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**Unformatted text preview: **MAT067 University of California, Davis Winter 2007 Solutions to Homework Set 2 As usual, we are using F to denote either R or C . 1. Solve the following systems of linear equations and characterize their solution set (unique solution, no solution, ....). Also write each system of linear equations as a single function f : R n R m for appropriate m, n . (a) System of 3 equations in the unknowns x, y, z, w x + 2 y 2 z + 3 w = 2 2 x + 4 y 3 z + 4 w = 5 5 x + 10 y 8 z + 11 w = 12 . Notice that 1( FirstEqn. ) + 2( Sec.Eqn. ) = ThirdEqn. Hence, the bottom equation does not add any extra information. 2( FirstEqn. )+ ( Sec.Eqn. ) gives that z 2 w = 1. Plugging z = 2 w + 1 into both equations gives x + 2 y w = 4 , 2 x + 4 y 2 w = 8 . The second equation is twice the first. Solving w in terms of x and y yields w = x + 2 y 4 . Now using the relation z = 2 w + 1 we obtain z = 2 w + 1 = 2 x + 4 y 7 . There are no more equations to use, so we get a solution for every pair of x and y . The set of solutions to the system of equations is { ( x, y, z, w ) | x, y R , w = x + 2 y 4 , z = 2 w + 1 = 2 x + 4 y 7 } , The system of 3 equations can be written in terms of a function f : R 4 R 3 f ( x, y, z, w ) = ( x + 2 y 2 z + 3 w, 2 x + 4 y 3 z + 4 w, 5 x + 10 y 8 z + 11 w ) such that f ( x, y, z, w ) = (2 , 5 , 12). The solution set is a plane in R 4 . 2 (b) System of 4 equations in the unknowns x, y, z x + 2 y 3 z = 4 x + 3 y + z = 11 2 x + 5 y 4 z = 13 2 x + 6 y + 2 z = 22 . Solution: The second and fourth equations just differ by a factor of 2 so we can throw out the fourth. ( FirstEqn. ) + ( Sec.Eqn. ) ( ThirdEqn. ) gives z = 1. Plugging in z = 1 gives x + 2 y = 7 x + 3 y = 10 2 x + 5 y = 17 . ( FirstEqn. )+( Sec.Eqn ) = ( ThirdEqn. ) so we eliminate the third. Finally, we are left with two equations and two unknowns. ( Sec.Eqn. ) ( FirstEqn. ) yields y = 3. Plugging back in, we get x = 1. Hence there is a unique solution ( x, y, z ) = (1 , 3 , 1)....

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