courses involve systems of equations in one way or another.

Another Tool in the Toolbox Solving Matrix Equations.4 Learning Goals In this lesson you will: Determine the inverse of a matrix. Use matrices to solve systems of equations. Key Terms multiplicative identity matrix multiplicative inverse of a matrix matrix equation coefficient matrix variable matrix constant matrix 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 a)(s b)(s c)) (a 1 b 1 c) s 5 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? 137

Problem 1 Identity and Inverse Matrices 1. Calculate each product AB. a. A 5 5 6 0 3 B 5 1 0 0 1 b. A 5 4 1 1 0 5 3 3 B 5 1 0 0 0 1 0 0 0 1 c. A 5 1 5 0 1 3 B 5 1 0 0 1. How does each product matrix AB relate to the original matrices A and B? 3. Describe the similarities and differences in each matrix B. 138 Chapter Linear Systems and Matrices

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 multiplicative identity matrix, 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. I 1 5 [1], I 5 1 0 0 1, I 3 5 1 0 0 0 1 0 0 0 1,, I 5 n 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 5 4? I 5 3 5 4 b. 3 6 4 3? I 5 3 6 4 3 1 9 c. 0 5 1 9 4 1? I 5 0 5 d. [5 0 3]? I 5 [5 0 3] 4 1.4 Solving Matrix Equations 139

Recall that for any real number n, the multiplicative inverse is n 1 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 multiplicative inverse of a matrix (or just inverse) of A is designated as A 1, and is a matrix such that A? A 1 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 3 1 6 1 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 1 1 6 1 Use matrix multiplication to determine why RJ s reasoning is incorrect. 140 Chapter Linear Systems and Matrices

8. Use matrix multiplication to determine whether the two matrices are inverses. Explain your reasoning. a. C 5 9 4 1 D 5 1 4 9 b. W 5 1 3 3 3 4 V 5 3 1 4.4 Solving Matrix Equations 141

c. J 5 10 4 1 K 5 1 1 5 1 3 d. G 5 0 4 1 5 1 11 14 0 3 1 H 0 1 1 0 3 4 14 Chapter Linear Systems and Matrices

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. 5 8 1 A 5 4 4 0 8 1 Step 1: Enter matrix A into the calculator, and then return to the home screen. MATRIX[A] 3 *3 [5 8 1 ] [4-4 ] [0 8-1 ] 3,3= -1 Step : Go back into the matrix menu to select matrix A. Then press the x 1 key, followed by enter. [A] -1-3 4 5 1-1.5-1.5 8-10 -13 Step 3: Convert the decimals to fractions, if necessary. The inverse of matrix A is A 1 5 3 4 5 1 5 4 3 8 10 13. 8-10 -13 Ans Frac -3 4 5 1-5 4-3 8-10 -13.4 Solving Matrix Equations 143

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. 1 5 T 5 3 6 15 ERR:SINGULAR MAA 0 4 1 1:Quit :Goto Use your knowledge of matrices to explain why this matrix is noninvertible. 144 Chapter Linear Systems and Matrices

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 9 6 1 1 1 b. T 5 4 0 5 3 4 c. L 5 4 5 0 1 3 1 3.5 0 d. S 5 5 3 1 0 8.4 Solving Matrix Equations 145

Problem Solving Systems with Matrices Matrices can be used to solve a system of equations. A system can be written as a matrix equation, 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 coefficient matrix and a variable matrix equal to a constant matrix. A coefficient matrix is 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 variable matrix is a matrix in one column that represents all of the variables in the system of equations. A constant matrix is 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: x y z 5 3 3x 1 y z 5 11 x y 1 z 5 8. 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. Coefficient matrix 1 3 1 x 1 1 y z 3 5 11 8 A? X 5 B 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. 146 Chapter Linear Systems and Matrices

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?. Write each system of equations as a matrix equation. x 1 y z 5 8 x y 1 z 5 1 a. x 1 3y 1 z 5 13 3x 1 y 1 z = 4 b. 5x 5 4y x 1 y z 5 1 3. Describe the error in Antoine s method. Antoine x 4y 5 5z 1 x 5 y 3z 3x 5y 3z 5 14 4 5 x 1 3 3 5 3 y z 5 1 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..4 Solving Matrix Equations 147

You can solve matrix equations using a process that is similar to the one you use to solve linear equations. Solving Linear Equations Description Solving Matrix Equations 3 x 5 8 x 1 y 5 6 x 1 3y 5 7 1 1 3 x y 5 6 7 A? X 5 B ( 3 ) ( 3 ) x 5 ( 3 ) 18 Multiply each side by the multiplicative inverse. A 1 5 3 5 1 5 1 5 5 3 5 1 5 1 5 1 5 1 3 x y 5 A 1? A? X 5 3 5 1 5 1 5 5 A 1? B 6 7 1x 5 7 x 5 7 A number times its multiplicative inverse equals the identity. A quantity multiplied by the identity equals itself. ( 3 ) 1 ( 1 5 )1 ( 3 5 ) 1 1 ( 1 5 )3 ( 1 5 ) 1 ( 5 ) 1 ( 1 ( 3 5 ) (6) 1 ( 1 5 )7 ( 1 5 )(6) 1 ( 5 ) 7 1 0 18 5 )1 1 ( 5 ) 3 ( 0 1 x y 5 5 )1 ( 7 5 ) ( 6 5 ) 1 ( 14 5 ) I? X 5 A 1? B x y 5 5 4 X 5 A 1? B x y 5 148 Chapter Linear Systems and Matrices

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..4 Solving Matrix Equations 149

5. Natalie attempts to solve the system of equations and claims that it is not possible. Natalie x z 5 1 5x 3z 5 3 x 1 y 1 z 5 5 1 0 1 5 0 3 x y z 5 1 3 5 1 1 3 1 0 1 0 1 5 1 0 1 0 1 5 0 3 x y 3 5 1 1 z 5 1 3 1 0 1 0 1 5 1 0 A 1? A? X 5 B? A 1 The product B? A 1 is not defined so there is no solution to the system. a. Describe the error in Natalie s method. 150 Chapter Linear Systems and Matrices

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..4 Solving Matrix Equations 151

Remember, you can use technology with matrices as a tool when solving matrix equations. x y z 5 3 Consider the system of equations 3x 1 y z 5 11 x y 1 z 5 8. Step 1: Write the system as a matrix equation. Step : Calculate the inverse of A. 1 3 1 x 1 1 y z 3 5 11 8. A? X 5 B Ans * [B] J 3 j -1 Step 3: Multiply the inverse by matrix B to calculate the solution to the system. Ans * [B] J 3 j -1 The solution to the matrix equation is x y z 5 3, or (, 3, 1). 11 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 y z 5 1 x y z 5 3x 1 6y 3z 5 3 1 1 1 3 6 3 x y A? X 5 B X 5 A 1? B z 5 1 ERR:SINGULAR MAA 1:Quit :Goto 3 Why might Anthony have gotten the error? 15 Chapter Linear Systems and Matrices

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. Many Solutions x 1 5y z 5 3 x 1 10y 4z 5 6 x 1 y 1 3z 5 1 1 5 10 4 x 1 1 3 y z 5 3 6 1 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. No Solution x 1 5y z 5 3 x 1 10y 4z 5 7 x 1 y 1 3z 5 1 1 5 10 4 x 1 1 3 y z 5 3 7 1 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..4 Solving Matrix Equations 153

8. Two of the equations in the system of three linear equations are given. 5x 3y 1 z 5 4 x 1 y 3z 5 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. x 3y 5 7 y 1 z 5 5 a. x 1 y 1 4z 5 17 Don t forget to fill the variables that have a coefficient of zero. 154 Chapter Linear Systems and Matrices

5x 1 y 1 3z 5 9 x y z 5 16 b. x 1 4y 1 z 5 30 x 4y 1 3z 5 7 x 1 3y 5z 519 c. 4x 1 y z 517.4 Solving Matrix Equations 155

x 3z 5 4 x 1 y 5z 5 1 d. 3y 4z 5 156 Chapter Linear Systems and Matrices

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 $5 each, and large fruit baskets sold for $3.50 each. The Key Club took in a total of $086.5, 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.. 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..4 Solving Matrix Equations 157

The Drama Club is putting on their fall play. Tickets cost $.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. 158 Chapter Linear Systems and Matrices

Chapter Summary Key Terms Gaussian elimination (.) matrix (matrices) (.3) dimensions (.3) square matrix (.3) matrix element (.3) scalar multiplication (.3) matrix multiplication (.3) multiplicative identity matrix (.4) multiplicative inverse of a square matrix (.4) matrix equation (.4) coefficient matrix (.4) variable matrix (.4) constant matrix (.4).1 Solve Systems of Two Linear Equations A 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 y 5 0 5x 1 y 5 Multiplying both sides of the equation 3x y 5 0 by gives 6x y 5 0. 6x y 5 0 5x 1 y 5 11x 5 11x 5 x 5 Substitute x 5 into one of the linear equations. 3x y 5 0 3() y 5 0 6 y 5 0 y 5 6 y 5 6 The solution to the system is (, 6). 159

.1 Solve Systems of Equations Involving One Linear and One Quadratic Equation A 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 x 1 1 y 5 x 3x 1 4 x 3x 1 4 5 x 1 1 x 4x 1 3 5 0 (x 3)(x 1) 5 0 x 3 5 0 or x 1 5 0 x 5 3 or x 5 1 Substitute x 5 3 into the linear equation. y 5 3 1 1 y 5 4 Substitute x 5 1 into the linear equation. y 5 1 1 1 y 5 The solutions to the system are (3, 4) and (1, ). y y 5 x 3x 1 4 8 6 4 8 6 4 4 y 5 x 1 1 6 (1, ) (3, 4) 0 4 6 8 x 8 160 Chapter Linear Systems and Matrices

. Solve Systems of Three Linear Equations in Three Variables by Using Substitution To 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 3x 1 y 1 5z 5 5 Use substitution to solve the system 4x 1 y 1 z 5 13 5x 1 3y 1 6z 5 4. Solve the first equation for the variable y. 3x 1 y 1 5z 5 5 y 5 3x 5z 5 Substitute this equation into the other two equations. 4x 1 y 1 z 5 13 5x 1 3y 1 6z 5 4 4x 1 (3x 5z 5) 1 z 5 13 5x 1 3(3x 5z 5) 1 6z 5 4 4x 1 6x 10z 10 1 z 5 13 5x 1 9x 15z 15 1 6z 5 4 x 9z 10 5 13 4x 9z 15 5 4 x 9z 5 3 4x 9z 5 19 Solve the resulting system of two equations in two variables. x 9z 5 3 4x 9z 5 19 1(x 9z 5 3) 4x 9z 5 19 x 1 9z 5 3 4x 9z 5 19 x 5 4 x 5 Substitute the x-value to determine the z-value. x 9z 5 3 () 9z 5 3 4 9z 5 3 9z 5 7 z 5 3 Substitute the x- and z-values to determine the y-value. y 5 3x 5z 5 y 5 3() 5(3) 5 y 5 6 1 15 5 y 5 4 The solution to the system is (x, y, z) 5 (, 4, 3). Chapter Summary 161

. Solve Systems of Three Linear Equations in Three Variables by Using Gaussian Elimination The 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 x 1 5y 6z 5 4 Solve the system x 4y 1 5z 5 1 5x 4y 1 z 5 1 using Gaussian elimination. Add the first equation and x 1 5y 6z 5 4 the second equation. Replace x 4y 1 5z 5 1 the second equation. y z 5 3 x 1 5y 6z 5 4 y z 5 3 5x 4y 1 z 5 1 Multiply the first equation by 5 5(x 1 5y 6z 5 4) and add it to the third equation. 5x 5y 1 30z 5 10 Replace the third equation. 5x 5y 1 30z 5 10 5x 4y 1 z 5 1 9y 1 3z 5 99 Multiply the second equation by 9 9(y z 5 3) and add it to the third equation. 9y 9z 5 87 Replace the third equation. 9y 9z 5 87 9y 1 3z 5 99 3z 5 1 x 1 5y 6z 5 4 y z 5 3 9y 1 3z 5 99 x 1 5y 6z 5 4 y z 5 3 3z 5 1 Multiply the third equation by 1 3 1 (3z 5 1) 3 and add it to the second equation. z 5 4 Replace the second equation. y z 5 3 z 5 4 y 5 1 Multiply the second equation by 5 5(y 5 1) and add it to the first equation. 5y 5 5 Replace the first equation. x 1 5y 6z 5 4 5y 5 5 x 6z 5 9 x 1 5y 6z 5 4 y 5 1 3z 5 1 x 6z 5 9 y 5 1 3z 5 1 Multiply the third equation by (3z 5 1) and add it to the first equation. 6z 5 4 Replace the first equation. x 6z 5 9 6z 5 4 x 5 5 x 5 5 y 5 1 3z 5 1 Multiply the third equation by 1. 1 3 3 (3z 5 1) x 5 5 Replace the third equation. z 5 4 y 5 1 z 5 4 The solution to the system is (x, y, z) 5 (5, 1, 4). 16 Chapter Linear Systems and Matrices

.3 Determine the Dimensions of a Matrix A 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. A 5 1 9 15 4 7 8 3 0 10 5 11 1 6 3 5 10 13 The dimensions of matrix A are 6 3 3 because it has 6 rows and 3 columns..3 Identify Specific Matrix Elements Each number in a matrix is a matrix element. Each matrix element is labeled using the notation a ij, where i represents the row the element is in and j represents the column the element is in. Therefore, the notation a ij, represents the matrix element in the ith row and the jth column. Example Determine the matrix element represented by the notation b 15. B 5 5. 1.3 7.0 4.3 9.1 11.4 3.7 6.6 0.5 1.9 5.4 8.6 Matrix element b 15 is 9.1 because it is the element in row 5 and column 1. Chapter Summary 163

.3 Perform Matrix Operations: Addition, Subtraction, and Multiplication Matrices 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 a pq 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 CD 5 (5)(6) 1 4(1) 1 (8) (5)() 1 4(10) 1 (4) 5 4 3 7 9 and D 5 3(6) 1 (7)(1) 1 9(8) 3() 1 (7)(10) 1 9(4) 5 18 97 100 6 1 10. 8 4.4 Determine the Inverse of a Matrix The multiplicative inverse of a square matrix A, labeled as A 1, is a matrix such that when matrix A is multiplied by it the result is the identity matrix. In symbols, A? A 1 5 I, where I is the matrix with the same dimensions as matrix A having 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 J 1 is the inverse of matrix J. 3 1 1 3 1 3 ()( )1 1() ( ) ( 1 ) 1 1(1) 5 1 J 5 1 J 4 3 1 5 1 4 3 (4)( 3 )1 3() (4) ( 1 ) 1 3(1) 5 3 1 1 1 6 6 1 3 5 1 0 0 1 Since the product matrix J? J 1 yields the identity matrix, J 1 is the inverse of matrix J. 164 Chapter Linear Systems and Matrices

.4 Use Matrices to Solve Systems of Equations A 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 A 1. 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. x 1 8y 1 3z 5 1 7x 5y 1 z 5 7 6x y 1 3z 5 17 8 3 7 5 1 6 3 x 1 y 7 z 5 17 A? X 5 B A 1 5 13 374 15 3 187 374 7 1 19 34 187 374 17 17 3 17 X 5 A 1? B x y 5 z 13 374 15 3 187 374 7 1 19 34 187 374 17 17 3 17 1 7 17 4 5 7 9 The solution to the system is (4, 7, 9). Chapter Summary 165

166 Chapter Linear Systems and Matrices