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 elimination. Then solve thesystem.3x + y + 2z = 31–2x + y + 2z = 12x + y + 2z = 25

A. x + y + 2z = 19y + 2z = 13z = –5; (11, 18, –5)

B. x + y + 2z = 19y + 2z = 13z = 5; (6, 3, 5)

C. x + y + 2z = 19y + 2z = 13z = 5; (3, 6, 5)

D. no solution

Over Lesson 6-1

Solve the system of equations.3x + 2y + 3z = 34x – 5y + 7z = 12x + 3y – 2z = 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.8x + 5y + 11z = 30–x – 4y + 2z = 32x – y + 5z = 12

A. no solution

B. (5 –2z, 2 + z, z)

C. (–5 + 2z, 2 – z, z)

D. (5 – 2z, –2 + z, z)

Over Lesson 6-1

Which of the following matrices is in row-echelon form?

A.

B.

C.

D.

You performed operations on matrices. (Lesson 0-5)

• Multiply matrices.

• Find determinants and inverses of 2 × 2 and 3 × 3 matrices.

• identity matrix

• inverse matrix

• inverse

• invertible

• singular matrix

• determinant

Multiply Matrices

A. Use matrices and to

find AB, if possible.

AB = Dimensions of A: 3 X 2,

Dimensions of B: 2 X 3

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 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

Answer:

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 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 each entry.

Multiply Matrices

Answer:

Finally, simplify each sum.

Use matrices A = and B = to find

AB, if possible.

A.

B.

C.

D.

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.

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 each team scored this season. You can use the product matrix to determine which team scored the most points.

Answer: Tigers

The Tigers scored the most points.

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

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 = 3x1 + 3x2 + 2x3 = 53x1 + x2 + x3 = 4

Write the system in the form, AX = B.

Solve a System of Linear Equations

Write the augmented matrix . Use Gauss-

Jordan elimination to solve the system.

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 Gauss-Jordan elimination on the augmented matrix to solve the system.2x1 – x2 + x3 = –1x1 + x2 – x3 = –2x1 – 2x2 + x3 = –2

A. ; (–1, 2, 3)

B. ; (1, –2, –3)

C. ; (–1, 2, 3)

D. ; (1, –2, –3)

Verify an Inverse Matrix

If A and B are inverse matrices, then AB = BA = I.

Determine whether and are inverse matrices.

Answer: yes; AB = BA = I2

Verify an Inverse Matrix

Because AB = BA = I, B = A–1 and A = B–1.

A. A

B. B

C. C

D. D

Which matrix below is the inverse of A

= ?

A. B. C. D.

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 in reduced row-echelon form.

Doubly Augmented Matrix

R1 + R2

–1R1

Inverse of a Matrix

R2 – 3R1 Row-echelon form R2

R1 + R2Reduced

row-echelon

form

A–1

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–1A = I.

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 .

Answer: singular

Inverse of a Matrix

Step 2 Apply elementary row operations to write the matrix in reduced row-echelon form.

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.

Doubly Augmented Matrix

Find A–1 when , if it exists. If A–1 does

not exist, write singular.

A.

B.

C.

D.

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.

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.

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–1B = .

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

A. 2;

B. –2;

C. 2;

D. 0; does not exist

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]

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 feature under the MATH menu to write the inverse using fractions, as shown below.

Determinant and Inverse of a 3 × 3 Matrix

Answer: –25;

Therefore, D–1 = .

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

A. –3; C. 3,

B. 3; D. 0; does not exist