Section 6.2 Matrix Algebra

Before Starting this Section, Review

  1. 1 Solving systems of equations (Sections 5.1 and 5.2)

Objectives

  1. 1 Define equality of two matrices.

  2. 2 Define matrix addition and scalar multiplication.

  3. 3 Define matrix multiplication.

  4. 4 Apply matrix multiplication to computer graphics.

Computer Graphics

The following fields illustrate the widespread and growing applications of computer graphics:

  1. The automobile industry. Both computer-aided design (CAD) and computer-aided manufacturing (CAM) have revolutionized the automotive industry. Many months before a new car is built, engineers design and construct a mathematical car—a wire-frame model that exists only in computer memory and on graphics display terminals. The mathematical model organizes and influences each step of the design and manufacture of the car.

  2. The entertainment industry. Computer graphics are used by the entertainment industry with spectacular success. They were used in AVATAR and other action movies, such as the latest James Bond thriller, for special effects, and they are used in the design and creation of video games.

  3. Military. From the beginning of the widespread use of computers (in the 1960s), computer graphics have been used by the military to prepare for or defend against war. Modern weaponry depends heavily on computer-generated images.

Engineers use matrix algebra techniques on graphic images to change their orientation of scale, to zoom in on them, or to switch between two- and three-dimensional views. In Example 10 and in the exercises, we examine how basic matrix algebra can be used to manipulate graphical images.

Equality of Matrices

  1. 1 Define equality of two matrices.

In Section 6.1, we defined a matrix as a rectangular array of numbers written within brackets. In this and the next sections, you will learn about the operations with and uses of matrices. A matrix may be represented in any of the following ways:

  1. A matrix may be denoted by capital letters: A, B, C, and so on.

  2. A matrix may be denoted by its representative (i, j)th entry: [aij], [bij], [cij],[aij], [bij], [cij], and so on.

  3. A matrix may be written as a rectangular array of numbers:

    A=[aij]=[a11a12⋯a1na21a22⋯a2n⋮⋮⋯⋮am1am2⋯amn]
    A=[aij]=⎡⎣⎢⎢⎢⎢a11a21⋮am1a12a22⋮am2⋯⋯⋯⋯a1na2n⋮amn⎤⎦⎥⎥⎥⎥

We now examine some algebraic operations and properties of matrices.

Example 1 Determining Equality of Matrices

Find x and y such that

[x+y23x−y]=[4232].
[x+y32x−y]=[4322].

Solution

Because equal matrices must have identical entries in corresponding positions, we have the following system of equations:

{x+y=4(1)x−y=1(2)
{x+y=4x−y=1(1)(2)

Solve this system: Add the two equations to obtain x=52.x=52. Substitute x=52x=52 in equation x−y=1x−y=1 and solve for y to obtain y=32.y=32. So x=52x=52 and y=32.y=32.

Practice Problem 1

  1. Find x and y such that

    [12x−yx+y5]=[1125].
    [1x+y2x−y5]=[1215].

Matrix Addition and Scalar Multiplication

  1. 2 Define matrix addition and scalar multiplication.

Matrix addition differs somewhat from real number addition. You can add any two real numbers, but you can add two matrices only if they have the same order.

The definition of matrix addition says that

  1. we can add two matrices of the same order simply by adding their corresponding entries, but

  2. the sum of two matrices of different orders is not defined.

Example 2 Adding Matrices

Let A=[−234−102], B=[64−798−5],A=⎡⎣⎢−2403−12⎤⎦⎥, B=⎡⎣⎢6−7849−5⎤⎦⎥, and C=[1234].C=[1324]. Find each sum if possible.

  1. A+BA+B

  2. A+CA+C

Solution

  1. Because A and B have the same order, A+BA+B is defined. We have

    A+B=[−234−102]+[64−798−5]=[−2+63+44−7−1+90+82−5]=[47−388−3].
    A+B=⎡⎣⎢−2403−12⎤⎦⎥+⎡⎣⎢6−7849−5⎤⎦⎥=⎡⎣⎢−2+64−70+83+4−1+92−5⎤⎦⎥=⎡⎣⎢4−3878−3⎤⎦⎥.

  2. A+CA+C is not defined because A and C do not have the same order.

Practice Problem 2

  1. Let A=[2−14509]A=[25−1049] and B=[−829736].B=[−872396]. Find A+B.A+B.

You can also multiply any matrix A by any real number c. The real number c is called a scalar, and the multiplication process is called scalar multiplication.

In other words, to multiply a matrix A by a real number c, you multiply each entry of A by c. When c=−1,c=−1, we frequently write −A−A instead of (−1)A.(−1)A. The matrix −A−A is called the negative of A. Further, the difference of two matrices, A−B,A−B, is defined to be the sum A+(−B).A+(−B).

Example 3 Performing Scalar Multiplication and Matrix Subtraction

Let A=[120−1312−14]A=⎡⎣⎢1−1223−1014⎤⎦⎥ and B=[21310−2−345].B=⎡⎣⎢21−31043−25⎤⎦⎥. Find the following.

  1. 3A

  2. 2B

  3. 3A−2B3A−2B

Solution

  1. 3A=3[120−1312−14]=[3(1)3(2)3(0)3(−1)3(3)3(1)3(2)3(−1)3(4)]=[360−3936−312]3A=3⎡⎣⎢1−1223−1014⎤⎦⎥=⎡⎣⎢3(1)3(−1)3(2)3(2)3(3)3(−1)3(0)3(1)3(4)⎤⎦⎥=⎡⎣⎢3−3669−30312⎤⎦⎥

  2. 2B=2[21310−2−345]=[2(2)2(1)2(3)2(1)2(0)2(−2)2(−3)2(4)2(5)]=[42620−4−6810]2B=2⎡⎣⎢21−31043−25⎤⎦⎥=⎡⎣⎢2(2)2(1)2(−3)2(1)2(0)2(4)2(3)2(−2)2(5)⎤⎦⎥=⎡⎣⎢42−62086−410⎤⎦⎥

  3. 3A−2B=3A+(−1)2B=[360−3936−312]+[−4−2−6−2046−8−10]Substitute from parts a and b,changing the sign of each entryin b.=[3+(−4)6+(−2)0+(−6)−3+(−2)9+03+46+6−3+(−8)12+(−10)]=[−14−6−59712−112]3A−2B=3A+(−1)2B=⎡⎣⎢3−3669−30312⎤⎦⎥+⎡⎣⎢−4−26−20−8−64−10⎤⎦⎥Substitute from parts a and b,changing the sign of each entryin b.=⎡⎣⎢3+(−4)−3+(−2)6+66+(−2)9+0−3+(−8)0+(−6)3+412+(−10)⎤⎦⎥=⎡⎣⎢−1−51249−11−672⎤⎦⎥

Practice Problem 3

  1. Let A=[7−4360−2]A=⎡⎣⎢730−46−2⎤⎦⎥ and B=[1−32258].B=⎡⎣⎢125−328⎤⎦⎥. Find 2A−3B.2A−3B.

An m×nm×n matrix whose entries are all equal to 0 is called the m×nm×n zero matrix and is denoted by 0mn.0mn. When m and n are understood from the context, we simply write 0. Because the entries of a matrix are real numbers, many of the algebraic properties of matrix addition and scalar multiplication are similar to the familiar properties of real numbers.

In the next example, we use these properties to solve an equation involving matrices. Such an equation is called a matrix equation.

Example 4 Solving a Matrix Equation

Solve the matrix equation 3A+2X=4B3A+2X=4B for X, where

A=[2046] and B=[1352].
A=[2406] and B=[1532].

Solution

3A+2X=4BOriginal equation2X=4B−3ASubtract 3A from both sides.X=12(4B−3A)Multiply both sides by 12; simplify=12(4[1352]−3[2046])Substitute for A and B.=12([412208]−[601218])Scalar multiplication=12[−2128−10]Matrix subtraction=[−164−5]Scalar multiplication
3A+2X2XX=======4B4B−3A12(4B−3A)12(4[1532]−3[2406])12([420128]−[612018])12[−2812−10][−146−5]Original equationSubtract 3A from both sides.Multiply both sides by 12; simplifySubstitute for A and B.Scalar multiplicationMatrix subtractionScalar multiplication

You should check that the matrix X=[−164−5]X=[−146−5] satisfies the given matrix equation.

Practice Problem 4

  1. Solve the matrix equation 5A+3X=2B5A+3X=2B for X, where

    A=[1−135] and B=[27−3−5].
    A=[13−15] and B=[2−37−5].

Matrix Multiplication

  1. 3 Define matrix multiplication.

You know that the sum of two matrices is defined only if both matrices are of the same order. In the case of multiplication, however, the general rule is that if A is an m×pm×p matrix and B is a p×np×n matrix, then the product AB is defined; otherwise, it is not defined.

Example 5 Determining the Order of the Product

For each pair of matrices A and B, state whether the product matrix AB is defined. If it is, find the order of AB.

  1. A=[135], B=[237]A=[135], B=⎡⎣⎢237⎤⎦⎥

  2. A=[23−1456], B=[21354−1−26]A=⎡⎣⎢2−15346⎤⎦⎥, B=[241−13−256]

  3. A=[1−12], B=[214022]A=⎡⎣⎢1−12⎤⎦⎥, B=[201242]

Solution

  1. Yes, the product AB is defined. A is a 1×31×3 matrix, and B is a 3×13×1 matrix; so the number of columns of A equals the number of rows of B. The product AB is a 1×11×1 matrix that has a single entry.

  2. Yes, AB is defined. A is a 3×23×2 matrix, and B is a 2×42×4 matrix; so the number of columns of A equals the number of rows of B. The product AB is a 3×43×4 matrix.

  3. No, AB is not defined. A is a 3×13×1 matrix, and B is a 2×32×3 matrix; so the number of columns of A does not equal the number of rows of B.

Practice Problem 5

  1. State whether the product matrix AB is defined. If AB is defined, find the order of the product matrix AB.

    A=[14702−2],B=[43−1]
    A=[10427−2],B=⎡⎣⎢43−1⎤⎦⎥

Before you learn how to multiply two matrices A and B, consider the next example.

Example 6 Calculating the Total Revenue from Car Sales

A car dealership sold 10 sports cars (S), 30 compacts (C), and 45 mini-compacts (M). The average price per car of type S, C, and M was 42, 24, and 17 (in thousands of dollars), respectively. Find the total revenue received by the dealership from the sale of these cars.

Solution

You can represent the number of cars of each type sold by a 1×31×3 matrix:

N=[SCM]=[103045].
N=[SCM]=[103045].

Using the price of each type of car, you can construct a 3×13×1 price matrix:

P=[422417].
P=⎡⎣⎢422417⎤⎦⎥.

Then the total revenue R received by the dealership from the sale of these cars last month is given by R=10(42)+30(24)+45(17)=1905R=10(42)+30(24)+45(17)=1905 (or $1,905,000).

We define the right side of this equation to be the product NP of the two matrices N and P. In matrix notation,

NP=[103045][422417]=10(42)+30(24)+45(17)=1905.
NP=[103045]⎡⎣⎢422417⎤⎦⎥=10(42)+30(24)+45(17)=1905.

Practice Problem 6

  1. In Example 6 , suppose the average price per car of types S, C, and M is 41, 26, and 19 (in thousands of dollars), respectively. Find the total revenue received by the dealership from the sale of these cars.

Example 6 leads to the following definition.

We make the following observations about this definition.

  1. Because A is a 1×n1×n matrix and B is an n×1n×1 matrix, the product AB is defined and is a 1×11×1 matrix.

  2. To find the product AB, multiply the leftmost entry of A and the top entry of B; then move to the right in A and down in B while multiplying corresponding (first, second, third, and so on) entries together; finally, add all of the resulting products.

Example 7 Finding the Product AB

Find AB, where A=[201−2]A=[201−2] and B=[31−14].B=⎡⎣⎢⎢⎢⎢31−14⎤⎦⎥⎥⎥⎥.

Solution

Because A has order 1×41×4 and B has order 4×1,4×1, AB is defined. The product rule gives

AB=[2(3)+0(1)+1(−1)+(−2)(4)]=[−3].
AB=[2(3)+0(1)+1(−1)+(−2)(4)]=[−3].

The product of the 1×41×4 matrix A and the 4×14×1 matrix B is therefore the 1×11×1 matrix [−3].[−3].

Practice Problem 7

  1. Find AB:

    A=[3−127]andB=[−2015].
    A=[3−127]andB=⎡⎣⎢⎢⎢⎢−2015⎤⎦⎥⎥⎥⎥.

In Example 7, you may think of the answer −3−3 as the 1×11×1 matrix [−3][−3] or simply as the number −3.−3. We will have occasion to make use of both points of view.

The rule for multiplying any two matrices can be reduced to the special case of multiplying a 1×p1×p matrix by a p×1p×1 matrix repeatedly.

Written in full, the matrix product AB=C in the definition is as follows:

Example 8 Finding the Product of Two Matrices

Find the products AB and BA, if possible.

A=[12−13]andB=[3−21123]

Solution

Because A is of order 2×2 and the order of B is 2×3, the product AB is defined and has order 2×3.

Let AB=C=[cij]. By the definition of the product AB, each entry cij of C is found by multiplying the ith row of A by the jth column of B.

Summarizing,

The product BA is not defined because B is of order 2×3 and A is of order 2×2; that is, the number of columns of B is not equal to the number of rows of A.

Practice Problem 8

  1. Find the products AB and BA if possible.

    A=[502−1]andB=[81−2604]

In Example 8, it is obvious that AB≠BA because BA is undefined. Even if both AB and BA are defined, AB may not equal BA, as the next example illustrates.

Example 9 Finding the Product of Two Matrices

Find the products AB and BA.

A=[23−10]andB=[3451]

Solution

AB=[23−10][3451]=[2(3)+3(5)2(4)+3(1)−1(3)+0(5)−1(4)+0(1)]=[2111−3−4]BA=[3451][23−10]=[3(2)+4(−1)3(3)+4(0)5(2)+1(−1)5(3)+1(0)]=[29915]

Practice Problem 9

  1. Find the products AB and BA.

    A=[7103]andB=[2−144]

In Example 9, observe that AB≠BA. Thus, in general, matrix multiplication is not commutative. Matrix multiplication, however, does share many of the properties of the multiplication of real numbers.

In the exercises, you will be asked to verify these properties and compare them to similar properties of real numbers. Also, integral powers of square matrices are defined exactly as for the real numbers. We write A2 to mean AA, A3 to mean AAA, and so on. We define A0 by

A0=In=[100⋯0010⋯0001⋯0⋮⋮⋮000⋯1].

The matrix In is called the identity matrix.

When the order of the matrix is clear from the context, we can drop the subscript n and write I for In. For any square matrix A, we have

IA=A and AI=A

So in matrix multiplication, the matrix I plays a role similar to that of the number 1 in the multiplication of real numbers.

Computer Graphics

  1. 4 Apply matrix multiplication to computer graphics.

Letters used for labels on a computer screen are very simple two-dimensional graphics. The next example illustrates how a letter (or a figure) is transformed when the coordinates of the points on the figure are all multiplied by the same matrix.

Example 10 Transforming a Letter

The capital letter L in Figure 6.2 is determined by six points (or vertices) P1 through P6. The coordinates of the six points can be stored in a data matrix D. We draw line segments connecting these vertices in order.

P1P2P3P4P5P6x-coordinatey-coordinate[044110001166]=D

Figure 6.2

If A=[10.2501], compute AD and graph the figure it represents.

Solution

The columns of the product matrix AD represent the transformed vertices of the letter L.

AD=[10.2501][044110001166]=[1⋅0+0.25⋅01⋅4+0.25⋅01⋅4+0.25⋅11⋅1+0.25⋅11⋅1+0.25⋅61⋅0+0.25⋅60⋅0+1⋅00⋅4+1⋅00⋅4+1⋅10⋅1+1⋅10⋅1+1⋅60⋅0+1⋅6]P′1P′2P′3P′4P′5P′6=[044.251.252.51.5001166]

Figure 6.3 shows the transformed vertices and the transformed figure “L” formed by these vertices.

Figure 6.3

Practice Problem 10

  1. In Example 10 , use the matrix A=[0110.25]. Graph the figure represented by the matrix AD.

If Figure 6.4 below, we present the most common transformations and their corresponding matrices.

Figure 6.4

It is important to observe that combining transformations is equivalent to multiplying their corresponding matrices and that the order in which we perform transformations matters. As an example, consider the square PQRS with vertices P=(0, 0), Q=(2, 0), R=(2, 2) and S=(0, 2).

If we first apply a rotation of 45° clockwise about the origin and then horizontal scaling by a factor of 2 to the square PQRS, we get a rhombus.

This sequence of transformations corresponds to the following matrix multiplication (notice that the matrix corresponding to the first transformation is in the second place and the matrix corresponding to the second transformation is in the first place in this multiplication). You should notice the similarity to how the composition of two functions is written:

[2001][√22√22−√22√22]=[√2√2−√22√22].

However, if we first perform scaling by a factor of 2 and then apply a rotation of 45° clockwise about the origin to the square PQRS, we get a rectangle.

This sequence of transformations corresponds to the following matrix multiplication:

[√22√22−√22√22][2001]=[√2√22−√2√22].

The above transformations provide us with an example of two matrices A and B such that AB≠BA.

Section 6.2 Exercises

Concepts and Vocabulary

  1. Two m×n matrices A=[aij] and B=[bij] are equal if                           for all i and j.

  2. Let A=[aij], B=[bij], and C=[cij]. If A+B=C, then cij=_ for all i and for all j.

  3. The product of a 1×n matrix A and an n×1 matrix B is a                           matrix.

  4. If A is an m×n matrix and B is an n×p matrix, then AB is defined and is a(n)                           matrix.

  5. True or False. If AB and BA are defined, then AB=BA.

  6. True or False. Any two square matrices can be multiplied.

  7. True or False. If A is a 2×2 matrix and A2=0, then A is the zero matrix.

  8. True or False. If A is the zero matrix and AB is defined, then AB is a zero matrix.

Building Skills

In Exercises 9–16, find the values of all variables.

  1. [2−3]=[xu]

  2. [−y2x]=[43]

  3. [2xy−3]=[2−13−3]

  4. [2−x1−23+y]=[31−2−3]

  5. [2x−3y−453x+y]=[1−457]

  6. [3x+y−17192]=[−12−17192x+3y]

  7. [x−y1243x−2y3565x−10y]=[−1124−13566]

  8. [x+y232x−y4342x+3y]=[−12325434−5]

In Exercises 17–24, find each of the following if possible.

  1. A+B

  2. A−B

  3. −3A

  4. 3A−2B

  5. (A+B)2

  6. A2−B2

  1. A=[1234], B=[−102−3]

  2. A=[131−12], B=[102−12]

  3. A=[23−45], B=[102314]

  4. A=[123−1−34], B=[102314]

  5. A=[40−1−252001], B=[3101−42213]

  6. A=[102210013], B=[3−1202114−1]

  7. A=[12−33452−10], B=[3101−42213]

  8. A=[102210013], B=[3−1202114−1]

In Exercises 25–32, solve each matrix equation for X, where

A=[23−11−24]andB=[−210234].

  1. A+X=B

  2. B+X=A

  3. 2X−A=B

  4. 2X−B=A

  5. 2X+3A=B

  6. 3X+2A=B

  7. 2A+3B+4X=0

  8. 3X−2A+5B=0

In Exercises 33–42, find each product if possible.

  1. AB

  2. BA

  1. A=[1234], B=[−2135]

  2. A=[321501], B=[13−2250]

  3. A=[2−10−312], B=[15−2340]

  4. A=[123456], B=[−1−3−6257]

  5. A=[235], B=[1−24]

  6. A=[12−1], B=[−302]

  7. A=[123], B=[12−103120−3]

  8. A=[4−6223012−3], B=[−201]

  9. A=[2011423−10], B=[310−120452]

  10. A=[3−1204−31−22], B=[2−50−12−1302]

  11. Let A=[3120431−22] and B=[25012−1302]. Verify that AB≠BA.

  12. Let A=[524−6−310−265] and B=[412−305−134]. Verify that AB=BA.

In Exercises 45–48, let

A=[1234], B=[2−335], and C=[0124].

  1. Verify the associative property of multiplication: (AB)C=A(BC).

  2. Verify the distributive property: A(B+C)=AB+AC.

  3. Verify the distributive property: (A+B)C=AC+BC.

  4. For any real number c, verify that c(AB)=(cA)(B)=A(cB).

Applying the Concepts

  1. Cost matrix. The Build-Rite Co. is building an apartment complex. The cost of purchasing and transporting specific amounts of steel, glass, and wood (in appropriate units) from two different locations is given by the following matrices:

    SteelGlassWoodA=[7318413]Cost of materialTransportation costB=[6220314]Cost of materialTransportation cost

    Find the matrix representing the total cost of material and transportation for steel, glass, and wood from both locations.

  2. Cost matrix. Repeat Exercise 49 if the order from location A is tripled and the order from location B is doubled.

  3. Stock purchase. Ms. Goodbyer plans to buy 100 shares of computer stock, 300 shares of oil stock, and 400 shares of automobile stock. The computer stock is selling for $60 a share, oil stock is selling for $38 a share, and automobile stock is selling for $17 a share. Use matrix multiplication to calculate the total cost of Ms. Goodbyer’s purchases.

  4. Product export. The International Export Corporation received an export order for three of its products, say, A, B, and C. The export order is (in thousands) for 50 of product A, 75 of product B, and 150 of product C. The material cost, labor, and profit (each in appropriate units) required to produce each unit of the product are given in the following table:

    MaterialLaborProfitProduct A Product B Product C [202201532525115]

    Use matrix multiplication to compute the total material cost, total labor, and total profit if the entire export order is filled.

  5. Executive compensation. The Multinational Oil Corporation pays its top executives a salary, a cash bonus, and shares of its stock annually. In 2017, the chairman of the board received $2.5 million in salary, a $1.5 million bonus, and 50,000 shares of stock; the president of the company received one-half the compensation of the chairman; and each of the four vice presidents was paid $100,000 in salary, a $150,000 bonus, and 5000 shares of stock.

    1. Express payments to these executives in salary, bonus, and stock as a 3×3 matrix.

    2. Express the number of executives of each rank as a column matrix.

    3. Use matrix multiplication to compute the total amount the company paid to each executive in each category in 2017.

  6. Calories and protein. The Browns (B) and Newgards (N) are neighboring families. The Brown family has two men, three women, and one child, and the Newgard family has one man, one woman, and two children. Both families are weight watchers. They have the same dietician, who recommends the following daily allowance of calories and proteins:

    Calories

    • male adults: 2400 calories

    • female adults: 1900 calories

    • children: 1800 calories

    Protein (in grams)

    • male adults: 55

    • female adults: 45

    • children: 33

    Represent the preceding information in matrix form. Use matrix multiplication to compute the total requirements of calories and proteins for each of the two families.

In Exercises 55–58, a matrix A is given. Graph the figure represented by the matrix AD, where D is the matrix representation of the letter L of Example 10.

  1. A=[100−1]

  2. A=[−1001]

  3. A=[100.251]

  4. A=[1−0.2501]

In Exercises 59–62, a matrix A is given. Graph the figure represented by the matrix AD, where D is the matrix representation of a quadrilateral PQRS with the vertices P=(0, 0), Q=(2, 0), R=(1, 2), and S=(0, 1).

  1. [2001]

  2. [−100−1]

  3. [0.60.8−0.80.6]

  4. [1101]

In Exercises 63–66, a figure represented by the matrix AD is given, where D is the matrix representation of a quadrilateral PQRS with vertices P=(0, 0), Q=(1, 0), R=(2, 1), and S=(1, 2). Find matrix A.

Beyond the Basics

  1. Let A=[0300], B=[2130], and C=[5430]. Verify that AB=AC. This example shows that AB=AC does not, in general, imply that B=C.

  2. Let A=[2−3−5−1451−3−4] and B=[−1351−3−5−135]. Verify that AB=0. This example shows that AB=0 does not imply that A=0 or B=0.

  3. Select appropriate 2×2 matrices A and B to show that, in general, (A+B)2≠A2+2AB+B2.

  4. Repeat Exercise 69 to show that, in general, A2−B2≠(A−B)(A+B).

  5. Let A=[1−232−413−52]. Show that 3A2−2A+I=[17−2315−133010−92221], where I=[100010001].

  6. Let A=[1322031−11]. Show that

    A3−3A2+A−I=[−314−35−285−76],

    where I is given in Exercise 71.

  7. If possible, find a matrix B such that [2312]B=[1001].

  8. If possible find a matrix B such that

    [1−2−24]B=[1001].

  9. Find x and y if

    (4[2−13102]−[12−12−34])[2−11]=[xy].

  10. Find x, y, and z if

    [32−149250−2][xyz]=[072].

  11. Let A=[0100], B=[1000]. Show that AB=0 but BA=[0100]≠0.

  12. Find matrices A and B such that

    A+B=[5209] and A−B=[360−1].

  13. Find matrices A and B such that

    2A−B=[6−60−421] and A+B=[303−21−4].

  14. Let A=[2−3−5−1451−3−4], B=[−1351−3−5−135]. Show that AB=0 and BA=0.

  15. Let A=[2−3−5−1451−3−4], B=[2−2−4−1341−2−3]. Show that (i) AB=A, (ii) BA=B, (iii) A2=A.

  16. Let A=[23−12], I=[1001].

    1. Show that A2=4A−7I.

    2. Use part (a) to show that A4=8A−63I.

    3. Use part (b) to show that A4=[−4724−8−47].

  17. Let A=[4−1−430−43−1−3]. Show that A2=I, where I is given in Exercise 71.

Critical Thinking/Discussion/Writing

  1. Let A be a 3×n matrix and B be a 5×m matrix.

    1. Under what conditions is AB defined? What is the order of AB when this product is defined?

    2. Repeat part (a) for BA.

  2. Let A=[3−10−231], B=[234], and C=[−12]. In which order should all three matrices be multiplied to produce a number?

  3. Market share. The Asma Corporation (A), Bronkial Brothers (B), and Coufmore Company (C) simultaneously decide to introduce a new cigarette at a time when each company has one-third of the market. During the year, the following occurs:

    1. A retains 40% of its customers and loses 30% to B and 30% to C.

    2. B retains 30% of its customers and loses 60% to A and 10% to C.

    3. C retains 30% of its customers and loses 60% to A and 10% to B.

    The foregoing information can be represented by the following transition matrix:

    ABCP=ABC[0.40.30.30.60.30.10.60.10.3].

    Write the initial market share as X=[131313].

    1. Compute XP2 and interpret your result.

    2. Compute XP3, XP4, and XP5; observe the pattern for the long run. Describe the long-term market share for each company.

Getting Ready for the Next Section

In Exercises 87 and 88, rewrite each expression without exponents.

  1. (12)−1

  2. (512)−1

In Exercises 89–94, solve each equation.

  1. x−1=18

  2. (x35)−1=7

  3. 14x−712=1112−54x

  4. 23x−35=43x+215

  5. 24x−1=34x+1

  6. 2x+1=1x−1

In Exercises 95–98, solve each system.

  1. {3x+5y=26x+10y=4

  2. {−9x+2y=85x+8y=32

  3. {3x+2y=66x+4y=−13

  4. {3x−4y=132x+5y=1

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