17.Matrices and Matrix Transformations (SC)

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17. MATRICES AND MATRIX TRANSFORMATIONS

MATRICES ⎛ 2 1 3 0 ⎞ A matrix is a rectangular array of numbers (or M = ⎜ 5 7 −6 8 ⎟ 3 × 4 symbols) enclosed in brackets either curved or ⎜ ⎟ ⎜ 9 −2 6 −3 ⎟ square. The constituents of a matrix are called entries ⎝ ⎠ or elements. A matrix is usually named by a letter for ⎛ 4 5 7 ⎞ convenience. Some examples are shown below. ⎜ −2 3 6 ⎟ N = ⎜ ⎟ 4 × 3 ⎜ −7 1 0 ⎟ æö314 ⎝⎜ 9 −5 8 ⎠⎟ X = ç÷ èø270- Note that the orders 3 × 4 and 4 × 3 are NOT the æö31426- same. A = ç÷ èø00105- Row matrices ⎡ a b ⎤ F = ⎢ ⎥ ⎣ c d ⎦ If a matrix is composed only of one row, then it is called a row matrix (regardless of its number of ⎡ 4 1 2 ⎤ elements). The matrices �, � and � are row matrices, ⎢ ⎥ Y = ⎢ 3 −1 5 ⎥ ⎢ 6 −2 10 ⎥ J = ()413 L =(30- 42) ⎣ ⎦ K = 21 ⎛ 1 0 ⎞ () I = ⎝⎜ 0 1 ⎠⎟ Column matrices

Rows and Columns If a matrix is composed of only one column, then it is called a column matrix (regardless of the number of The elements of a matrix are arranged in rows and elements). The matrices �, � and � are column columns. Elements that are written from left to right matrices. (horizontally) are called rows. Elements that are written from top to bottom (vertically) are called ⎛ ⎞ columns. The first row is called ‘row 1’, the second ⎛ ⎞ 6 4 ⎜ ⎟ ‘row 2’, and so on. The first column is called ⎛ 1 ⎞ ⎜ ⎟ 7 P = ⎜ ⎟ Q = 0 R=⎜ ⎟ ‘column 1, the second ‘column 2’, and so on. ⎝ 2 ⎠ ⎜ ⎟ ⎜ 8 ⎟ ⎝⎜ 3 ⎠⎟ ⎝⎜ 9 ⎠⎟ ⎛ 3 1 4 −1 ⎞ Row 1 ⎜ ⎟ Square matrices M = ⎜ 0 11 −5 8 ⎟ Row 2

⎜ 2 1 6 −4 ⎟ Row 3 ⎝ ⎠ If a matrix has the same number of rows as the Col 1 Col 2 Col 3 Col 4 number of columns, then it is called square. For example, a matrix that has 6 rows and 6 columns is a ⎛ 3 1 4 −1 ⎞ square matrix. We may describe such a matrix as ⎜ ⎟ being square of order 6 or simply a 66´ matrix. An N = ⎜www.faspassmaths.com0 11 − 5 8 ⎟ mm´ matrix is a square matrix of order m. Q and S ⎜ 2 1 6 − 4 ⎟ ⎝ ⎠ are square matrices.

Order of a matrix æö11- æö124 Q = ç÷ ç÷ èø42 S =ç÷--112 The order of a matrix is written as � × �, where m ç÷ represents the number of rows and n represents the 22´ èø0011 number of columns. A matrix of order 4 × 3, consists 33´ of 4 rows and 3 columns while a matrix of order 3 × 4 consists of 3 rows and 4 columns.

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The position of elements in a matrix Diagonal matrices

Every single entry in a matrix has a specific position A diagonal matrix is a square matrix whose non- that can be uniquely described. In describing the diagonal elements are zero. T and V are diagonal

position, we use the notation aij where the subscript, matrices.

i, refers to the row number and the subscript j refers

to the column number of the element. ⎛ 3 0 0 ⎞ Since each position is unique, no two entries can ⎛ 2 0 ⎞ V = ⎜ 0 −5 0 ⎟ T = ⎜ ⎟ ⎜ ⎟ have the same row and column number. The symbol ⎝ 0 5 ⎠ ⎜ 0 0 7 ⎟ a represents an element. ⎝ ⎠

a belongs to the ith row and the jth column. ij a belongs to the 4th row and the 2nd column. Zero matrix 42 a belongs to the 2nd row and the 4th column. 24 If all the elements of any matrix are zero(s), then the For the matrix, A, defined below, we can assign each matrix is called a zero matrix. Some examples are element its unique position shown below. Notice shown below. A zero matrix can be of any order.

a ≠ a . For example, a ≠ a . ij ji 13 31 ⎡ 0 0 ⎤ ⎡ 0 0 0 ⎤ ⎡ 0 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 0 0 0 0 0 ⎛ 4 1 3 2 ⎞ ⎣ ⎦ ⎣ ⎦ ⎜ ⎟ A = 0 −11 6 5 ⎜ ⎟ ⎜ 5 10 2 1 ⎟ ⎝ − − ⎠ Operations on Matrices

Col 1 Col 2 Col 3 Col 4 In performing operations on matrices, there are some restrictions. Unlike numbers, one cannot always add, ⎛ 4 1 3 2 ⎞ 11 12 13 14 Row 1 subtract or multiply any two matrices. In fact, a ⎜ ⎟ division of two matrices is not even possible. A = ⎜ 021 −1122 623 524 ⎟ Row 2 ⎜ ⎟ 5 −10 − 2 1 Row 3 ⎝ 31 32 33 34 ⎠ Addition

Diagonal elements Matrices can only be added if they are of the same order. This is done by adding or subtracting Diagonal elements are the elements positioned along corresponding entries. The resulting matrix will also the diagonal of a square matrix. A and B are square be of the same order.

matrices and have diagonal elements of the type aii . éù12- 1 éù042- A = êú B = êú Square matrix with three diagonal 43- 5 -163 elements ëû ëû 23´ 23´ æö3111 4 ç÷ A = 20 6 éù1211 12-- 1 13 é 011 42 12 13 ù ç÷22 AB+=êú+ê ú ç÷ 43-- 5 1 63 èø--11 5 833 ëû21 22 23ë 21 22 23 û www.faspassmaths.com23´´23 Square matrix with two diagonal éù()()10++() 2-- 4() 12+ elements AB+=êú11 12 13 41+()- ()() 36+ 53 + ëûêú()21 22 23 æö1411 23´ B = ç÷06 èø22 éù121- AB+=êú ëû398 23´

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Subtraction Solution Matrices can only be subtracted if they are of the æöæö21 6 3 same order. This is done by subtracting (i) 33A ==ç÷ç÷ corresponding entries. The resulting matrix will also èøèø34 912 be of the same order. æöæö412--- 821 (ii) -22B =-ç÷ç÷= éù12- 1 073-- 0146 A = êú èøèø ëû43- 5 23´ éù042- B = Equal matrices êú-163 ëû 23´ Two matrices are equal if: 1. They are of the same order éù1211 12-- 1 13 é 011 42 12 13 ù AB-=êú-ê ú 2. Their corresponding entries are equal. ëû4321 22-- 5 23ë 1 21 63 22 23 û 23´´23 Both conditions must be satisfied before we can deduce that the matrices are equal. Conversely, if éù()()10-----() 2 4() 12 AB-=êú11 12 13 matrices are equal then we can deduce that they must 41--() ()() 36 - 53 - be of the same order and that their corresponding ëûêú()21 22 23 entries are equal. Consider the matrices S and T. They 23´ are of the same order and their corresponding entries éù163- are the same. Therefore, ST= . AB-=êú ëû532- æö13 æö 13 23´ ç÷ ç÷ ST=46- = 46- ç÷ ç÷ ç÷ ç÷ Rule for addition and subtraction of matrices èø20 èø 20

If two matrices have the same order, the matrices Example 2 are said to be conformable to addition and éù42x éù46 subtraction. We obtain the resulting matrix by Given that A = êú B = êú and adding or subtracting corresponding elements. ëû--y 2 ëû22- AB= , find the value of x and of y.

Scalar multiplication Solution If AB= , then A and B are of the same order and If a matrix is multiplied by a scalar then each element their corresponding entries are the same. of the matrix is multiplied by the scalar. The resulting Equating corresponding entries: 26x = -y =2 matrix is of the same order. æöabc \x =3 \y =-2 A = ç÷ èødef www.faspassmaths.com æöka kb kc Example 3 kA = ç÷,k is a scalar èøkd ke kf æö12- æö01 Given that A = ç÷ and B = ç÷ and èø13 èø43 Example 1 æö12x æö21 (i) If A = ç÷, then calculate 3� AB-2 =ç÷y , find the value of x, y and z. èø34 ç÷-z èø2 æö412- (ii) If B = ç÷, then calculate −2� èø073-

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Solution We first need to obtain a single matrix for A – 2B. Order of A Order of B Order of C æö12x 2´3 3´2 2´2 AB-2 =ç÷y 3´4 4´2 3´2 ç÷-z èø2 4´2 2´3 4´3 4´3 3´2 4´2 æö12x æöæö12- 01 ç÷ ç÷ç÷-2 =y If two matrices cannot be multiplied, they are said to èøèø13 43ç÷-z èø2 be non-conformable to multiplication. For example, the product YX is not possible since the number of

æö12x columns of Y (2) ≠ the number of rows of X, (4) æöæö12- 02ç÷ ç÷ç÷-=y èøèø13 86ç÷-z YX´ÞThis cannot be done. èø2 33´´2 4 æö12x æö14- ç÷ ç÷= y We will now illustrate how two matrices can be èø--73ç÷-z èø2 multiplied using the matrices A and B below. Equating corresponding entries æö1232 -42=x y -3 =-z æö14- 1 ç÷ -7 = A = ç÷ and B =ç÷---1111 24x =- 2 -z =-3 èø20- 3 ç÷ y èø0462 -4 =-7 z = 3 x = 2 We wish to determine the matrix AB. 2 y =-72 x =-2 () y =-14 First, check to see if this is possible. AB´= C

2424´´33 ´ The Identity or Unit matrix Based on the above rule, the matrix product �� exists If all the diagonal elements of a diagonal matrix are and the product, � will be a 2´4 matrix. equal to one, then it is called the unit or identity matrix and is denoted by U or I only. We write out the structure of the answer, that is, what C looks like. We set up a 2´4 matrix, using the A unit matrix of A unit matrix of order 3 symbol e, to represent an element. order 2 æö100 æö10 ç÷ C1 C2 C3 C4 I = U = ç÷010 ç÷01 ⎛ ⎞ èø ç÷001 e11 e12 e13 e14 R1 èø C = ⎜ ⎟ ⎝ e21 e22 e23 e24 ⎠ R2

Matrix multiplication We then set up the multiplication in the correct order:

Unlike addition and subtraction, the order of two æö1232 matrices need not be the same for multiplication. If A æö14- 1 ç÷ www.faspassmaths.comAB´=ç÷´--ç÷1111 - and B are two matrices, for AB´ to be possible, then èø20- 3 ç÷ the number of columns of A must be equal to the èø0462 number of rows of B. We write this symbolically as: 24´´33

A ´B =C Each element in the product is calculated by mm´´n n p ´ pmultiplying corresponding elements, for example, to compute e13, we multiply corresponding elements of Row 1 from the first matrix, A, with elements of The table below list some examples of this principle, Column 3 from the second matrix, B. for 2 matrices A and B.

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Each element in the product is the sum of three terms. Commutative Property For example, � is the product of the elements of Row 1 and Column 1. The commutative law of multiplication is not obeyed 1 in matrices. Consider two matrices �, and �, such �� = (1 4 −1) −1. that: 0 Notice that there are 3 elements in each row and 3 AB´= C elements in each column. Hence, there is a one-one correspondence between the elements. We simply 444´´22 ´4 match corresponding elements in order and multiply BA´= D each pair. The sum of the three products is the value 222´´44 ´2 of the element. The computation is completed below. The product, �� = � and � is a 4´4 matrix. e11 =11´+ 4´-()() 1+- 1 ´ 0= 1- 4+ 0 =- 3 The product �� = � and � is a 2´2 matrix. e12 =12´+ 4´-()() 1+- 1 ´ 4= 2-- 4 4=- 6 Since the order of � and � is not the same, C and D cannot be equal, C¹D, and it follows that ��¹��. e =1341´+´+- 1 ´ 6= 34 + - 6= 1 13 () e21 =210´+´-()() 1+-´ 3 0=++= 2 0 0 2 We can conclude that matrix multiplication is not communicative. e =220´+´-()() 1+-´ 3 4=+ 40128- =- 22 e23 =2301´+´+()-´ 3 6=+ 601812- =- Example 5 e =220´+´-()()() 1+-´- 3 2=++= 40610 æö1 24 Given that A =()11- and B = ç÷ . æö--36 10 èø2 \´AB=ç÷ Calculate (i) AB (ii) BA èø281210-- Comment on the products.

Example 4 Solution æö11- (i) ABC´= æö210- ç÷ Given that P = ç÷ and Q = ç÷41, 11´´Þ2211´ èø141 ç÷ èø01 Ce= ()11 find PQ. ABC´=

Solution æö1 ()()11-´ç÷=e11 First, we check the order of the resulting matrix: èø2 PQR´= e11 =11´+()- 1 ´ 2= 1- 2=- 1 22´´Þ33 22´ \AB =-1 Now set up the multiplication, equating it to the () structure of the product (ii) BAD´= PQR´= 22´´Þ11 22´ æö11- ee æö210- æöee æö11 12 ´ç÷41=11 12 D = ç÷ ç÷ç÷ç÷ èøee21 22 èø141ç÷èøee21 22 èø01 æö1 æöee www.faspassmaths.com´-()11=11 12 Multiply corresponding elements as follows: ç÷2 ç÷ee èø èø21 22 e11 =21´+()-´ 1 4+ 00´=- 2 � = 1 × 1 = 1 e =21´-()()+-´ 11013+´=- � = 1 × −1 = −1 12 � = 2 × 1 = 2 e21 =1´ 1+ 4´ 4+ 2´ 0= 17 � = 2 × −1 = −2 e22 =1141215´-() +´+´= 1 −1 �� = 2 −2 The products �� ≠ ��. This illustrates the non- æö--23 \PQ =ç÷ commutative property of matrix multiplication. èø17 5

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Example 6 We can apply the same concept. æö-10 æö31- Given that A = , B = and The additive inverse of a matrix M is -M because ç÷23 ç÷04 èø èø M + − M = Zero Matrix æö2ab- ↑ ↑ ↑ AB = ç÷d , calculate the values of a, b, c and d. ç÷c Matrix Inverse Identity èø3

Solution The null or zero matrix is the identity element for the æöæö--10 3 1 addition of matrices because adding the zero matrix AB´=ç÷ç÷´ to any matrix leaves the matrix unchanged. 23 0 4 � � 0 0 � � èøèø + = 22´´22 � � 0 0 � �

2´2 ¬¾¾® æöee AB´= 11 12 We will now recall the terms multiplicative identity ç÷ee èø21 22 elements and multiplicative inverses in relation to e11 =()-1300 ´+ ´ =- 3 numbers. e =()()-11041 ´- + ´ = 12 The multiplicative inverse of a number, N is � e21 =2330´+´= 6 because � × � = 1 e22 =2´-() 1+ 3´ 4= 10

æö-31 ↑ ↑ ↑ \AB =ç÷ èø6 10 Number Inverse Identity Element æö2ab- ç÷æö-31 The identity element for multiplication of d = ç÷ numbers is 1 because multiplying any number ç÷c èø6 10 èø3 by 1 leaves the number unchanged. If N Equating corresponding elements: represents any number, then � × 1 = � -32=a -b =1 By d = 10 23a =- b =-1 inspection 3 c = 6 We can apply the same concept to matrices, whereby, 3 d =10´ 3 a =- 2 d = 30 The multiplicative inverse of a matrix, � is

� because

� × � = � Identity elements and inverses - 2×2 matrices

↑ ↑ ↑ We have encountered the terms additive identity Matrix Inverse Identity Matrix elements and additive inverses in relation to numbers. The identity element for multiplication of matrices is The additive inverse of a number N is -N because the identity matrix, I because multiplying any matrix N + − N = Zero by I leaves the matrix unchanged. If M represents www.faspassmaths.com ↑ ↑ ↑ any 2×2 matrix, then � � 1 0 � � Number Inverse Identity × = � � 0 1 � �

Zero is called the identity element for addition because adding zero to any number leaves the To find the additive inverse of a matrix, we simply number unchanged. If � represents any number, reverse the directions of all its elements. However, to then N + 0 = N find the multiplicative inverse of a matrix requires a number of steps. We will now illustrate how to carry these steps in relation to 2×2 matrices only.

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Inverse of a 2×2 matrix Example 8 Show that the matrix, N is singular where The inverse of a matrix requires applying a set of æö64 N = . procedures and is not as simple as finding the inverse ç÷32 of a number. In addition, all matrices do not have èø inverses. To determine if a matrix has an inverse we must examine its determinant. Solution N =()()62´-´ 43 The determinant of a 2×2 matrix =12- 12 æöab If M = then MMadbcor det =-. = 0 ç÷cd èø NN=0 Þ is singular and has no inverse. If detMM or= 0, then the matrix does not have an inverse and is singular.

After establishing that the matrix is non-singular we Example 9 proceed to find its adjunct. æö4 a Given that A = ç÷ and that A is singular, find The adjunct of a 2×2 matrix èø23- The adjunct of a 2x2 matrix is found by a. rearranging the elements on the main diagonal and changing the signs of the elements on the other Solution diagonal. If A is singular then A = 0 . æöab If M = , then the adjunct of M, is ()()43´- -a ´ 20= ç÷cd èø --12 2a = 0 � −� −� � -212a = a =-6 The inverse of a 2×2 matrix æöab Property of the inverse If , M = ç÷and Aadbc=-, where èøcd ad-¹ bc 0 , then Only square matrices have inverses. The product of a matrix and its inverse is the identity inverse. This is true regardless of its order. �= (adjoint of A) || 1 � −� Example 10 � = �� − �� −� � æö21- Given that Q = ç÷, find � . Show that èø43-

� × � = �. Example 7

æö62 -1 Solution Given that B = ç÷ , find B . èø34 Det Q = (2 × −3) − (−1 × 4) = −6 − (−4) = −2 www.faspassmaths.comDeterminant is non-zero, Q has an inverse Solution 1 3 1 −3 1 − First, examine the value � = = 2 2 -1 1 æö42-() −2 −4 2 of the determinant. B = ç÷ 2 −1 18 èø-()36 detB =()() 6´-´ 4 2 3 3 1 æö42 2 −1 − =24- 6 - � × � = 2 2 ç÷18 18 4 −3 = 18 = ç÷ 2 −1 ç÷36 3 − 1 −1 + 1 1 0 Since det B is not zero, B ç÷- � × � = = èø18 18 6 − 6 −2 + 3 0 1 has an inverse.

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Using matrices to solve a pair of simultaneous Pre-multiply the matrix equation by A-1 on both equations sides to isolate the unknown matrix. æö53 æö53 To solve a pair of simultaneous equations using the ç÷47 47 æöæö7- 3 x ç÷47 47 æö11 matrix method, we must first convert the pair of ç÷ç÷ç÷= ç÷ ç÷ ç÷4 7èøèø 4 5y ç÷ 4 7 èø 13 equations to the matrix form. ç÷-- ç÷ èø47 47 èø47 47 This is done by extracting the coefficients of the variables to form a 2´2 matrix. This matrix of The left-hand side simplifies to: � � � � � × � = � = coefficients multiplied by the matrix � should � � � produce the left-hand side of the matrix equation, as shown in the examples below. 3. Simplify the right-hand side: æö53 ç÷ 32xy+= 8 431xy-= 47 47 æö11 æöe11 ç÷= 56xy-= 4 ç÷ç÷ 57xy+= 12 ç÷4713èøèøe21 ç÷- The matrix equation is: The matrix equation is èø47 47 æöæöæö32x 8 æöæöæö43- x 1 21´2 2´12´ ç÷ç÷ç÷= ç÷ç÷ç÷= èøèøèø56- y 4 èøèøèø57y 12 æöæö53 e11 =ç÷ç÷´11+´ 13= 2 èøèø47 47

æöæö47 Using a matrix method, we will solve for x and y in: e21 =ç÷ç÷- ´11+ ´ 13= 1 7xy- 3= 11 …(1) èøèø47 47

4xy+= 5 13 …(2) Equating LHS to RHS gives

æöx æö2 1. Rewrite as a matrix equation: = ç÷y ç÷1 æöæöæö73- x 11 èø èø \= Equating corresponding elements, ç÷ç÷ç÷45y 13 èøèøèø xy==2 and 1 This is of the form �� = �, where �, � and � are Summary of steps in solving simultaneous matrices with equations using the matrix method. 7 −3 � 11 � = , � = and � = 4 5 � 13 1. Express the pair of equations as a matrix � equation. 2. Isolate � = � 2. Find A-1 , the inverse of A, which is the In our study of simple equations, we used inverses to 22´ matrix in the matrix equation isolate the variable and solve the equation. We do 3. Pre-multiply both sides of the equation by likewise when solving matrix equations. To isolate � A-1 the unknown matrix, �, we need to determine the , this simplifies to: � = � � inverse of the matrix, �. 4. Simplify the right-hand side to obtain a æö73- A =()()75´--´ 34 2 × 1 matrix. Let A = ç÷, now 5. Equate corresponding elements to obtain 45 = 47 èø the values of the unknowns. www.faspassmaths.com So, the inverse of A is 1 æö53--() It should now be noted that so far, we have A-1 = ç÷ encountered three methods to solve a pair of 47 èø-()47 simultaneous equations. These are: æö53 1. Algebraic methods (elimination and substitution) ç÷47 47 2. Graphical method = ç÷ ç÷47 3. Matrix method ç÷- èø47 47 In some cases, we are asked to use a specific method and so it is necessary to be familiar with all three

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com methods. Of course, when we have a choice we select producing three sizes of T-shirts may sell different the one that is most efficient and easy for us to apply. quantities of each size on a particular day. They may record their sales in a table after 2 days as such: Example 10 Express the equations Small Medium Large 2� + 5� = 6 Day 1 8 12 3 3� + 4� = 8 Day 2 6 10 5 in the form �� = �, where �, � and � are matrices. Hence, solve for � and � using the matrix method. Their prices can also be recorded in another table, such as: Solution The equations in a matrix form are: Small Medium Large æ 52 öæ x ö æ6ö Cost $15 $20 $25 ç ÷ç ÷ = ç ÷ ç 43 ÷ç y÷ ç8÷ è øè ø è ø AX = B If we wish to determine the total sales for each day, ´ A-1 we can set up a pair of matrices whose product will produce these totals. -1 ´´ = -1 ´ BAXAA ´ = -1BAXI We can represent the matrix of quantities sold as -1 follows: Each cell refers to the number of T-shirts = BAX sold on a particular day in a given size. Two possibilities exist, a 2 × 3 or a 3 × 1. Det A = 2 × 4 − 5 × 3 = −7 4 5 8 6 æ ö 8 12 3 ç - ÷ or 12 10 A-1 = ç 7 7 ÷ 6 10 5 3 2 6 5 ç - ÷ 7 7 è ø We can also represent the cost of each size as a 4 5 æ ö matrix, either a 3 × 1 or a 1 × 3 matrix can be used. æ x ö ç- ÷æ6ö ç ÷ = ç 7 7 ÷ç ÷ This would look like: ç y÷ 3 2 ç8÷ è ø ç - ÷è ø è 7 7 ø 15 æ æ 4 ö æ 5 ö ö (15 20 25) or 20 ç ç ´- 6÷ + ç ´ 8÷ ÷ 7 7 25 = ç è ø è ø ÷ ç æ 3 ö æ 2 ö÷ Assuming we wish to calculate the total sales on each ç ç ´ 6÷ + ç ´- 8÷÷ è è 7 ø è 7 øø day, we can set up a multiplication of two matrices æ 24 40 ö such that the product yields the total. ç- + ÷ 7 7 = ç ÷ Using our rules for matrix multiplication, we can ç 18 16 ÷ ç - ÷ come up with the following product: è 7 7 ø æ 2 ö 15 ç2 ÷ 8 12 3 20. = ç 7 ÷ 6 10 5 2 25 www.faspassmaths.comç ÷ è 7 ø This product is conformable to multiplication Hence, � = 2 , � = because a 2 × 3 multiplied by a 3 × 1 results in a 2 × 1 matrix. We now set up the multiplication.

15 Using matrices to solve word problems 8 12 3 20 6 10 5 A single matrix can capture an array of numbers and 25 it is sometimes convenient to use such arrays to 8 × 15 + 12 × 20 + 3 × 25 = represent information. For example, a company 6 × 15 + 10 × 20 + 5 × 25

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com 120 + 240 + 75 435 = = needed to know the geometric properties of the 90 + 200 + 125 415 transformations so that we can locate their images.

This is interpreted as: In this section, we will learn how to perform The total sales of T-shirts for day 1 is $435. geometric transformations using matrices – this The total sales of T-shirts for day 2 is $415 method relies on algebraic rather than geometric techniques and can be quite effective. In fact, we do Example 11 not need to use a Cartesian Plane to locate the image, A business makes toy trucks and toy cars. The we simply calculate the coordinates of the image following table is used in calculating the cost of points using the appropriate matrix. manufacturing each toy.

Labour Wood Paint Matrices for Translation (Hours) (Blocks) (Tins) Trucks 6 8 3 When we performed a translation, we used a column Cars 3 4 2 matrix to locate the image. For example, to determine Boats 5 7 1 the coordinates of the image of P(5, 4) under a translation of 3 units parallel to the X-axis, we Labour costs $80 per hour, wood costs $10 per block represent both the object point and translation as and paint costs $20 per tin. column vectors. Then we added vectors to obtain the Using matrix multiplication, calculate the cost of image point. manufacturing each toy. 5 3 8 + = 4 0 4 Solution The matrix for the quantities of toys manufactured is 6 8 3 Object Point Translation vector Image Point 3 4 2 5 7 1 80 Hence, the coordinates of the image of P is (8, 4). The cost matrix is written as follows: 20. Note that for the purpose of computation, we write 10 The required product is: the coordinates of (x, y) as a column matrix.

6 8 3 80 240 + 160 + 30 Translation matrices are written as 2×1 matrices, also 3 4 2 20 = 240 + 80 + 20 called column vectors because a translation is really a 5 7 1 10 400 + 140 + 10 vector. To perform any of the other geometric 3 × 3 3 × 1 3 × 1 transformations, we use 2× 2 matrices. 430 = 340 Deriving Matrices for Transformations 560 We now interpret the product as The matrices for performing reflections, rotations and

Cost of producing trucks is $430, dilations are all 2× 2 matrices. To derive these Cost of producingwww.faspassmaths.com cars $340 and matrices, we use a general principle that holds for Cost of producing boats $560. transforming an object point (�, �) to its corresponding image point (�, �′).

This general principle is stated below. MATRIX TRANSFORMATIONS In general, the image of a point (x, y), under a � � transformation, defined by a 2× 2 matrix , � � In our study of transformations so far, we examined is the point (�, �′), where different types of movements and their effect on � � � �′ = , plane shapes. To perform these movements, we � � � �′

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Our knowledge of the geometric properties of these Reflection in the X-axis transformations will be applied to determine the matrices for each transformation. (1, 0 ) ⟶ (1, 0) (0, 1 ) ⟶ (0, −1) Hence, Hence, Consider a unit square, drawn in the first quadrant � � 1 1 � � 0 0 = = whose coordinates are (0, 0), (1, 0), (1, 1) and (0, 1). � � 0 0 � � 1 −1 Under a given transformation each point (x, y) will Multiplying Multiplying � 1 � 0 move to (�, �′) where, = = � 0 � −1 � � � � Therefore Therefore = � � � � � = 1 and � = 0 � = 0 and � = −1

1 0 We can calculate the values of �, �, �, and � by The matrix for reflection in the X-axis is substituting two pairs of object and image points in 0 −1 the above matrix equation.

Unit square y Reflection in the Y-axis

(1, 0 ) ⟶ (−1, 0) (0, 1 ) ⟶ (0, 1) D(0,1) C(1,1) Hence, Hence, � � 1 −1 � � 0 0 = = � � 0 0 � � 1 1 Multiplying Multiplying � −1 � 0 A(0,0) B(1, 0) x = = � 0 � 1 Therefore Therefore � = −1 and � = 0 � = 0 and � = 1

−1 0 The matrix for reflection in the Y-axis is 0 1

Matrices for Reflection We will start with the matrix for reflection in the � axis. Under this reflection, the unit square will flip so Reflection in the line � = � that it is now in the fourth quadrant. (1, 0 ) ⟶ (0, 1) (0, 1 ) ⟶ (1, 0) Unit square and Hence, Hence, � � 1 0 � � 0 1 its image after a = = reflection in the � � 0 1 � � 1 0 Multiplying Multiplying -axis. � � 0 � 1 ( ) = = 1, 0 ⟶ (1, 0) � 1 � 0 (0, 1) ⟶ (0, −1) Therefore Therefore (1, 1) ⟶ (1, −1) � = 0 and � = 1 � = 1 and � = 0 (0, 0) ⟶ (0, 0)

www.faspassmaths.com0 1 The matrix for reflection in the X-axis is 1 0

For convenience, we select the two points (1, 0) and (0, 1), This is repeated for each of the transformations performed below.

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(1, 0 ) ⟶ (0, −1) (0, 1 ) ⟶ (−1, 0) (1, 0 ) ⟶ (−1, 0) (0, 1 ) ⟶ (0, −1) Hence, Hence, Hence, Hence, � � 1 0 � � 0 −1 � � 1 −1 � � 0 0 = = = = � � 0 −1 � � 1 0 � � 0 0 � � 1 −1 Multiplying Multiplying Multiplying Multiplying � 0 � −1 � −1 � 0 = = = = � −1 � 0 � 0 � −1 Therefore Therefore Therefore Therefore � = 0 and � = −1 � = −1 and � = 0 � = −1 and � = 0 � = 0 and � = −1 The matrix for reflection in the line � = −� is The matrix for a rotation about 0 −1 −1 0 the origin through 1800 is −1 0 0 −1 Note that for a 1800 rotation, there is no need to specify the direction since clockwise or anticlockwise Matrices for Rotation turns will produce the same image.

The matrices derived for rotation are defined for an anticlockwise rotation about the origin. In each case, General matrix for rotation � � the matrix represents an anticlockwise � � The above matrices are confined to angles that are rotation about the origin. Also, we should note that an multiples of 900 degrees. If one has to perform a anticlockwise rotation of 900 is the same as a rotation through other angles, there is a general clockwise rotation of 2700 so we need not derive matrix for rotation of �about the origin in an matrices for clockwise rotations. anticlockwise direction where � can be any angle. The matrix for an anticlockwise rotation through �about the origin is Anticlockwise rotation about the origin through 900 �� −�� (1, 0 ) ⟶ (0, 1) (0, 1 ) ⟶ (−1, 0) �� Hence, Hence, � � 1 0 � � −1 −1 = = � � 0 1 � � 0 0 Matrix for Dilation (enlargement) Multiplying Multiplying � 0 � −1 � � = = Let represent the matrix for a dilation. We � 1 � 0 � � Therefore Therefore can now derive the general matrix for a dilation with � = 0 and � = 1 � = −1 and � = 0 scale factor, k, about the origin. Recall that under a The matrix for an anticlockwise rotation about dilation, a point (x, y) is mapped onto (kx, ky). So, the 0 −1 the origin through 900 is 1 0 unit square will have sides that are � units in length.

Enlargement with center origin, scale factor, k Anticlockwise rotation about the origin through 2700 (1, 0 ) ⟶ (�, 0) (0, 1 ) ⟶ (0, �) (1, 0 ) ⟶ (0, −1) (0, 1 ) ⟶ (1, 0) Hence, Hence, www.faspassmaths.com� � 1 � � � 0 0 Hence, Hence, = = � � 1 0 � � 0 1 � � 0 0 � � 1 � = = Multiplying Multiplying � � 0 −1 � � 1 0 � � � 0 Multiplying Multiplying = = � 0 � 1 � 0 � � = = Therefore Therefore � −1 � 0 Therefore Therefore � = � and � = 0 � = 0 and � = � � = 0 and � = −1 � = 1 and � = 0 The matrix for enlargement with centre origin, � 0 The matrix for an anticlockwise rotation about scale factor, k is 0 1 0 � the origin through 2700 is −1 0

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Example 12 Example 14 The triangle PQR has coordinates P(3, 5), Q(6, 5) and R(6, 7). The triangle LMN has coordinates L(1, 4), M(3, 4) Determine the coordinates of the image of triangle and N(1, 6). Determine the coordinates of the image PQR under a reflection in the �- axis of triangle LMN under an enlargement with center (0, 0) with scale factor 3.

Solution Solution −1 0 Matrix for reflection in the Y axis is . Matrix for enlargement with center (0, 0), k=3 is 0 1 3 0 To calculate the image of P, Q and R under the . 0 3 reflection, pre-multiply as follows: To calculate the image of P, Q and R under the enlargement, pre-multiply as follows: −1 0 3 −3 = , hence 0 1 5 5 3 0 1 3 = , hence �(3,5) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(−3,5) 0 3 4 12 �(1,4) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(3,12) −1 0 6 −6 = , hence 0 1 5 5 3 0 3 9 = , hence �(6,5) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(−6,5) 0 3 4 12 �(3,4) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(9,12) −1 0 6 −6 = , hence 0 1 7 7 3 0 1 3 = , hence �(6,7) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(−6,7) 0 3 6 18 �(1,6) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(3,18) Under the reflection in the Y-axis, triangle PQR is mapped onto triangle �′�′�′ where Under the reflection in the Y-axis, triangle LMN is ( ) ( ) ( ) � −3,5 , � −6,5 and � −6,7 . mapped onto triangle �′�′�′ where �(3,12), �(9,12) and �(3,18). Example 13 The triangle ABC has coordinates A(2, 4), B(7, 4) and C(2, 7). Determine the coordinates of the image of triangle ABC under a rotation about the origin Matrices for combined transformations through 1800 It is possible to obtain a single matrix to describe a Solution combination of two transformations. Once we know Matrix for rotation about the origin through 1800 is the 2×2 matrix for each of the transformations we −1 0 can multiply them to obtain this single matrix using . 0 −1 the following rule. To calculate the image of P, Q and R under the rotation, pre-multiply as follows: If A and B are both 2×2 matrices representing two geometric transformations, then the matrix product −1 0 2 −2 = , hence AB represents the combined transformation B 0 −1 4 −4 followed by A. �(3,5) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(−2, −4) −1 0 7 www.faspassmaths.com−7 = , hence Example 14 0 −1 4 −4 The transformation, M is a reflection in the line � = �(6,5) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(−7, −5) −�. The transformation N, is an enlargement, centre −1 0 2 −2 = , hence origin, k= 2. 0 −1 7 −7 (i) Write down the 2×2 matrices for M and N. �(6,7) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ �(−2, −7) (ii) The matrix, P represents the combined transformation, M followed by N. Determine the Under the reflection in the Y-axis, triangle PQR is matrix P. mapped onto triangle �′�′�′ where (iii) Determine the coordinates of the image of (−3, �(−2, −4), �(−7, −5) and �(−2, −7). 4) under P.

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Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Solution Solution – Example 15, part (iii) (i) M is a reflection in the line � = −�, The transformation M represents a rotation of 90º 0 −1 M= . clockwise about O. This is deduced by applying −1 0 geometrical properties of rotation of having N is an enlargement, center origin, � = 2, knowledge of the matrix for the rotation. 2 0 N= . 0 2 (3, 5)

(ii) The combined transformation, M followed by

N is NM. 2 0 0 −1 � = �� = 0 2 −1 0 0 −2 = −2 0 (5, −3)

(iii) The image of (−3, 4) under P: 0 −2 −3 −8 = Alternative Method for Example 15 Part (i) −2 0 4 6 The image of (−3, 4) is (−8, 6). ´ = VVM ¢ Simultaneous equations for

�(3, 5) ⟶ �(5, −3) æ ba öæ3ö æ 5 ö ç ÷ç ÷ = ç ÷ Example 15 è dc øè5ø è- 3ø 3� + 5� = 5 … (1) � � Under a matrix transformation, � = , the æ + 53 ba ö æ 5 ö 3� + 5� = −3 … (2) � � \ç ÷ = ç ÷ points � and � are mapped onto �′ and �′ such è + 53 dc ø è- 3ø that: � × � = �′ Simultaneous equations for �(3, 5) ⟶ �(5, −3) æ ba öæ7ö æ 2 ö �(7, 2) ⟶ �(2, −7) �(7, 2) ⟶ �(2, −7) ç ÷ç ÷ = ç ÷ 7� + 2� = 2 … (3) è dc øè2ø è- 7ø (i) Determine the values of �, �, �, and �. 7� + 2� = −7 … (4) (ii) State the coordinates of Z such that æ + 27 ba ö æ 2 ö \ç ÷ = ç ÷ �(�, �) ⟶ �(5, 1) under the è + 27 dc ø è- 7ø transformation, �. Consider equations (1) and (3) (iii) Describe fully the geometric transformation, Equation (1) × 7 21� + 35� = 35 … (5) � Equation (3) × −3 −21� − 6� = −6 … (6) Equation (5) + (6) 29� = 29 � = 1 Solution- parts (i) and (ii) Substitute � = 1 into equation (1) (i) We note that under the transformation, � a a () =+ 5153 point reverses its coordinates and changes the sign of the y coordinate. By inspection, the matrix for a = 03 this transformation M is derived by consideration \a = 0 of the following products: Consider equations (2) and (4) 0 1 3 5 = Equation (2) × 7 21� + 35� = −21 … (7) −1 0 5 −3 Equation (4) × −3 −21� − 6� = 21 … (8) 0 1 7 2 = Equation (7) + (8) 29� = 0 −1 0 2 −7 � = 0 www.faspassmaths.com� � 0 1 Substitute � = 0 into equation (2) � = = � � −1 0 c () =+ -3053 c = -33 (ii) (�, �) ⟶ �(5, 1) 0 1 � 5 \c = -1 = −1 0 � 1 Therefore, � 5 = æ ba ö æ 10 ö −� 1 ç ÷ = ç ÷ � = 5, � = −1 è dc ø è - 01 ø (�, �) = (−1, 5)

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