Practice Midterm key

Practice Midterm key - Math 4 Winter 2008(Instructor...

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Math 4 Winter 2008 (Instructor: Professor R. C. Reilly) Answer Key for Practice Midterm #1 (1) (a) Let I 3 denote the 3-by-3 identity matrix. We use Gauss-Jordan Elimination on the matrix [ A | I 3 ] to obtain A - 1 . You should be able to figure out the row operations being used, so I won’t list them out here. (I do list them out in some of the other problems.) [ A | I 3 ] = 2 - 1 4 1 0 0 0 6 0 0 1 0 - 1 2 3 0 0 1 - 1 2 3 0 0 1 0 6 0 0 1 0 2 - 1 4 1 0 0 - 1 2 3 0 0 1 0 6 0 0 1 0 0 3 10 1 0 2 - 1 2 3 0 0 1 0 3 10 1 0 2 0 6 0 0 1 0 - 1 2 3 0 0 1 0 3 10 1 0 2 0 0 - 20 - 2 1 - 4 - 1 2 3 0 0 1 0 3 10 1 0 2 0 0 1 1 / 10 - 1 / 20 1 / 5 - 1 2 0 - 3 / 10 3 / 20 2 / 5 0 3 0 0 1 / 2 0 0 0 1 1 / 10 - 1 / 20 1 / 5 - 1 2 0 - 3 / 10 3 / 20 2 / 5 0 1 0 0 1 / 6 0 0 0 1 1 / 10 - 1 / 20 1 / 5 - 1 0 0 - 3 / 10 - 11 / 60 2 / 5 0 1 0 0 1 / 6 0 0 0 1 1 / 10 - 1 / 20 1 / 5 1 0 0 3 / 10 11 / 60 - 2 / 5 0 1 0 0 1 / 6 0 0 0 1 1 / 10 - 1 / 20 1 / 5 From this one reads off A - 1 = 3 / 10 11 / 60 - 2 / 5 0 1 / 6 0 1 / 10 - 1 / 20 1 / 5 (b) We learned in the lecture that when A - 1 exists, then the solution to A x = b is A - 1 b . In this case that means x = 3 / 10 11 / 60 - 2 / 5 0 1 / 6 0 1 / 10 - 1 / 20 1 / 5 3 - 1 0 = 43 / 60 - 1 / 6 7 / 20 (2) Let us transform the augmented matrix C = [ A | b ] = - 1 2 4 - 2 4 3 - 6 5 1 3 to reduced row-echelon form by the usual elementary row operations: - 1 2 4 - 2 4 3 - 6 5 1 3 (1) - 1 2 4 - 2 4 0 0 17 - 5 15 (2) - 1 2 4 - 2 4 0 0 1 - 5 / 17 15 / 17 (3) - 1 2 0 - 14 / 17 8 / 17 0 0 1 - 5 / 17 15 / 17 (4) 1 - 2 0 14 / 17 - 8 / 17 0 0 1 - 5 / 17 15 / 17 The steps in the preceding row-reduction can be summarized as follows: Step (1) : R 2 R 2 + 3 R 1 . Note: By the end of this step it is clear that the original system is consistent and that there will be two free variables in the final solution, namely x 2 and x 4 . Step (2) : R 2 R 2 / 17 Step (3) : R 1 = R 1 + ( - 4) R 2 . This is the first part of the ‘Jordan’ half of ’Gauss-Jordan Elimination’. Step (4) : R 1 ( - 1) R 1 Note: This completes the ‘Jordan’ half; the final matrix is in reduced row-echelon form. When one translates the final matrix into equations, one gets x 1 = - 8 17 + 2 x 2 - 14 17 x 4 , x 3 = 15 17 + 5 17 x 4 . 1
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Thus, when put into the ‘standard form’, the general solution is x = x 1 x 2 x 3 x 4 = - 8 17
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