1. Splash Screen

2. Concept

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

4. Example 1 A.–2 B.2 C.6 D.1

5. Concept

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

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

8. Example 2 Use Diagonals Step 2Find the product of the elements of the diagonals. 90–4 Step 3Find the sum of each group. 9 + 0 + (–4) = 5 1 + 0 + 12 = 13 1012

9. 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

10. Example 2 A.–79 B.–81 C.81 D.79

11. Concept

12. 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

13. Example 3 Use Determinants Diagonal Rule Sum of products of diagonals 0 + (–3) + 4 = 1–18 + 0 + 2 = –16

14. 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.

15. 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)?

16. Concept

17. 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.

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

19. 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 ? ? 20 + 8 = 28Simplify. 28 = 28  ? 12 – 4 = 8Simplify. 8 = 8 

20. 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

21. Concept

22. 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

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

24. –(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  0.5 + (–1) + 2(2)= 3.5 ? ? 0.5 – 1 + 4= 3.5 3.5= 3.5 

25. 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

26. Homework Section 7 (pg 195): 27 – 57 odd, 54, 56, & 58(19 problems) 62 – 65 all(4 problems)

27. End of the Lesson