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Unformatted text preview: Math 311 first practice exam answers
1. We need to solve the system 1 2 0 6 0 = t 1 + u 3 + 0 s 0 0 1 1 for s, t, u. In matrix form, we can write this problem as 1 2 0 s 6 0 1 3 t = 0 . 0 0 1 u 1 Thus u = 1 so u = 1, t  3u = 0 so t = 3u = 3, s  2t = 6 so s = 2t  6 = 0. Thus the unique point of intersection is the origin 0 0 . 0 2. Using row reduction, 1 0 0 2 4 8 1 0 0 , 0 1 0 , 0 0 1 1 3 7 0 4 8 1 2 0 1 0 0 , 0 1 0 , 1 3 7 0 1 1 0 1 2 1/4 1/2 0 1 0 0 , 0 1 0 , 0 0 1 3/4 5/2 1 7/4 11/2 2 0 1 0 1 0 0 , 0 1 0 , 3/4 5/2 1 0 0 1 so the inverse of the original matrix is 0 1 0 7/4 11/2 2 . 3/4 5/2 1 1 3. Using cofactor expansion, 2 1 1 3 1 1 1 det 0 3 1 = 2 +2 = 2(3 + 2) + 2(1  3) = 10  8 = 2. 2 1 3 1 2 2 1 Using row reduction: 2 1 1 2 1 1 3 1 det 0 3 1 = det 0 3 1 = 2 = 2. 1 0 2 2 1 0 1 0 4. (a) AB is undefined. (b) 11 11 BA = 6 7 . 2 3 (c) A2 is undefined. (d) 1 4 10 B 2 = 0 1 4 . 0 0 1 5. The system is homogeneous. Using row reduction gives 1 1 2 0 1 2 4 1 1 1 2 0 0 1 2 1 1 0 0 1 . 0 1 2 1 The first two columns are reduced, so we set z = s, w = t, and get x = t, y = 2s + t. So the solutions are 1 0 t x 1 2 y 2s + t = z s = s 1 + t 0 t 0 1 w and so form a 2plane. 6. Let x be any solution of the system Ax = 0. Then for any matrix B, BAx = B(Ax) = B0 = 0. Thus the same x is also a solution of the system BAx = 0. So this system also has more than one solution. Alternatively: if the equation Ax = 0 has more than one solution, then det A = 0. So for any matrix B, det(BA) = det B det A = 0. This in turn means that the equation (BA)x = 0 has more than one solution. ...
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 Spring '08
 Anshelvich
 Math, Linear Algebra

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