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Unformatted text preview: MATH 235/W08 Review of Linear Algebra I SOLUTIONS to Assignment 1 Hand in questions 2,6,10,11,13,15,17. 1. Solve the following system of equations for x and y : 2 x 2 2 xy + y 2 = 17 x 2 + 2 xy + 3 y 2 = 20 x 2 + 4 xy + 2 y 2 = 7 BEGIN SOLUTION: This problem illustrates the difficulty in solving nonlinear equations compared to solving linear equations. MATLAB has several special routines e.g. the GaussNewton method. However, if we assume that the Prof. made this problem simple , then there should be an integer solution. Therefore, we can use MATLAB to test out solutions, e.g. the following mfile actually finds two solutions. (Is it easy to see that there are two solutions?) a=[2 1;1 1]; b=[1 1;1 3]; c=[1 2;2 2]; r=[17;20;7]; for x=10:10, for y=10:10, z=[x;y]; s=[ z*a*z z*b*z z*c*z]; if r==s, disp([solution found with [x y] equal to: ,num2str([x y])]); keyboard end end end END SOLUTION. 1 2. Solve A x = b and y A = c where A = 1 1 1 2 1 3 1 2 , x = x 1 x 2 x 3 , b = 2 3 1 , y = ( y 1 ,y 2 ,y 3 ) , c = (5 , 3 , 7) . BEGIN SOLUTION: (a) We pivot (row elimination) to solve Ax = b 1 1 1  2 2 1 3  3 1 2  1 1 1 1  2 1 1  1 1 1  1 1 2  1 1 1  1  Therefore, the general solution for Ax = b is x 1 = 1 + 2 x 3 , x 2 = 1 + x 3 , x 3 arbitrary (b) To solve yA = c , we solve A T y T = c T . Therefore, we pivot using A T 1 2 1  5 1 1  3 1 3 2  7 1 2 1  5 1 1  2 1 1  2 1 1  1 1 1  2  Therefore, the general solution for yA = c is y 1 = 1 + y 3 , y 2 = 2 y 3 , y 3 arbitrary END SOLUTION. 3. A system of linear equations in x 1 ,x 2 ,x 3 has the augmented matrix 1 1 b 1 1 a 1 a b a , where a and b are real numbers. Determine values of a and b , if they exist, such that the system has: 2 (a) no solution; (b) a unique solution; (c) infinitely many solutions. For the values of a and b that you find in part (c) above, give the general solution to the system of equations. BEGIN SOLUTION: We pivot (row elimination) to solve Ax = r 1 1  b 1 1  a 1 a b  a 1 1  b 1 1  a b a b 1  a b 1 1  b 1 1  a b a + b 1  a b a 2 + ab Therefore, the key to the solutions is in the last two elements of the last row. ( a + b 1  a b a 2 + ab ) (a) The system has no solution if and only if a + b 1 = 0 and a b a 2 + ab = ( a b )(1 a ) negationslash = 0 ....
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This note was uploaded on 04/27/2010 for the course MATH 235 taught by Professor Celmin during the Winter '08 term at Waterloo.
 Winter '08
 CELMIN
 Linear Algebra, Algebra, Equations

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