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FinalTestReviewSpring2011

FinalTestReviewSpring2011 - MATH 441.001 Instr K Ciesielski...

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MATH 441.001 NAME (print): Instr. K. Ciesielski Spring 2011 FINAL TEST Review Final Test will start with: “Solve the following exercises. Show your work. (No credit will be given for an answer with no supporting work shown.)” Remember, That Final Test is comprehensive! This review is based on a combination of the actual tests I have given you, as well as in the in-class reviews. Some problems that I consider especially important to study for the final are marked by read rectangle. 1
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MATH 441.001 NAME (print): Instr. K. Ciesielski Spring 2011 TEST # 1 Solve the following exercises. Show your work. (No credit will be given for an answer with no supporting work shown.) Ex. 1. Evaluate (a) 1 2 3 1 4 1 = (9 pts) Solution: = [1 8 3] = [ 10] (b) 1 5 1 1 2 3 = (9 pts) Solution: = 1 2 3 5 10 15 1 2 3 (c) ( AB ) 1 , if A 1 = 1 2 1 1 6 3 0 1 2 and B 1 = 1 1 1 0 1 1 0 0 1 . Do not calculate A or B . (13 pts) Solution: ( AB ) 1 = B 1 A 1 = 1 1 1 0 1 1 0 0 1 1 2 1 1 6 3 0 1 2 = 0 9 6 1 7 5 0 1 2 1
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Ex. 2. Find the matrix A for which A x y z t = x + y y + z z t x t (10 pts) Solution: A = 1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 1 Ex. 3. Using block multiplication, evaluate 1 2 1 2 3 4 3 4 1 2 1 2 3 4 3 4 5 0 5 0 1 2 1 2 5 0 5 0 1 2 1 2 (13 pts) Solution: 1 2 1 2 3 4 3 4 1 2 1 2 3 4 3 4 5 0 5 0 1 2 1 2 5 0 5 0 1 2 1 2 = A A A A B B B B = AB + AB AB + AB AB + AB AB + AB , where A = 1 2 3 4 and B = 5 0 1 2 . Since AB = 1 2 3 4 5 0 1 2 = 7 4 19 8 and AB + AB = 7 4 19 8 + 7 4 19 8 = 14 8 38 16 , the final answer is 14 8 14 8 38 16 38 16 14 8 14 8 38 16 38 16 Ex. 4. Using Gauss-Jordan elimination, find the inverse of the matrix A = 1 1 0 0 1 1 2 1 0 (13 pts) Solution: [ A : I ] = 1 1 0 1 0 0 0 1 1 0 1 0 2 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 1 0 2 0 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 2 1 1 1 0 0 1 0 1 0 1 0 2 0 1 0 0 1 2 1 1 1 0 0 1 0 1 0 1 0 2 0 1 0 0 1 2 1 1 , so A 1 = 1 0 1 2 0 1 2 1 1 2
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Ex. 5. (a) Use Guassian elimination to represent the linear system A x = b in an upper triangular form U x = c , where A = 1 1 1 2 2 2 3 3 3 , x = x y z , and b = 2 0 6 . (12 pts) Solution: [ A : b ] = 1 1 1 2 2 2 2 0 3 3 3 6 2 r 1 1 1 1 2 0 4 0 4 3 3 3 6 3 r 1 1 1 1 2 0 4 0 4 0 6 6 0 3 2 r 2 1 1 1 2 0 4 0 4 0 0 6 6 So, in U x = c we have U = 1 1 1 0 4 0 0 0 6
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