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**Unformatted text preview: **MATH 51 FINAL EXAM SOLUTIONS (AUTUMN 2000) 1. Let u 1 = 1- 1 3 u 2 = 1 1- 2 u 3 = 3- 1 4 (a) (6 points) Find the dimension of span( u 1 , u 2 , u 3 ). Solution. Let A = 1 1 3- 1 1- 1 3- 2 4 Then rref( A ) = 1 0 2 0 1 1 0 0 0 , so dim(span( u 1 , u 2 , u 3 )) = dim( C ( A )) = 2. (b) (8 points) Find all vectors v which are simultaneously orthogonal (i.e. perpen- dicular) to all three vectors u 1 , u 2 and u 3 . Solution. The set of all such vectors is the null space of B = 1- 1 3 1 1- 2 3- 1 4 . Since rref( B ) = 1 0 1 2 0 1- 5 2 0 0 , it follows that N ( B ) = span - 1 5 2 . 2. (10 points) Suppose B = ( x,y ) is a point on the circle of radius 1 centered at the origin. That is, x and y satisfy x 2 + y 2 = 1. Let A = (- 1 , 0), C = (1 , 0) and assume y 6 = 0 (so that B is not equal to A or C ). A B C 1 Use dot products to show that angle ABC is a right angle. Solution. The vector from B to A is v 1 = - 1- x- y and the vector from B to C is v 2 = 1- x- y . Thus v 1 v 2 = (- 1- x )(1- x ) + y 2 =- 1 + x 2 + y 2 = 0 since x 2 + y 2 = 1. Therefore v 1 and v 2 are orthogonal. Since y 6 = 0, these vectors are nonzero, so the angle between them is / 2. 3. Suppose A is a 5 5 matrix with rref( A ) = 1 0- 1 4 0 0 1 2 3 0 0 0 0 1 0 0 0 0 0 0 0 0 For each part below, give the answer when possible. Otherwise answer not enough information. (a) (2 points) Find a basis for N ( A ). Solution. 1- 2 1 , - 4- 3 1 (b) (2 points) Find dim( N ( A )). Solution. 2. (c) (2 points) Find a basis for C ( A ). Solution. Not enough information. (d) (2 points) Find dim( C ( A )). Solution. 3. (e) (2 points) Find the rank of A . Solution. 3. (f) (2 points) Find a vector b R 5 such that A x = b has no solutions. Solution. Not enough information. 2 (g) (2 points) Are there vectors b R 5 such that A x = b has exactly one solution? Solution. No. There are free variables. (h) (2 points) Find the eigenvalues of A . Solution. Not enough information. 4. Let A = 1 1 0 2 1 2 2 0 3 (a) (5 points) Compute det( A )....

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