lesson menu five-minute check (over lesson 6-1) then/now new vocabulary key concept: matrix...
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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
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