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Unformatted text preview: ECE 220, Test 2 Solutions 27 October 2005 The file test2_f05_sol.m in the class locker contains Matlab scripts for all the problems. —————————————————————— Problem 1. (10 points) If it is possible, compute the matrix product C = B T A , if A and B are defined by A =  1 2 1 1 1 3 2 1 2 , B = 2 1 1 4 2 2 , If it is not possible state why. Solution: The matrix B T is a 2 × 3 matrix; the matrix A is a 3 × 3 matrix. The inner dimensions agree (both equal to 3) and therefore the product C = B T A is welldefined. We have C = B T A = " 2 1 2 1 4 2 # ·  1 2 1 1 1 3 2 1 2 = " 1 7 5 1 8 15 # So, finally, C = 1 7 5 1 8 15 1 Problem 2. (15 points) Determine, if possible, values of a and b such that the system below has no solutions. If that is not possible, state clearly why. " 1 a 1 2 #" x 1 x 2 # = " 3 b # Solution: Start with the array: " 1 a 3 1 2 b # Subtract row 1 from row 2 and substitute for row 2. We have " 1 a 3 0 2 a b 3 # The second row of this matrix represents the equation · x 1 + (2 a ) · x 2 = b 3 . If a 6 = 2, the value x 2 = ( b 3) / (2 a ) is a solution for the variable x 2 , regardless of what the value of b is. Then, from the first row, a solution for x 1 will be x 1 = 3 a ( b 3) / (2 a ) . In this case (i.e., when a 6 = 2), then, the system would have a unique solution. So, consider next the case a = 2. In this case, the second row of this matrix represents the equation · x 1 + 0 · x 2 = b 3 . (1) If b = 3, equation 1 reads · x 1 + 0 · x 2 = 0 . Clearly, in this case, (i.e., when a = 2 and b = 3), there are infinite solutions. So, consider next the case when a = 2 and b 6 = 3. Equation 1 reads 0 = 0 · x 1 + 0 · x 2 = b 3 6 = 0 . Clearly, in this case, (i.e., when a = 2 and b 6 = 3), there are no solutions. So finally, 2 a = 2 , b 6 = 3 . 3 Problem 3. (10 points) Solve the following system of equations using Gaussian elimina tion. 1 1 2 0 1 3 1 0 1 x 1 x 2 x 3 = 5 10 1 Solution: Start with the array: 1 1 2 5 0 1 3 10 1 0 1 1 Subtract row 1 from row 3 and substitute for row 3. We have 1 1 2 5 1 3 10 1 1 4 Subtract row 2 from row 1 and substitute for row 1. We have 1 1 5 1 3 10 1 1 4 Add row 2 to row 3 and substitute for row 3. We have 1 0 1 5 0 1 3 10 0 0 2 6 Divide row 3 by 2. We have...
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 Summer '08
 NILSON
 Complex number, total solution, Numerical ordinary differential equations

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