This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
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.
 Fall '08
 Strang

Click to edit the document details