pset3sol

pset3sol - 18.06 Problem Set 3 Solutions Please note that...

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Unformatted text preview: 18.06 Problem Set 3 Solutions Please note that the book problems listed below are out of the 4th edition. Please make sure to check that you are doing the correct problems. Problem 1: Do problem 12 from section 3.3.(10pts) Solution If a matrix A has rank r, then the (dimension of the column space) = (dimension of the row space) = r. To find an invertible submatrix S , we need to find r linearly independent rows and r linearly independent columns. The choice of these columns and rows need not be unique - however the question suggests that we take the pivot rows and pivot columns. For matrix A , after one row and column reduction: A = 1 2 3 1 2 4 1 2 3 0 0 1 1 2 0 0 0 1 (1) Here the 1 st and 3 rd columns are linearly independent, while the 1 st and 2 nd rows are also linearly independent. Hence rank A = 2. An appropriate submatrix is: S A = 1 3 1 4 (2) Note also that S A = 2 3 2 4 (3) would work, however S A = 1 2 1 2 (4) would not! For matrix B : B = 1 2 3 2 4 6 1 2 3 0 0 0 (5) Hence, rank B = 1: S B = ( 1 ) (6) For matrix C , the rows and columns are already reduce - we only need to permute them to obtain echelon form. C = 0 1 0 0 0 0 0 0 1 (7) 1 rank C = 2, deleting the first column and middle row: S C = 1 0 0 1 (8) Problem 2: Do problem 21 from section 3.4.(10pts) Solution a) x + y + z = 4 is equivalent to the matrix equation: A x = 4 (9) A = ( 1 1 1 ) (10) We seek the most general solution in the form x = x p + x n , where x n is a homoge- neous solution to: A x n = 0 (11) First note that rank A is r = 1, while the size of A is m = 1 and n = 3. Therefore A has infinitely many solutions. Moreover, there are n- r = 2 linearly independent homogeneous solutions. To find the general solution, parameterize the null space as follows. Let y = c 1 , z = c 2 (we need two constants since the dimension of the null space is 2). Write x = 4- y- z , so that x = 4- c 1- c 2 . In vector form: x = 4- c 1- c 2 c 1 c 2 = 4 + c 1 - 1 1 + c 2 - 1 1 (12) We can check that this is a solution for any constants c 1 and c 2 . Specifically: x p = 4 (13) is a particular solution A x p = 4, while x n = c 1 - 1 1 + c 2 - 1 1 (14) is the homogeneous solution A x n = 0 (Note that the null space of A is spanned by the two linearly independent vectors (- 1 , 1 , 0) T and (- 1 , , 1) T ) 2 b) We have: A = 1 1 1 1- 1 1 (15) A...
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Fall '08 term at MIT.

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pset3sol - 18.06 Problem Set 3 Solutions Please note that...

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