 d2x x−3+2 x+4e1 (x−5) 2+2 x−5f3x (x−4) 2+2 x−4 g1 x−3−2 x−3h2 x−3−5 x+4i2x x−3+3x x+3 j1 (x−5) 2−2 x−5k2x (x−6) 3−2 (x−6) 2 l2x+3 x−4−2x−4 x−3 4Express each of the following as a single fraction. Example 19 a√ 1−x+2 √1−xb2 √x−4+2 3 c3 √x+4+2 √x+4d3 √x+4+√ x+4 e3x 3 √x+4−3x 2√x+4 f3x 3 2√ x+3+3x 2√x+3 5Simplify each of the following. Example 20 a( 6x−3) 13−( 6x−3) −23 b( 2x+3) 13−2x(2x+3) −23 c( 3−x) 13−2x( 3−x) −23 2.8 Literal equations A literal equation inxis an equation whose solution will be expressed in terms of pronumerals rather than numbers. 2x+5=7 is an equation whose solution is the number 1. In the literal equationax+b=c, the solution isx=c−b a. Literal equations are solved in the same way as solving numerical equations or transposing formulas. Essentially, the literal equation is transposed to makexthe subject.

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 48 Essential Advanced General Mathematics Example 21 Solve the following forx. apx−q=rbax+b=cx+d ca x=b 2x+c Solution a px−q=r px=r+q x=r+q pb ax+b=cx+d ax−cx=d−b x(a−c)=d−b x=d−b a−cca x=b 2x+c Multiply both sides by lowest common denominator (2x) 2a=b+2xc 2a−b=2xc 2a−b 2c=x Simultaneous literal equations Simultaneous literal equations are solved by the same methods that are used for solving simultaneous equations, i.e. substitution and elimination. Example 22 Solve each of the following pairs of simultaneous equations forxandy. ay=ax+c y=bx+dbax−y=c x+by=d Solution a ax+c=bx+d ∴ ax−bx=d−c x(a−b)=d−c x=d−c a−b and therefore y=a d−c a−b +c =ad−bc a−bb ax−y=c... 1 x+by=d... 2 Multiply 1 byb abx−by=cb... 1 Add 1and 2 abx+x=cb+d x(ab+1)=cb+d x=cb+d ab+1 Substitute in 1 a cb+d ab+1 −y=c ∴ y=a cb+d ab+1 −c =ad−c ab+1

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 Chapter 2 — Algebra I 49 Exercise2H 1Solve each of the following forx. Example 21 aax+n=mbax+b=bxcax b+c=0 dpx=qx+5emx+n=nx−m f1 x+a=b x gb x−a=2b x+ahx m+n=x n+m i−b(ax+b)=a(bx−a)jp 2(1−x)−2pq x=q 2(1+x) kx a−1=x b+2lx a−b+2x a+b=1 a2−b 2 mp−qx t+p=qx−t pn1 x+a+1 x+2a=2 x+3a 2For the simultaneous equationsax+by=pandbx–ay=q, show thatx=ap+bq a2+b 2and y=bp−aq a2+b 2. 3For the simultaneous equationsx a+y b=1 andx b+y a=1,show thatx=y=ab a+b. 4Solve each of the following pairs of simultaneous equations forxandy. Example 22 aax+y=c x+by=dbax−by=a 2 bx−ay=b 2 cax+by=t ax−by=sdax+by=a 2+2ab−b 2 bx+ay=a 2+b 2 e(a+b)x+cy=bc (b+c)y+ax=−abf3(x−a)−2(y+a)=5−4a 2(x+a)+3(y−a)=4a−1 5Writesin terms ofaonly in the following pairs of equations. as=ah h=2a+1bs=ah h=a(2+h) cas=a+h h+ah=1das=s+h ah=a +h es=h 2+ah h=3a 2 fas=a+2h h=a−s gs=2+ah+h 2 h=a−1 ah3s−ah=a 2 as+2h=3a

P1: FXS/ABE P2: FXS 9780521740494c02.xml CUAU033-EVANS August 21, 2009 15:24 50 Essential Advanced General Mathematics Using the TI-Nspire with algebra In this section a demonstration of the basic algebra properties of the TI-Nspire is provided. To access these, open aCalculatorapplication ( 1 ) and select the Algebramenu ( b 3 ). The three main commands are solve, factor and expand. Solve( ) This command is used to solve equations, simultaneous equations and some inequalities. An approximate (decimal) answer can be obtained by pressing / enter or by including a decimal number in the expression. The following screens illustrate its use. Factor( ) This command is used for factorisation. Factorisation over the rational numbers is obtained bynotspecifying the variable, whereas factorisation over the real numbers is obtained by specifying the variable. The following screens illustrate its use.

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