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Unformatted text preview: THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  SPRING SESSION 2005 110.201  LINEAR ALGEBRA. Examiner: Professor C. Consani Duration: 3 HOURS (9am12noon), May 12, 2005. No calculators allowed. Total Marks = 100 Student Name : TA Name & Session : 1. 2. 3. 4. 5. 6. 7. 8. Total 1. [10 marks] Consider the matrix A = 2 1 0 4 2 1 1 2 4 2 3 2 . 1a. [2 marks] Compute the reduced rowechelon form of A . 1b. [2 marks] Determine the rank of A . 1c. [2 marks] Determine a basis of the column space of A . 1d. [2 marks] Determine a basis of the nullspace of A . 1e. [2 marks] For what value(s) of r R is the following system solvable Ax = 2 3 r ? Sol. [1a.] A = 2 1 0 4 2 1 1 2 4 2 3 2 2 1 0 4 0 0 1 2 4 2 3 2 2 1 0 4 0 0 1 2 0 0 3 6 2 1 0 4 0 0 1 2 0 0 0 1 1 / 2 0 2 1 2 rref( A ) = 1 1 / 2 0 2 1 2 . [1b.] It follows from the description of rref( A ) that rk( A ) = 2. [1c.] A basis of the column space of A is given by the vectors 2 2 4 and 1 3 corre sponding to the pivot columns in A . [1d.] There are two nontrivial relations among the columns of A . Let denote by v 1 ,...,v 4 the first,...,fourth column of A . We have v 1 2 v 2 = 0 , 2 v 1 2 v 3 v 4 = 0 . Hence, a basis of the nullspace of A is given by the vectors 1 2 and 2 2 1 . [1e.] Consider the complete matrix B associated to the system B = 2 1 0 4  2 2 1 1 2  3 4 2 3 2  r 2 1 0 4  2 0 0 1 2  1 0 0 3 6  r 4 2 1 0 4  2 0 0 1 2  1 0 0 0  r 7 The system is solvable if and only if r = 7. 2. [15 marks] Consider the matrix A = 1 4 1 1 1 1 . 2a. [5 marks] Give a factorization A = QR , where R is an uppertriangular matrix and Q is a matrix with orthonormal columns. 2b. [5 marks] Find the least square solution to the system Ax = b , for b = 4 8 6 . 2c. [5 marks] The projection matrix P = A ( A T A ) 1 A T projects all vectors onto the column space of A . Find a vector q , not in the column space of A such that Pq = 1 4 4 ....
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 Spring '05
 CONSANI
 Linear Algebra, Algebra

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