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Chapter 2 Section 2 Solving a System of Linear Equations II.

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Presentation on theme: "Chapter 2 Section 2 Solving a System of Linear Equations II."— Presentation transcript:

1 Chapter 2 Section 2 Solving a System of Linear Equations II

2 Recall from Algebra A system of linear equations has one of the following results: 1.A unique solution 2.An infinite number of solutions 3.No solution

3 A Unique Solution That each variable is equated to a number. The examples and exercises in Chapter 2 Section 1 are systems of equations that have a unique solution.

4 Exercise 11 (page 76) x + 2y = 5 3x – y = 1 – x + 3y = 5 1 2 5 3 – 1 1 – 1 3 5 rref 1 2 5 3 – 1 1 – 1 3 5 1 0 1 0 1 2 0 0 0,, 1x + 0y = 1 0x + 1y = 2 0x + 0y = 0 x = 1 y = 2 0 = 0 x = 1 and y = 2 (A unique solution)

5 An Infinite Number of Solutions If there are (1) no inconsistent equations (we’ll talk about these in the case of no solutions) and (2) there is/are variable(s) that are not assigned a value, then there are an infinite number of solutions The trick here is to get the resulting rref matrix and convert the rref matrix back into equation notation.

6 Exercise 15 (page 76) x + y + z = – 1 2x + 3y + 2z = 3 2x + y + 2z = – 7 1 1 1 – 1 2 3 2 3 2 1 2 – 7 rref, 1x + 0y + 1z = – 6 0x + 1y + 0z = 5 0x + 0y + 0z = 0 x + z = – 6 y = 5 0 = 0 1 1 1 – 1 2 3 2 3 2 1 2 – 7 1 0 1 – 6 0 1 0 5 0 0 0 0

7 Exercise 15 continued x + z = – 6 y = 5 0 = 0 x = – z – 6 y = 5 z = any value Answer: x = – z – 6, y = 5, and z = any value (This is a case of an infinite number of solutions) (Note that 0 = 0 is a true statement)

8 Exercise 19 (page 76) x + y – 2z + 2w = 5 2x + 1y – 4z + w = 5 3x + 4y – 6z + 9w = 20 4x + 4y – 8z + 8w = 20 rref, 1x + 0y – 2z – 1w = 0 0x + 1y + 0z + 3w = 5 0x + 0y + 0z + 0w = 0 x – 2z – w = 0 y + 3w = 5 0 = 0 1 0 – 2 – 1 0 0 1 0 3 5 0 0 0 0 0 1 1 – 2 2 5 2 1 – 4 1 5 3 4 – 6 9 20 4 4 – 8 8 20 1 1 – 2 2 5 2 1 – 4 1 5 3 4 – 6 9 20 4 4 – 8 8 20

9 Exercise 19 continued x = 2z + w y = – 3w + 5 z = any value w = any value Answer: x = 2z – w, y = – 3w + 5, z = any value, and w = any value (This is a case of an infinite number of solutions) x – 2z – w = 0 y + 3w = 5 0 = 0 (Note that 0 = 0 are true statements)

10 No Solution The trick here is to look for an inconsistent row. (i.e. 0 = 1 row)

11 Exercise 39 (page 77) 4x – 3y + 2z = 3 – 7x + 5y = 2 – 10x + 7y + 2z = 4 4 –3 2 3 – 7 5 0 2 –10 7 2 4 rref, 1x + 0y – 10z = 0 0x + 1y – 14z = 0 0x + 0y + 0z = 1 x – 10z = 0 y – 14z = 0 0 = 1 1 0 – 10 0 0 1 – 14 0 0 0 0 1 4 –3 2 3 – 7 5 0 2 –10 7 2 4

12 Exercise 39 continued Answer: There is no solution to the system of equations Note that 0 = 1 is an inconsistent equation (i.e. a false statement), thus the system has no solution. x – 10z = 0 y – 14z = 0 0 = 1


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