EXAM 01-solutions - Version 088 EXAM 01 gilbert(53415 This...

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Version 088 – EXAM 01 – gilbert – (53415) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points When P is the plane in R 3 given in vector form by x = 3 3 1 + s 1 2 1 + t 3 8 5 , determine where P intersects the z -axis. 1. z = 4 2. z = 3 3. z = 6 4. z = 2 5. z = 5 correct Explanation: Since the z -axis consists of points in R 3 with x = y = 0, P intersects the z -axis when 0 0 z = 3 3 1 + s 1 2 1 + t 3 8 5 for some choice of s and t , i.e. , when s 1 2 1 + t 3 8 5 = 3 3 z 1 is consistent as a vector equation in s, t . This will be true if and only if the associated aug- mented matrix A = 1 3 3 2 8 3 1 5 z 1 is row equivalent to an echelon matrix whose entries in the last row are all zero. Now by row reduction in the first column of A , we obtain A 1 3 3 0 2 9 0 2 z 4 , and after row reduction in the second column A 1 3 3 0 2 9 0 0 z + 5 . Consequently, P intersects the z -axis at z = 5 . 002 10.0points If matrices A and B are row equivalent, they have the same reduced echelon form. True or False? 1. TRUE correct 2. FALSE Explanation: If A is row equivalent to B , then A can be transformed by elementary row operations first into B and then further transformed into the reduced echelon form U of B . Since the reduced echelon form of A is unique, it must be U . Consequently, the statement is TRUE . 003 10.0points Determine b when a = bracketleftbigg 4 3 bracketrightbigg , b = bracketleftbigg a b bracketrightbigg , c = bracketleftbigg 1 2 bracketrightbigg , and comp a b = 1 , proj c b = 4 5 c . 1. b = 3
Version 088 – EXAM 01 – gilbert – (53415) 2 2. b = 1 correct 3. b = 2 4. b = 4 5. b = 5 Explanation: Since comp a b = a · b bardbl a bardbl , proj c b = parenleftbigg b · c bardbl c bardbl 2 parenrightbigg c , we see that 4 a + 3 b radicalbig ( 4) 2 + (3) 2 = 1 , while a + 2 b (1) 2 + (2) 2 = 4 5 . Thus 4 a + 3 b = 5 , a + 2 b = 4 . Consequently, solving for b gives b = 1 . 004 10.0points Determine the Reduced Row Echelon Form of the matrix A = 1 1 2 3 2 9 2 1 13 . 1. rref( A ) = 1 1 5 0 1 3 0 0 0 2. rref( A ) = 1 0 0 0 1 0 0 0 0 3. rref( A ) = 1 0 0 0 1 0 0 0 1 4. rref( A ) = 1 0 5 0 1 3 0 0 0 correct 5. rref( A ) = 1 2 5 0 1 3 0 0 0 Explanation: After performing elementary row opera- tions downwards on the columns we see that A 1 1 2 0 1 3 0 3 9 1 1 2 0 1 3 0 0 0 . Next perform elementary row operations up- wards. Then A 1 0 5 0 1 3 0 0 0 . Consequently, rref( A ) = 1 0 5 0 1 3 0 0 0 . 005 10.0points Under what conditions on b 1 , b 2 does the equation bracketleftbigg 3 2 12 8 bracketrightbigg x = bracketleftbigg b 1 b 2 bracketrightbigg have a solution in R 2 ? 1. b 2 4 b 1 = 0 2. b 2 + 4 b 1 = 0 correct 3.
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