Lesson Menu Five-Minute Check (over Lesson 3–6) CCSS Then/Now New Vocabulary Key Concept: Second-Order Determinant Example 1: Second-Order Determinant Key Concept: Diagonal Rule Example 2: Use Diagonals Key Concept: Area of a Triangle Example 3: Real-World Example: Use Determinants Key Concept: Cramer’s Rule Example 4: Solve a System of Two Equations Key Concept: Cramer’s Rule for a System of Three Equations Example 5: Solve a System of Three Equations

CCSS Content Standards A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context Mathematical Practices 7 Look for and make use of structure.

Then/Now You solved systems of equations algebraically. Evaluate determinants. Solve systems of linear equations by using Cramer’s Rule.

Vocabulary determinant second-order determinant third-order determinant diagonal rule Cramer’s Rule coefficient matrix

Concept

Example 1 Second-Order Determinant Definition of determinant Multiply. = 4Simplify. Answer: Evaluate

Example 1 Second-Order Determinant Definition of determinant Multiply. = 4Simplify. Answer: 4 Evaluate

Example 1 A.–2 B.2 C.6 D.1

Example 1 A.–2 B.2 C.6 D.1

Concept

Example 2 Use Diagonals Step 1Rewrite the first two columns to the right of the determinant.

Example 2 Use Diagonals Step 2Find the product of the elements of the diagonals. 90–4

Example 2 Use Diagonals Step 2Find the product of the elements of the diagonals. 90–4 Step 3Find the sum of each group (–4) = =

Example 2 Use Diagonals Answer: Step 4Subtract the sum of the second group from the sum of the first group. 5 –13 = –8

Example 2 Use Diagonals Answer: The value of the determinant is –8. Step 4Subtract the sum of the second group from the sum of the first group. 5 –13 = –8

Example 2 A.–79 B.–81 C.81 D.79

Example 2 A.–79 B.–81 C.81 D.79

Concept

Example 3 Use Determinants SURVEYING A surveying crew located three points on a map that formed the vertices of a triangular area. A coordinate grid in which one unit equals 10 miles is placed over the map so that the vertices are located at (0, –1), (–2, –6), and (3, –2). Use a determinant to find the area of the triangle. Area Formula

Example 3 Use Determinants Diagonal Rule Sum of products of diagonals 0 + (–3) + 4 = 1– = –16

Example 3 Use Determinants Answer: Area of triangle. Simplify.

Example 3 Use Determinants Answer: Remember that 1 unit equals 10 inches, so 1 square unit = 10 × 10 or 100 square miles. Thus, the area is 8.5 × 100 or 850 square miles. Area of triangle. Simplify.

Example 3 A.10 units 2 B.5 units 2 C.2 units 2 D.0.5 units 2 What is the area of a triangle whose vertices are located at (2, 3), (–2, 2), and (0, 0)?

Example 3 A.10 units 2 B.5 units 2 C.2 units 2 D.0.5 units 2 What is the area of a triangle whose vertices are located at (2, 3), (–2, 2), and (0, 0)?

Concept

Example 4 Solve a System of Two Equations Use Cramer’s Rule to solve the system of equations. 5x + 4y = 28 3x – 2y = 8 Cramer’s RuleSubstitute values.

Example 4 Solve a System of Two Equations Multiply. Add and subtract. = 4Simplify.= 2 Answer: Evaluate.

Example 4 Solve a System of Two Equations Multiply. Add and subtract. = 4Simplify.= 2 Answer: The solution of the system is (4, 2). Evaluate.

Example 4 Solve a System of Two Equations Check 5(4) + 4(2) = 28x = 4, y = 2 ? 3(4) – 2(2) = 8x = 4, y = 2 ? ? = 28Simplify. 28 = 28  ? 12 – 4 = 8Simplify. 8 = 8 

Example 4 A.(3, 5) B.(–3, 7) C.(9, 3) D.(9, –3) Use Cramer’s Rule to solve the system of equations. 2x + 6y = 36 5x + 3y = 54

Example 4 A.(3, 5) B.(–3, 7) C.(9, 3) D.(9, –3) Use Cramer’s Rule to solve the system of equations. 2x + 6y = 36 5x + 3y = 54

Concept

Example 5 Solve a System of Three Equations Solve the system by using Cramer’s Rule. 2x + y – z = –2 –x + 2y + z = –0.5 x + y + 2z = 3.5

Example 5 Solve a System of Three Equations Answer: ======

Example 5 Solve a System of Three Equations Answer: The solution of the system is (0.5, –1, 2). ======

–(0.5) + 2(–1) + 2= –0.5 ? ? –0.5 – 2 + 2= –0.5 –0.5= –0.5  Example 5 Solve a System of Three Equations Check 2(0.5) + (–1) – 2= –2 ? ? 1 – 1 – 2= –2 –2= –2  (–1) + 2(2)= 3.5 ? ? 0.5 – 1 + 4= = 3.5 

Example 5 A.(3, –5, 2) B.(3, 1, –22) C.(2, –5, 5) D.(–3, 0, 0) Solve the system by using Cramer’s Rule. 3x + 4y + z = –9 x + 2y + 3z = –1 –2x + 5y –6z = –43

Example 5 A.(3, –5, 2) B.(3, 1, –22) C.(2, –5, 5) D.(–3, 0, 0) Solve the system by using Cramer’s Rule. 3x + 4y + z = –9 x + 2y + 3z = –1 –2x + 5y –6z = –43