hw1sols - Math 215 Spring 2003—HW#1 1.1#13 Solve the...

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Unformatted text preview: Math 215 Spring 2003—HW #1 § 1.1, #13: Solve the system x 1- 3 x 3 = 8 2 x 1 + 2 x 2 + 9 x 3 = 7 x 2 + 5 x 3 =- 2 We write the augmented matrix: 1 0- 3 8 2 2 9 7 0 1 5- 2 Now replace Row 2 by Row 2 plus- 2 times Row 1: 1 0- 3 8 0 2 15- 9 0 1 5- 2 Now replace Row 3 by Row 3 plus- 1 / 2 times Row 2: 1 0- 3 8 0 2 15- 9 0 0- 5 2 5 2 The last row translates into the equation- 5 2 x 3 = 5 2 , so that x 3 =- 1. The second row translates into the equation 2 x 2 + 15 x 3 =- 9, or 2 x 2- 15 =- 9, or 2 x 2 = 6, or x 2 = 3. The first row translates into the equation x 1- 3 x 3 = 8, or x 1 + 3 = 8, or x 1 = 5. So, ( x 1 , x 2 , x 3 ) = (5 , 3 ,- 1). § 1.1, #16: Determine if the following system is consistent x 1- 2 x 4 =- 3 2 x 2 + 2 x 3 = 0 x 3 + 3 x 4 = 1- 2 x 1 + 3 x 2 + 2 x 3 + x 4 = 5 We write the augmented matrix: 1 0 0- 2- 3 2 2 0 1 3 1- 2 3 2 1 5 Now replace Row 4 by Row 4 plus 2 times Row 1: 1 0 0- 2- 3 0 2 2 0 0 1 3 1 0 3 2- 3- 1 Now replace Row 4 by Row 4 plus- 3 / 2 times Row 2: 1 0- 2- 3 0 2 2 0 0 1 3 1 0 0- 1- 3- 1 Now replace Row 4 by Row 4 plus Row 3: 1 0 0- 2- 3 0 2 2 0 0 1 3 1 0 0 0 We have reached row echelon form, and there are no pivots in the last column, so, yes, the system is consistent (True, this is a § 1.2 method, but give me a break). We could let x 4 be anything we want, then use the third row to find x 3 , then the second row to find x 2 , and then the first row to find x 1 . § 1.1, # 19: Determine the value(s) of h such that the following matrix is the augmented matrix of a consistent linear system: 1 h 4 3 6 8 We replace Row 2 by Row 2 +- 3 times Row 1: 1 h 4 0 6- 3 h- 4 Now, if 6- 3 h 6 = 0, then the last line says that (6- 3 h ) x 2 =- 4, so that x 2 =- 4 6- 3 h , and then the first line will allow us to solve for x 1 , so the system is consistent. On the other hand, if 6- 3 h = 0, then the second line says 0 =- 4, which is not true, so the system is inconsistent. Thus, the system is consistent if h 6 = 2. § 1.1, # 28: Suppose a 6 = 0 and the system below is consistent for all possible values of f and g . What can you say about the coefficients a, b, c, d ? ax 1 + bx 2 = f cx 1 + dx 2 = g We write the augmented matrix: a b f c d g We replace Row 2 by Row 2 plus- c/a times Row 1: a b f d- cb/a g- fc/a The only way for the system to be inconsistent is if the last row is [00nonzero]. Thus, as long as d- cb/a 6 = 0, we know that the system is consistent. If d- cb/a = 0, then there any many values of ( f, g ) for which the system is inconsistent, namely any f, g with g- c/a 6 = 0. So, given what we’re told, we can say that d- cb/a 6 = 0, or, ad- bc 6 = 0....
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.

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hw1sols - Math 215 Spring 2003—HW#1 1.1#13 Solve the...

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