Five-Minute Check (over Lesson 6-1) Then/Now New Vocabulary Key Concept: Matrix Multiplication Example 1: Multiply Matrices Key Concept: Properties of Matrix Multiplication Example 2: Real-World Example: Multiply Matrices Key Concept: Identity Matrix Example 3: Solve a System of Linear Equations Key Concept: Inverse of a Square Matrix Example 4: Verify an Inverse Matrix Example 5: Inverse of a Matrix Concept Summary: Finding the Inverse of a Square Matrix Theorem 6.1 Inverse and Determinant of a 2 × 2 Matrix Example 6: Determinant and Inverse of a 2 × 2 Matrix Theorem 6.2 Determinant of a 3 × 3 Matrix Example 7: Determinant and Inverse of a 3 × 3 Matrix Lesson Menu

Write the system of equations in triangular form using Gaussian elimination. Then solve the system. 3x + y + 2z = 31 –2x + y + 2z = 1 2x + y + 2z = 25 A. x + y + 2z = 19 y + 2z = 13 z = –5; (11, 18, –5) B. x + y + 2z = 19 y + 2z = 13 z = 5; (6, 3, 5) C. x + y + 2z = 19 y + 2z = 13 z = 5; (3, 6, 5) D. no solution 5–Minute Check 1

Solve the system of equations Solve the system of equations. 3x + 2y + 3z = 3 4x – 5y + 7z = 1 2x + 3y – 2z = 6 A. (0, 0, 1) B. (–2, 0, 3) C. (2, 0, –1) D. no solution 5–Minute Check 2

Solve the system of equations Solve the system of equations. 8x + 5y + 11z = 30 –x – 4y + 2z = 3 2x – y + 5z = 12 A. no solution B. (5 –2z, 2 + z, z) C. (–5 + 2z, 2 – z, z) D. (5 – 2z, –2 + z, z) 5–Minute Check 3

Which of the following matrices is in row-echelon form? B. C. D. 5–Minute Check 4

You performed operations on matrices. (Lesson 0-5) Multiply matrices. Find determinants and inverses of 2 × 2 and 3 × 3 matrices. Then/Now

identity matrix inverse matrix inverse invertible singular matrix determinant Vocabulary

Key Concept 1

A. Use matrices and to find AB, if possible. Multiply Matrices A. Use matrices and to find AB, if possible. AB = Dimensions of A: 3 X 2, Dimensions of B: 2 X 3 Example 1

Multiply Matrices A is a 3 X 2 matrix and B is a 2 X 3 matrix. Because the number of columns for A is equal to the number of rows for B, the product AB exists. To find the first entry in AB, write the sum of the products of the entries in row 1 of A and in column 1 of B. Example 1

Follow the same procedure to find the entry for row 1, column 2 of AB. Multiply Matrices Follow the same procedure to find the entry for row 1, column 2 of AB. Continue multiplying each row by each column to find the sum for each entry. Example 1

Finally, simplify each sum. Multiply Matrices Finally, simplify each sum. Example 1

Multiply Matrices Answer: Example 1

B. Use matrices and to find BA, if possible. Multiply Matrices B. Use matrices and to find BA, if possible. Dimensions of B: 2 X 3, Dimensions of A: 3 X 2 B is a 2 X 3 matrix and A is a 3 X 2 matrix. Because the number of columns for B is equal to the number of rows for A, the product BA exists. Example 1

Multiply Matrices To find the first entry in BA, write the sum of the products of the entries in row 1 of B and in column 1 of A. Follow this same procedure to find the entry for row 1, column 2 of BA. Example 1

Multiply Matrices Continue multiplying each row by each column to find the sum for each entry. Example 1

Finally, simplify each sum. Multiply Matrices Finally, simplify each sum. Answer: Example 1

Use matrices A = and B = to find AB, if possible. Example 1

Key Concept 2

Multiply Matrices FOOTBALL The number of touchdowns (TD), field goals (FG), points after touchdown (PAT), and two-point conversions (2EP) for the three top teams in the high school league for this season is shown in the table below. The other table shows the number of points each type of score is worth. Use the information to determine the team that scored the most points. Example 2

Multiply Matrices Let matrix X represent the Team/Score matrix, and let matrix Y represent the Score/Points matrix. Then find the product XY. Example 2

The Tigers scored the most points. Multiply Matrices The product XY represents the teams and the total number of points each team scored this season. You can use the product matrix to determine which team scored the most points. The Tigers scored the most points. Answer: Tigers Example 2

CAR SALES A car dealership sells four types of vehicles; compact cars (CC), full size cars (FS), trucks (T), and sports utility vehicles (SUV). The number of each vehicle sold during one recent month is shown in the table below. The other table shows the selling price for each of the vehicles. Which vehicle brought in the greatest revenue during the month? Example 2

D. sports utility vehicles A. compact cars B. full size cars C. trucks D. sports utility vehicles Example 2

Key Concept 3

Write the system in the form, AX = B. Solve a System of Linear Equations Write the system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve for X. 2x1 + 2x2 + 3x3 = 3 x1 + 3x2 + 2x3 = 5 3x1 + x2 + x3 = 4 Write the system in the form, AX = B. Example 3

Solve a System of Linear Equations Write the augmented matrix . Use Gauss-Jordan elimination to solve the system. Example 3

Therefore, the solution of the system of equations is (1, 2, –1). Solve a System of Linear Equations Therefore, the solution of the system of equations is (1, 2, –1). Answer: ; (1, 2, –1) Example 3

Write the system of equations as a matrix equation, AX = B Write the system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the augmented matrix to solve the system. 2x1 – x2 + x3 = –1 x1 + x2 – x3 = –2 x1 – 2x2 + x3 = –2 Example 3

A. ; (–1, 2, 3) B. ; (1, –2, –3) C. ; (–1, 2, 3) D. ; (1, –2, –3) Example 3

Key Concept 4

Determine whether and are inverse matrices. Verify an Inverse Matrix Determine whether and are inverse matrices. If A and B are inverse matrices, then AB = BA = I. Example 4

Because AB = BA = I, B = A–1 and A = B–1. Verify an Inverse Matrix Because AB = BA = I, B = A–1 and A = B–1. Answer: yes; AB = BA = I2 Example 4

Which matrix below is the inverse of A = ? A. B. C. D. B. B C. C D. D Example 4

Step 1 Create the doubly augmented matrix . Inverse of a Matrix A. Find A–1 when , if it exists. If A–1 does not exist, write singular. Step 1 Create the doubly augmented matrix . Example 5

Doubly Augmented Matrix Inverse of a Matrix Step 2 Apply elementary row operations to write the matrix in reduced row-echelon form. Doubly Augmented Matrix R1 + R2  –1R1  Example 5

R2 – 3R1 Row- echelon form R2 R1 + R2 Reduced A–1 row-echelon form Inverse of a Matrix R2 – 3R1 Row- echelon form R2 R1 + R2 Reduced row-echelon form A–1 Example 5

Inverse of a Matrix The first two columns are the identity matrix. Therefore, A is invertible and A–1 = . Answer: Example 5

Check Confirm that AA–1 = A–1A = I. Inverse of a Matrix Check Confirm that AA–1 = A–1A = I. Example 5

Step 1 Create the doubly augmented matrix . Inverse of a Matrix B. Find A–1 when , if it exists. If A–1 does not exist, write singular. Step 1 Create the doubly augmented matrix . Example 5

Doubly Augmented Matrix Inverse of a Matrix Step 2 Apply elementary row operations to write the matrix in reduced row-echelon form. Doubly Augmented Matrix 3R2 + R1 Notice that it is impossible to obtain the identity matrix I on the left side of the doubly augmented matrix. Therefore, A is singular. Answer: singular Example 5

Find A–1 when , if it exists. If A–1 does not exist, write singular. B. C. D. Example 5

Concept Summary 6

Key Concept 6

det (A) = a = –5, b = 10, c = 4, and d = –8 Determinant and Inverse of a 2 × 2 Matrix A. Find the determinant of . Then find the inverse of the matrix, if it exists. det (A) = a = –5, b = 10, c = 4, and d = –8 = (–5)(–8) – 10(4) or 0 ad – bc Answer: Because det(A) = 0, A is not invertible. Therefore, A–1 does not exist. Example 6

det (A) = a = –2, b = 4, c = –4, and d = 6 Determinant and Inverse of a 2 × 2 Matrix B. Find the determinant of . Then find the inverse of the matrix, if it exists. det (A) = a = –2, b = 4, c = –4, and d = 6 =(–2)(6) – (4)(–4) or 4 ad – bc Because det(A) ≠ 0, A is invertible. Apply the formula for the inverse of a 2 × 2 matrix. Example 6

Scalar multiplication Determinant and Inverse of a 2 × 2 Matrix B–1 Inverse of 2 × 2 matrix a = –2, b = 4, c = –4, and d = 6 Scalar multiplication Answer: 4; Example 6

CHECK BB–1= B–1B = . Determinant and Inverse of a 2 × 2 Matrix Example 6

Find the determinant of . Then find its inverse, if it exists. B. –2; C. 2; D. 0; does not exist Example 6

Key Concept 7

Find the determinant of . Then find D–1, if it exists. Determinant and Inverse of a 3 × 3 Matrix Find the determinant of . Then find D–1, if it exists. det(D) = 3[(–1)(5) – 4(2)] – [(–2)(5) –4(1)] + 0[(–2)(2) – (–1)1] Example 7

Determinant and Inverse of a 3 × 3 Matrix Because det(D) does not equal zero, D–1 exists. Use a graphing calculator to find D–1. Example 7

Determinant and Inverse of a 3 × 3 Matrix You can use the >Frac feature under the MATH menu to write the inverse using fractions, as shown below. Example 7

Therefore, D–1 = . Answer: –25; Determinant and Inverse of a 3 × 3 Matrix Therefore, D–1 = . Answer: –25; Example 7

Find the determinant of . Then find A–1, if it exists. C. 3, B. 3; D. 0; does not exist Example 7