Math254Mid1Sols

Math254Mid1Sols - MIDTERM 1 - SOLUTIONS MATH 254 - SUMMER...

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MIDTERM 1 - SOLUTIONS MATH 254 - SUMMER 2002 - KUNIYUKI CHAPTERS 1, 2, 3 GRADED OUT OF 75 POINTS ¥ 2 = 150 POINTS TOTAL 1) Use either Gaussian elimination with back-substitution or Gauss-Jordan elimination to solve the following system. You must use matrices, as we have done in class. Write your answer as an ordered triple of the form xxx 123 ,, () . 441 45 0 31 2 9 0 39 12 3 23 3 xx x x -+ = += = Ï Ì Ô Ô Ó Ô Ô (15 points) The given equations are already in Standard Form. Write the corresponding augmented matrix and use elementary row operations (EROs) to try to obtain an upper triangular matrix with "1"s along the main diagonal and "0"s below. 44 1 4 03 1 2 11 3 50 90 9 - - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ We need a "1" in the upper left corner. Let's switch Row 1 and Row 3. RR 13 ´ 3 031 2 41 4 9 90 50 - - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ 4 Now, turn the boldfaced " 4 " into a "0" by adding (-4) times Row 1 to Row 3. Old R 3 4- 4 1 4 5 0 (-4)• R 1 -4 4 -12 -36 New R 3 002 1 4 New matrix: 3 2 00 2 9 90 14 - È Î Í Í Í ˘ ˚ ˙ ˙ ˙
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We need “1”s along the main diagonal, so divide Row 2 through by 3, and divide Row 3 through by 2. 11 3 014 001 9 30 7 - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ Write the corresponding system. xx x x 12 3 23 3 39 43 0 7 -+ = += = Ï Ì Ô Ó Ô We have x 3 . Now, back-substitute to solve for x 2 , then x 1 . x 3 7 = x x x 2 2 2 0 47 30 28 30 2 + () = = x x x x x 3 1 1 1 1 79 22 19 19 9 10 = -() + () = -+ = =- Solution set: - () {} 10 2 7 ,, . You could check this in the original system. Note: If we had used Gauss-Jordan Elimination, we would have continued to apply EROs until we obtained the augmented matrix: 100 010 10 2 7 - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ The solution can then be read off on the right-hand-side.
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This note was uploaded on 09/08/2011 for the course MATH 254 taught by Professor Howard during the Spring '09 term at Mesa CC.

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Math254Mid1Sols - MIDTERM 1 - SOLUTIONS MATH 254 - SUMMER...

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