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

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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 0 2 2 0 0 0 0 1 3 1 - 2 3 2 1 5

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Now replace Row 4 by Row 4 plus 2 times Row 1: 1 0 0 - 2 - 3 0 2 2 0 0 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 0 - 2 - 3 0 2 2 0 0 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 0 0 1 3 1 0 0 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 0 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. § 1.1, # 33. For the plate shown in the book, let T 1 , T 2 , T 3 , T 4 denote the temperatures at the four interior nodes of the mesh. The temperature at each node is the average of the four nearest nodes. Write a system of four equations for T 1 , T 2 , T 3 , T 4 . We know that T 1 is the average of 20, 10, T 2 , and T 4 , so that T 1 = 20 + 10 + T 2 + T 4 4 or 4 T 1 = 30 + T 2 + T 4 or 4 T 1 - T 2 - T 4 = 30 We know that T 2 is the average of 20, 40, T 1 , and T 3 , so that T 2 = 20 + 40 + T 1 + T 3 4 or 4 T 2 = 60 + T 1 + T 3 or - T 1 + 4 T 2 - T 3 = 60

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We know that T 3 is the average of 30, 40, T 2 , and T 4 , so that T 3 = 30 + 40 + T 2 + T 4
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