soln1 - MATH 235/W08 Review of Linear Algebra I SOLUTIONS...

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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 Gauss-Newton 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 m-file 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.

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soln1 - MATH 235/W08 Review of Linear Algebra I SOLUTIONS...

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