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05aut-m1sols

# 05aut-m1sols - MATH 51 MIDTERM Name Numeric Student ID...

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MATH 51 MIDTERM October 20, 2005 Name: Numeric Student ID: Instructor’s Name: I agree to abide by the terms of the honor code: Signature: Instructions: Print your name, student ID number and instructor’s name in the space provided. During the test you may not use notes, books or calculators. Read each question carefully and show all your work ; full credit cannot be obtained without sufficient justification for your answer unless explicitly stated otherwise. Underline your final answer to each question. There are ?? questions. You have 90 minutes to do all the problems. Question Score Maximum 1 10 2 15 3 10 4 10 5 10 6 10 7 10 8 10 Total 85

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Question 1 of ?? , Page 2 of ?? Solutions 1. Solve the following system of equations in the variables x, y, z, w : x - y - z + w = 5 y - z + 2 w = 8 2 x - y - 3 z + 4 w = 18 If a solution exists, express your answer in parametric form. Solution: Writing this in matrix form, we obtain 1 - 1 - 1 1 0 1 - 1 2 2 - 1 - 3 4 5 8 18 . We will solve this system of equations by performing Gaussian elimination. We proceed as follows. First, subtract twice the top row from the third row. This operation yields 1 - 1 - 1 1 0 1 - 1 2 0 1 - 1 2 5 8 8 . Now subtract the second row from the third row, which yields 1 - 1 - 1 1 0 1 - 1 2 0 0 0 0 5 8 0 . At this point, we can read off the solution. First, the variables z and w are free. Next, y = z - 2 w and x = ( z - 2 w ) + z - w = 2 z - 3 w . Expressed parametrically, this is 2 z - 3 w z - 2 w z w z, w R or z 2 1 1 0 + w - 3 - 2 0 1 z, w R .
Question 2 of ?? , Page 3 of ?? Solutions 2. Let A be the matrix 1 1 0 1 4 1 2 1 1 6 0 1 1 1 3 2 2 0 1 7 . (a) Find a basis for the nullspace of A . Solution: For (a), we transform the matrix A to its RRE form: 1 1 0 1 4 1 2 1 1 6 0 1 1 1 3 2 2 0 1 7 7→ 1 1 0 1 4 0 1 1 0 2 0 1 1 1 3 0 0 0 - 1 - 1 7→ 1 0 - 1 1 2 0 1 1 0 2 0 0 0 1 1 0 0 0 - 1 - 1 7→ 1 0 - 1 0 1 0 1 1 0 2 0 0 0 1 1 0 0 0 0 0 . There are therefore two free variables x 3 and x 5 and three fixed variables x 1 , x 2 and x 4 . The homogeneous linear system associated to the RRE form is then: x 1 - x 3 + x 5 = 0 x 2 + x 3 + 2 x 5 = 0 x 4 + x 5 = 0 . Solving this, one obtains: x 1 x 2 x 3 x 4 x 5 = x 3 - x 5 - x 3 - 2 x 5 x 3 - x 5 x 5 = x 3 1 - 1 1 0 0 + x 5 - 1 - 2 0 - 1 1 .

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05aut-m1sols - MATH 51 MIDTERM Name Numeric Student ID...

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