MATH311 practice-ans1

MATH311 practice-ans1 - Math 311 first practice exam...

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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 2-plane. 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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MATH311 practice-ans1 - Math 311 first practice exam...

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