Another Tool 2.4 in the Toolboxhhspreapalgebra2.weebly.com/uploads/1/7/0/2/... · 146 Chapter 2 Linear Systems and Matrices 2 Problem 2 Solving Systems with Matrices Matrices can

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LEArnIng gOALS

Many problems that you have encountered in earlier grades and in different courses involve systems of equations in one way or another.One idea you may not have seen before is called Heron’s Formula. This is a formula used to calculate the area of a triangle. In the formula shown, the side lengths of a triangle are represented by the variables a, b, and c.

A 5 √______________________

(s(s 2 a)(s 2 b)(s 2 c))

s 5 (a 1 b 1 c)

___________ 2

This amounts to a system of equations you can use to determine the area of a triangle.

What other problems have you encountered before that you could use matrices to help you solve?

KEy TErMS

• multiplicative identity matrix• multiplicative inverse of a matrix• matrix equation• coefficient matrix• variable matrix• constant matrix

In this lesson you will:

• Determine the inverse of a matrix .• Use matrices to solve systems of equations .

Another Tool in the ToolboxSolving Matrix Equations

2.4

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138 Chapter 2 Linear Systems and Matrices

2

Problem 1 Identity and Inverse Matrices

1. Calculate each product AB .

a. A 5   25 6 0 3 B 5   1 0 0 1

b. A 5   4 22 1 21 0 5 3 23 2 B 5  

1 0 0 0 1 0

0 0 1

c. A 5   1 5 0 21 22 3

B 5   1 0 0 1

2. How does each product matrix AB relate to the original matrices A and B?

3. Describe the similarities and differences in each matrix B .

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2.4 Solving Matrix Equations 139

Previously, you learned that for any real number n, the multiplicative identity is 1 because n? 1 5 n . Similarly, a multiplicative identity exists for matrices as well .

The multiplicativeidentitymatrix, I, is a square matrix such that for any matrix A, AI 5 A .

The multiplicative identity matrix is a square matrix whose elements are 0s and 1s . The 1s are arranged diagonally from upper left to lower right as shown .

I1 5 [1], I2 5    1 0 0 1 , I3 5

     1 0 0 0 1 0 0 0 1 , ∙ ∙ ∙ , In 5   1 0 ∙ ∙ ∙ 0 0 1 ∙ ∙ ∙ 0

∙ ∙ ∙

∙ ∙ ∙

∙ ∙ ∙

0 0 ∙ ∙ ∙ 1

∙ ∙

∙

4. Determine the 5 3 5 multiplicative identity matrix .

5. If E is an n 3 m matrix, then what must the dimensions be for its multiplicative identity matrix?

6. For each equation, determine the appropriate multiplicative identity matrix . Explain your reasoning .

a.   3 22 5 24 ? I 5    3 22 5 24 b.   2 23 6 22 4 3 ? I 5  

2 23 6 22 4 3

c.   1 9 0 25 4 1

? I 5     1 9 0 25 4 1

d. [5 22 0 3] ? I 5 [5 22 0 3]

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Recall that for any real number n, the multiplicative inverse is 1 __ n because the product of n and 1 __ n is the multiplicative identity, or 1 . Similarly, when a matrix is multiplied by its inverse matrix, their product is the identity matrix .

The multiplicativeinverseofamatrix(or just inverse) of A is designated as A21, and is a matrix such that A ? A21 5 1 . Non-square matrices do not have inverses .

7. RJ used his knowledge of multiplicative inverses to make an assumption about the inverse of matrix R .

rJ

R 5   2

__ 3 1

6 21

Matrix S is the inverse of matrix R because each matrix element in matrix S is the reciprocal of the corresponding matrix element in matrix R.

S 5   3

__ 2 1

1 __ 6 21

Use matrix multiplication to determine why RJ’s reasoning is incorrect .

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8. Use matrix multiplication to determine whether the two matrices are inverses . Explain your reasoning .

a. C5   9 22 24 1 D 5   1 2 4 9

b. W 5   1 __ 2 2 __ 3 23 24

V 5

  2 3 2  

3 __ 2

1 __ 4


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c. J 5   10 24 22 1

K5   1 __ 2 2 1 5

d. G 5   1 2 3 0 24 1 0 3 21 H5   1 11 14 0 21 21

0 23 24

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9. Compare the inverse matrices in Question 8 . State the similarities . Can you determine a pattern to calculate an inverse of a matrix?

Calculating the inverse of a matrix by hand is a complicated process and is the focus of another course . However, we can use technology to help us to determine the inverse of a matrix, if it exists .

Calculate the inverse of matrix A .

A 5   5 8 1

4 24 2

0 8 21

Step1: Enter matrix Ainto the calculator, and then return to the home screen .

Step2: Go back into the matrix menu to select matrix A . Then press the x21 key, followed by enter .

Step3: Convert the decimals to fractions, if necessary .

The inverse of matrix A is A21 5  

23 4 5 1 2  

5 __ 4 2  

3 __ 2

8 210 213

.

MATRIX[A] 3 *3

[4 -4 2 ]

3,3= -1

[5 8 1 ]

[0 8 -1 ]

[A]-1

-3 4 51 -1.25 -1.58 -10 -13

8 -10 -13

-3 4 51 - -54

32

8 -10 -13Ans Frac


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Not all square matrices have inverses, however . The matrix is noninvertible, that is, the inverse of a matrix does not exist, when any row of a matrix is a multiple of another row, or any column is a multiple of another column .

10. Cameron attempted to calculate the inverse of matrix T and his calculator returned an error message that claimed it was a singular, or noninvertible, matrix .

T 5     21 22        5 23     6   15     0     4         1

ERR:SINGULAR MAA

2:Goto1:Quit

Use your knowledge of matrices to explain why this matrix is noninvertible .

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11. Use a graphing calculator to calculate the inverse of each matrix, if it exists . Use matrix multiplication to verify your answer .

a. M 5   29 22 6 1 b. T 5   1 2 21 4 0 5

3 2 4

c. L 5   4 5 0 21 23 2 22 22 .5 0

d. S5   1 2 3 2 5 3 1 0 8


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Problem 2 Solving Systems with Matrices

Matrices can be used to solve a system of equations . A system can be written as a matrixequation, or an equation with matrices .

Consider a system of three linear equations in three variables, written in standard form: Ax 1 By 1 Cz 5 D . The system can be written as a matrix equation A ? X 5 B, by writing it as the product of a coefficientmatrix and a variablematrix equal to a constantmatrix .

A coefficientmatrixis a square matrix that consists of each coefficient of each equation in the system of equations, in order, when they are written in standard form . A variablematrixis a matrix in one column that represents all of the variables in the system of equations . A constantmatrixis a matrix in one column that represents each of the constants in the system of equations .

In a system with n equations, the coefficient matrix will be an n 3 n matrix, the variable matrix will be an n 3 1 matrix, and the constant matrix will be an n x 1 matrix .

Consider the system of three linear equations in three variables:

  2x 2 y 2 2z 5 3

3x 1 y 2 2z 5 11

22x 2 y 1 z 5 28   .

Each equation is written in standard form Ax 1 By 1 Cz 5 D .

In the matrix equation A? X 5 B, A is a 3 3 3 matrix consisting of each coefficient, X is a 3 3 1 matrix representing all of the variables in the system, and B is a 3 3 1 matrix consisting of each constant term .

  2 21 22 3 1 22 22 21 1

  x y z5   3 11

28 A ? X 5 BCoefficient

matrix

Variable matrix

Coefficient matrix

1. Verify that the matrix equation is an equivalent representation of the system of equations .

a. Use matrix multiplication to calculate A ? X .

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b. Write the system of equations that the matrix equation represents . Justify your reasoning .

c. Is the matrix equation an equivalent representation of the system of equations?

2. Write each system of equations as a matrix equation .

a.   x 1 2y22z 5 28

2x2 y 1 z 5 21

2x1 3y1 2z5 13

b.   3x 1 2y 1 2z = 4

5x 5 24y

2x 1 2y2 z 5 12  

3. Describe the error in Antoine’s method .

Antoine

  2x 2 4y 5 5z 2 1 2x 5 2y 2 3z

3x 2 5y 2 3z 5 14  

  2 24 5 21 2 23 3 25 23  

x y z 5     

21 0

14


Be sure to check that each

equation is of the form Ax 1 By 1 Cz 5 D before you write it in

matrix form.

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You can solve matrix equations using a process that is similar to the one you use to solve linear equations .

SolvingLinearEquations

Description SolvingMatrixEquations

2 __ 3 x 5 8   2x 1 y 5 26 x 1 3y 5 7  

  2 1 1 3   x y 5   26 7

A ? X 5 B

( 3 __ 2 ) ( 2 __ 3 ) x 5 ( 3 __ 2 ) 18 Multiply each side by the multiplicative inverse .

A21 5   3 __ 5 2  1 __ 5

2  1 __ 5

2 __ 5

  3 __ 5 2  1 __ 5

2  1 __ 5

2 __ 5   2 1 1 3   x y5 

3 __ 5 2  1 __ 5

2  1 __ 5

2 __ 5   26 7

A21 ? A ? X 5 A21 ? B

1x 5 27 A number times its multiplicative inverse equals the identity .

  ( 3 __ 2 ) 2 1 ( 2  1 __ 5 ) 1 ( 3 __ 5 ) 1 1 ( 2  

1 __ 5 ) 3 ( 2  1 __ 5 ) 2 1 ( 2 __ 5 ) 1 ( 2  

1 __ 5 ) 1 1 ( 2 __ 5 ) 3   x y5

  ( 3 __ 5 ) (26) 1 ( 2  1 __ 5 ) 7

( 2  1 __ 5 ) (26) 1 ( 2 __ 5 ) 7

  1 0 0 1   x y 5   ( 2  

18 ___ 5 ) 1 ( 2  7 __ 5 )

( 6 __ 5 ) 1 ( 14 ___ 5 )

I ? X 5 A21 ?    B

x 5 27 A quantity multiplied by the identity equals itself .

 x__ y 5   25 ___ 4 X 5 A21 ? B

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4. Analyze the worked example .

a. In what ways are solving linear equations and solving matrix equations similar?

b. Describe in words how to calculate the solution to a matrix equation .


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5. Natalie attempts to solve the system of equations and claims that it is not possible .

natalie

 

x 2 z 5 215x 2 3z 5 32x 1 2y 1 z 5 5  

  1 0 215 0 2321 2 1

  x y z 5   21

3

5

 

2  3 __ 2

1 __ 2

0

1 __ 2

0 1 __ 2

2  5 __ 2 1 __ 2

0   1 0 21 5 0 2321 2 1

 

x y

z 5   21

3

5

 

2  3 __ 2

1 __ 2

0

1 __ 2

0 1 __ 2

2  5 __ 2 1 __ 2

0

A21 ? A ? X 5 B ? A21

The product B ? A21 is not defined so there is no solution to the system.

a. Describe the error in Natalie’s method .

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b. What conclusion can you draw from Natalie’s error .

c. Calculate the solution to the system of equations .

d. Use substitution to verify the solution to the system of equations .


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Remember, you can use technology with matrices as a tool when solving matrix equations .

Consider the system of equations   2x 2 y 2 2z 5 3

3x 1 y 2 2z 5 11

22x 2 y 1 z 5 28   .

Step1: Write the system as a matrix equation .   2 21 22 3 1 22 22 21 1

  x y z5   3 11

28 .

A ?    X 5 B

Step2: Calculate the inverse of A .

Step3: Multiply the inverse by matrix B to calculate the solution to the system .

The solution to the matrix equation is  x y z5  

2 3

211 , or (2, 3, 21) .

Ans*[B]23-1

J j

Ans*[B]23-1

J j

6. Anthony is asked to solve a system of three linear equations using technology with matrices and he gets an error message .

Anthony

  x 1 2y 2 z 5 1

2x 2 2y 2 z 5 2 3x 1 6y 2 3z 5 3  

  1 2 21 2 22 21 3 6 23   x y z 5  

1 2 3 A ? X 5 B

X 5 A21 ? B

ERR:SINGULAR MAA

2:Goto1:Quit

Why might Anthony have gotten the error?

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Recall that when any row of a matrix is a multiple of another row, or any column is a multiple of another column, the inverse of the matrix does not exist . When the coefficient matrix is noninvertible, the system is not able to be solved . In this case, the system will either have many solutions or no solution .

When using technology, the calculator will return the same error message for both cases, because in both cases, the inverse of the coefficient matrix does not exist . You can use the constant matrix to differentiate between a system with many solutions and a system with no solution .

ManySolutions

  x 1 5y 2 2z 5 3

2x 1 10y 2 4z 5 6 x 1    y 1 3z 5 21

 

  1 5 22 2 10 24 1 1 3   x y z5  

3 6 21

If each variable in one equation is a multiple of another equation, and the constants are a multiple by the same factor, the system will have many solutions .

NoSolution

  x 1 5y 2 2z 5 3

2x 1 10y 2 4z 5 7 x 1    y 1 3z 5 21

 

  1 5 22 2 10 24 1 1 3   x y z5  

3 7 21

If each variable in one equation is a multiple of another equation, and the constants are not a multiple by the same factor, the system will have no solution .

7. Determine whether the system of equations in Question 6 has no solution or many solutions .


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8. Two of the equations in the system of three linear equations are given .

  5x2 3y1 z5 24 x1 2y2 3z5 0   .a. Write a third equation that would produce a system with no solution .

Explain your reasoning .

b. Write a third equation that would produce a system with many solutions .

9. Write each system of equations as a matrix equation, if it exists . Then calculate the solution to each system of linear equations by using technology with matrices .

a.   2x2 3y 5 7

y 1 z 5 25

x 1 2y1 4z5 217  

Don’t forget to fill the variables

that have a coefficient of zero.

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b.   5x1 y1 3z5 9

2x2 2y2  z 5 216

2x1 4y1 2z5 230  

c.   x 2 4y 1 3z 5 27

2x1 3y2 5z519

4x1 y2z 517  


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d.   2x 23z 5 4

2x 1 y 2 5z 5 21

3y2 4z 5 2  

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Problem 3 Show Us Your Stuff!

Your school’s Key Club decided to sell fruit baskets to raise money for a local charity. The club sold a total of 80 fruit baskets. There were three different types of fruit baskets. Small fruit baskets sold for $15.75 each, medium fruit baskets sold for $25 each, and large fruit baskets sold for $32.50 each. The Key Club took in a total of $2086.25, and they sold twice as many large baskets as small baskets.

1. Formulate a system of three linear equations in three variables to represent this problem situation. Be sure to define your variables.

2. Solve the system of three linear equations using technology with matrices. Write your answer in terms of the problem situation.

3. Describe the solution in terms of the problem situation.


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The Drama Club is putting on their fall play. Tickets cost $2.00 for children, $5.00 for adults, and $3.00 for senior citizens. On the first weekend, they sold a total of 450 tickets and the ticket sales totaled $1800. The following weekend, they sold twice as many of each type of ticket and their ticket sales totaled $3600.

4. Formulate a system of three linear equations in three variables to represent the problem situation. Be sure to define your variables.

5. Solve the system of equations using technology with matrices.

6. Describe the solution mathematically, as well as in terms of the problem situation.

Be prepared to share your methods and solutions.

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KEy TErMS

• Gaussian elimination (2 .2)• matrix (matrices) (2 .3)• dimensions (2 .3)• square matrix (2 .3)• matrix element (2 .3)• scalar multiplication (2 .3)

• matrix multiplication (2 .3)• multiplicative identity

matrix (2 .4)• multiplicative inverse of a

square matrix (2 .4)• matrix equation (2 .4)

• coefficient matrix (2 .4)• variable matrix (2 .4)• constant matrix (2 .4)

Solve Systems of Two Linear EquationsA system of two linear equations can be solved algebraically by using either the substitution method or the linear combinations method . Systems of two linear equations can have one solution, no solution, or an infinite number of solutions .

Example

Solve the system of linear equations algebraically . Use either the substitution method or the linear combinations method .

  3x 2 y 5 0 5x1 2y5 22   Multiplying both sides of the equation 3x 2 y 5 0 by 2 gives 6x 2 2y 5 0 .

6x2 2y5 0 5x 1 2y 5 2211x 5 22

11x5 22x5 2

Substitute x 5 2 into one of the linear equations .

3x2 y5 03(2) 2 y 5 0

6 2 y5 02y5 26

y5 6

The solution to the system is (2, 6) .

2.1

Chapter 2 Summary

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Solve Systems of Equations Involving One Linear and One Quadratic EquationA system of two equations involving one linear equation and one quadratic equation can be solved using methods similar to those for solving a system of two linear equations . These systems can have one solution, two solutions, or no solutions . The solution(s) can be verified by graphing the equations on the same coordinate grid and then calculating the point(s) of intersection .

Example

Solve the system of two equations in two variables algebraically . Then verify the solution graphically .

 y 5 x1 1 y5 x 2 2 3x1 4   x 2 2 3x1 4 5 x 1 1x 2 2 4x1 3 5 0

(x2 3)(x2 1) 5 0

x2 35 0 or x2 1 5 0x 5 3 or x5 1

Substitute x 5 3 into the linear equation .

y 5 3 1 1y5 4

Substitute x 5 1 into the linear equation .

y5 1 1 1y5 2

The solutions to the system are (3, 4) and (1, 2) .

28 26 24 2222

24

20 4 6 8

28

26

8

6

4

2(1, 2)

(3, 4)

y 5 x 1 1

y 5 x2 2 3x 1 4

x

y

2.1

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Chapter 2 Summary 161

Solve Systems of Three Linear Equations in Three Variables by Using SubstitutionTo solve a system of three linear equations using substitution, the first step is to solve for one variable in one of the equations . Then substitute this expression for that variable in the other two equations . The two new equations will then have only two unknown variables and can be solved using either substitution or linear combinations .

Example

Use substitution to solve the system     23x1 y1 5z5 25

24x1 2y1 z 5 13

25x1 3y1 6z5 4   .

Solve the first equation for the variable y . 23x 1 y1 5z 5 25 y 5 3x 2 5z 2 5

Substitute this equation into the other two equations .24x1 2y 1 z 5 13 25x1 3y1 6z5 4

24x1 2(3x2 5z2 5) 1 z5 13 25x1 3(3x2 5z2 5) 1 6z 5 424x1 6x2 10z2 101 z5 13 25x1 9x2 15z2 15 1 6z5 4

2x 2 9z2 10 5 13 4x2 9z2 15 5 42x2 9z5 23 4x2 9z5 19

Solve the resulting system of two equations in two variables .

  2x2 9z5 23 4x2 9z5 19  

21(2x 2 9z 5 23)4x2 9z5 19

22x1 9z5 2234x2 9z5 19

2x5 24x 5 22

Substitute the x-value to determine the z-value .

2x2 9z5 232(22) 2 9z5 23

24 2 9z5 2329z5 27

z5 23

Substitute the x- and z-values to determine the y-value .

y5 3x2 5z2 5y5 3(22) 2 5(23) 2 5y5 26 1 15 2 5y5 4

The solution to the system is (x, y, z) 5 (22, 4, 23) .

2.2

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Solve Systems of Three Linear Equations in Three Variables by Using gaussian EliminationThe goal of Gaussian elimination is to use linear combinations to isolate one variable for each equation . When using this method, you can:

• swap the positions of two equations .• multiply an equation by a nonzero constant .• multiply an equation by a nonzero constant and add to result to one of the other equations .

Example

Solve the system  x 1 5y 2 6z 5 24

2x 2 4y 1 5z 5 221

5x 2 4y 1 2z 5 21

  using Gaussian elimination .

Add the first equation and x 1 5y 2 6z 5 242x 2 4y 1 5z 5 221

y 2 z 5 3

→

 

x 1 5y 2 6z 5 24y 2 z 5 3

5x 2 4y 1 2z 5 21   the second equation . Replace the second equation .

Multiply the first equation by 25 25(x1 5y2 6z5 24)

 x 1 5y 2 6z 5 24

y 2 z 5 3229y 1 32z 5 299  

and add it to the third equation . 25x2 25y1 30z5 2120 → Replace the third equation .  25x2 25y1 30z 5 2120

5x2 4y1 2z 5 21  229y1 32z 5 299

Multiply the second equation by 29 29(y2 z5 3)

 x 1 5y 2 6z 5 24

y 2 z 5 33z 5 212  

and add it to the third equation . 29y 2 29z 5 87 → Replace the third equation . 29y2 29z5 87

229y1 32z5 2993z5 212

Multiply the third equation by 1 __ 3 1 __

3 (3z5 212)

and add it to the second equation . z5 24 Replace the second equation . y2 z5 3

z5 24y 5 21

→

 x 1 5y 2 6z 5 24

y 5 213z 5 212  

Multiply the second equation by 25 25(y5 21)and add it to the first equation . 25y5 5Replace the first equation . x 1 5y2 6z5 24

25y 5 5x 2 6z 5 29

→

 x 2 6z 5 29

y 5 213z 5 212  

Multiply the third equation by 2 2(3z 5 212) and add it to the first equation . 6z 5 224 Replace the first equation . x2 6z5 29

6z5 224x 5 5

→

 

x 5 5y 5 21

3z 5 212  

Multiply the third equation by 1 __ 3 . 1 __

3 (3z 5 212)

 x 5 5y 5 21z 5 24  

Replace the third equation . z 5 24 →

The solution to the system is (x, y, z) 5 (5, 21, 24) .

2.2

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Determine the Dimensions of a MatrixA matrix is an array of numbers composed of rows and columns . The dimensions of the matrix are the number of rows by the number of columns . The dimensions of a matrix with a rows and b columns are written as a 3 b .

Example

Determine the dimensions of the matrix .

A5 

 21 2 9

215 4 7

8 23 0

10 5 211

1 26 3

25 210 13

The dimensions of matrix A are 6 3 3 because it has 6 rows and 3 columns .

Identify Specific Matrix ElementsEach number in a matrix is a matrix element . Each matrix element is labeled using the notation aij, where i represents the row the element is in and j represents the column the element is in . Therefore, the notation aij, represents the matrix element in the ith row and the jth column .

Example

Determine the matrix element represented by the notation b15 .

B 5  

5 .2 21 .3 7 .0 24 .3 29 .1 11 .4

3 .7 26 .6 0 .5 1 .9 25 .4 28 .6

Matrix element b15 is 29 .1 because it is the element in row 5 and column 1 .

2.3

2.3

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Perform Matrix Operations: Addition, Subtraction, and MultiplicationMatrices that have the same dimensions can be added or subtracted . The matrix elements in the corresponding location are added or subtracted . A matrix can also be multiplied by a constant, a process called scalar multiplication . Each element of the matrix is multiplied by the constant . Two matrices can be multiplied only if the number of columns in the first matrix is the same as the number of rows in the second matrix . An element apq of the product matrix is determined by multiplying each element in row p of the first matrix with an element from column q in the second matrix and calculating the sum of these products .

Example

Calculate the product CD of the matrices C 5  

25 4 2

3 27 9 and D 5

 

6 2

21 10

8 24

.

CD5   

(25)(6) 1 4(21) 1 2(8) (25)(2) 1 4(10) 1 2(24)

3(6) 1 (27)(21) 1 9(8) 3(2) 1 (27)(10) 1 9(24)

5   

218 22

97 2100

Determine the Inverse of a MatrixThe multiplicative inverse of a square matrix A, labeled as A21, is a matrix such that when matrix A is multiplied by it the result is the identity matrix . In symbols, A ? A21 5 I, where I is the matrix with the same dimensions as matrix Ahaving all zeros except for a diagonal from the upper left to the lower right where the elements are all ones . Technology can be used to find the inverse of a square matrix . Not every square matrix has an inverse, and non-square matrices do not have inverses .

Example

Use matrix multiplication to verify that matrix J21 is the inverse of matrix J .

J5   

22 1

24 3             J21 5

 

2  3 __ 2

1 __ 2

22 1

 

22 1

24 3  

 

2  3 __ 2

1 __ 2

22 1

5  (22) ( 2  3 __ 2 ) 1 1(22) ( 22) ( 1 __ 2 ) 1 1(1)

(24) ( 2  3 __ 2 ) 1 3(22) (24) ( 1 __ 2 ) 1 3(1) 5 

 

3 2 2 21 1 1

6 2 6 22 1 3

5 

 

1   0

0   1

Since the product matrix J ? J21 yields the identity matrix, J21 is the inverse of matrix J .

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Use Matrices to Solve Systems of EquationsA system of linear equations can be written as a matrix equation, in the form A ? X 5 B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix . To solve the matrix equation A ? X 5 B, multiply both sides by the inverse of matrix A, or A21 .

Example

Write the following system of equations as a matrix equation . Then calculate the solution to the system of linear equations, if it exists, by using technology with matrices .

  2x 1 8y1 3z5 221

7x2 5y1 z5 72

26x2 2y1 3z5 17  

 

2 8 3

7 25 1

26 22 3

 x

y

z 5  

 221

72

17

A ?        X 5 B

A21 5

  13 ____ 374

15 ____ 187

2  23 ____ 374

27 ____ 324

2  12 ____ 187 2  19 ____ 374

2 ___ 17

2 ___ 17

3 ___ 17

X 5 A21 ? B

 x

y

z

 5

  13 ____ 374

15 ____ 187

2  23 ____ 374

27 ____ 324

2  12 ____ 187 2  19 ____ 374

2 ___ 17

2 ___ 17

3 ___ 17

 

 221

72

17

5 

 

4

27

9

The solution to the system is (4, 27, 9) .

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Another Tool 2.4 in the Toolboxhhspreapalgebra2.weebly.com/uploads/1/7/0/2/... · 146 Chapter 2 Linear Systems and Matrices 2 Problem 2 Solving Systems with Matrices Matrices can