LESSON 6 2 Matrix Multiplication Inverses and Determinants

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LESSON 6– 2 Matrix Multiplication, Inverses, and Determinants

LESSON 6– 2 Matrix Multiplication, Inverses, and Determinants

Five-Minute Check (over Lesson 6 -1) Then/Now New Vocabulary Key Concept: Matrix Multiplication Example

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

Over Lesson 6 -1 Write the system of equations in triangular form using Gaussian

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

Over Lesson 6 -1 Solve the system of equations. 3 x + 2 y

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

Over Lesson 6 -1 Solve the system of equations. 8 x + 5 y

Over Lesson 6 -1 Solve the system of equations. 8 x + 5 y + 11 z = 30 –x – 4 y + 2 z = 3 2 x – y + 5 z = 12 A. no solution B. (5 – 2 z, 2 + z, z) C. (– 5 + 2 z, 2 – z, z) D. (5 – 2 z, – 2 + z, z)

Over Lesson 6 -1 Which of the following matrices is in row-echelon form? A.

Over Lesson 6 -1 Which of the following matrices is in row-echelon form? A. C. B. D.

You performed operations on matrices. (Lesson 0 -5) • Multiply matrices. • Find determinants

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

 • identity matrix • inverse • invertible • singular matrix • determinant

• identity matrix • inverse • invertible • singular matrix • determinant

Multiply Matrices A. Use matrices and find AB, if possible. AB = Dimensions of

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

Multiply Matrices A is a 3 X 2 matrix and B is a 2

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.

Multiply Matrices Follow the same procedure to find the entry for row 1, column

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.

Multiply Matrices Finally, simplify each sum.

Multiply Matrices Finally, simplify each sum.

Multiply Matrices Answer:

Multiply Matrices Answer:

Multiply Matrices B. Use matrices and to find BA, if possible. Dimensions of B:

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.

Multiply Matrices To find the first entry in BA, write the sum of the

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.

Multiply Matrices Continue multiplying each row by each column to find the sum for

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

Multiply Matrices Finally, simplify each sum. Answer:

Multiply Matrices Finally, simplify each sum. Answer:

Use matrices A = AB, if possible. A. B. C. D. and B =

Use matrices A = AB, if possible. A. B. C. D. and B = to find

Multiply Matrices FOOTBALL The number of touchdowns (TD), field goals (FG), points after touchdown

Multiply Matrices FOOTBALL The number of touchdowns (TD), field goals (FG), points after touchdown (PAT), and twopoint conversions (2 EP) 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.

Multiply Matrices Let matrix X represent the Team/Score matrix, and let matrix Y represent

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

Multiply Matrices The product XY represents the teams and the total number of 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

CAR SALES A car dealership sells four types of vehicles; compact cars (CC), full

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?

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

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

Solve a System of Linear Equations Write the system of equations as a matrix

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. 2 x 1 + 2 x 2 + 3 x 3 = 3 x 1 + 3 x 2 + 2 x 3 = 5 3 x 1 + x 2 + x 3 = 4 Write the system in the form, AX = B.

Solve a System of Linear Equations Write the augmented matrix elimination to solve the

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

Solve a System of Linear Equations Therefore, the solution of the system of equations

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

Write the system of equations as a matrix equation, AX = B. Then use

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

Verify an Inverse Matrix Determine whether inverse matrices. and If A and B are

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

Verify an Inverse Matrix Because AB = BA = I, B = A– 1

Verify an Inverse Matrix Because AB = BA = I, B = A– 1 and A = B– 1. Answer: yes; AB = BA = I 2

Which matrix below is the inverse of A = A. A. A B. B

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

Inverse of a Matrix A. Find A– 1 when , if it exists. If

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 .

Inverse of a Matrix Step 2 Apply elementary row operations to write the matrix

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

Inverse of a Matrix R 2 – 3 R 1 Rowechelon form R 2

Inverse of a Matrix R 2 – 3 R 1 Rowechelon form R 2 R 1 + R 2 Reduced row-echelon form A– 1

Inverse of a Matrix The first two columns are the identity matrix. Therefore, A

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

Inverse of a Matrix Check Confirm that AA– 1 = A– 1 A =

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

Inverse of a Matrix B. Find A– 1 when , if it exists. If

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 .

Inverse of a Matrix Step 2 Apply elementary row operations to write the matrix

Inverse of a Matrix Step 2 Apply elementary row operations to write the matrix in reduced row-echelon form. Doubly Augmented Matrix 3 R 2 + R 1 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

Find A– 1 when not exist, write singular. A. B. C. D. , if

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

Determinant and Inverse of a 2 × 2 Matrix A. Find the determinant of

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) = = (– 5)(– 8) – 10(4) or 0 a = – 5, b = 10, c = 4, and d = – 8 ad – bc Answer: Because det(A) = 0, A is not invertible. Therefore, A– 1 does not exist.

Determinant and Inverse of a 2 × 2 Matrix B. Find the determinant of

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

Determinant and Inverse of a 2 × 2 Matrix B– 1 Inverse of 2

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;

Determinant and Inverse of a 2 × 2 Matrix CHECK BB– 1= B– 1

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

Find the determinant of inverse, if it exists. . Then find its A. 2;

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

Determinant and Inverse of a 3 × 3 Matrix Find the determinant of D–

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

Determinant and Inverse of a 3 × 3 Matrix Because det(D) does not equal

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.

Determinant and Inverse of a 3 × 3 Matrix You can use the >Frac

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.

Determinant and Inverse of a 3 × 3 Matrix Therefore, D– 1 = Answer:

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

Find the determinant of A– 1, if it exists. . Then find A. –

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

LESSON 6– 2 Matrix Multiplication, Inverses, and Determinants

LESSON 6– 2 Matrix Multiplication, Inverses, and Determinants