mat67-2011-midterm-solutions

# mat67-2011-midterm-solutions - Solutions to Midterm Exam...

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Solutions to Midterm Exam Math 67 UC Davis, Fall 2011 Problem 1 (a) Find the general form of the solution of the following system of linear equations: - x 1 + x 2 - 5 x 3 + 2 x 4 = 0 2 x 1 - x 2 + 7 x 3 - 24 x 4 = 0 x 1 + x 2 - x 3 - 2 x 4 = 0 - x 1 + 2 x 2 - 8 x 3 + 2 x 4 = 0 Solution. Write the coefficient matrix and perform Gaussian elimination to get the matrix to reduced row-echelon form: - 1 1 - 5 2 2 - 1 7 - 24 1 1 - 1 - 2 - 1 2 - 8 2 R 1 ← - R 1 R 2 R 2 - 2 R 1 R 3 R 3 - R 1 R 4 R 4 + R 1 ----------→ 1 - 1 5 - 2 0 1 - 3 - 20 0 2 - 6 0 0 1 - 3 0 R 1 R 1 + R 2 R 3 R 3 - 2 R 2 R 4 R 4 - R 2 ----------→ 1 0 2 - 22 0 1 - 3 - 20 0 0 0 40 0 0 0 20 R 3 1 40 R 3 R 4 R 4 - 20 R 3 -----------→ 1 0 2 0 0 1 - 3 0 0 0 0 1 0 0 0 0 From the RREF we see that the solution set can be written as { ( - 2 x 3 , 3 x 3 , x 3 , 0) : x 3 R } = { x 3 ( - 2 , 3 , 1 , 0) : x 3 R } = span { ( - 2 , 3 , 1 , 0) } (b) If there is a solution except the trivial solution x 1 = x 2 = x 3 = x 4 = 0, write explicitly one other solution of the system. That is, find some specific numbers x 1 , x 2 , x 3 , x 4 , not all of them zero, which solve the system. Solution. The vector ( - 2 , 3 , 1 , 0) (which is obtained from the general solution by setting x 3 = 1) is a nonzero solution. (c) The set of solutions is a subspace of R 4 . What is its dimension? Explain how you know. Solution. The solution set was represented as the subspace spanned by the single vector ( - 2 , 3 , 1 , 0), so its dimension is 1 (the vector { ( - 2 , 3 , 1 , 0) } is a basis).

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