pset3-s09-soln

# pset3-s09-soln - 18.06 Problem Set 3 Solution Due Wednesday...

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18.06 Problem Set 3 Solution Due Wednesday, 25 February 2009 at 4 pm in 2-106. Total: 160 points. Problem 1: Consider the matrix A = 1 2 1 4 1 2 6 3 11 1 1 4 2 7 0 (a) Reduce A to echelon form U , ﬁnd a special solution for each free variable, and hence describe all solutions to Ax = 0. (b) By further row operations on U , ﬁnd the reduced echelon form R . (c) True or false: N ( R ) = N ( U )? (d) True or false: C ( A ) = C ( U )? Solution (25 points = 10+5+5+5) (a) Use Gaussian elimination. A = 1 2 1 4 1 2 6 3 11 1 1 4 2 7 0 ; 1 2 1 4 1 0 2 1 3 - 1 0 2 1 3 - 1 ; 1 2 1 4 1 0 2 1 3 - 1 0 0 0 0 0 = U Free variables are x 3 ,x 4 ,x 5 . When x 3 = 1 ,x 4 = 0 ,x 5 = 0 we get a special solution 0 - 1 / 2 1 0 0 . When x 3 = 0 ,x 4 = 1 ,x 5 = 0 we get a special solution - 1 - 3 / 2 0 1 0 . 1

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When x 3 = 0 ,x 4 = 0 ,x 5 = 1 we get a special solution - 2 1 / 2 0 0 1 . Hence the solution to Ax = 0 is x = x 3 0 - 1 / 2 1 0 0 + x 4 - 1 - 3 / 2 0 1 0 + x 5 - 2 1 / 2 0 0 1 for x 3 ,x 4 ,x 5 R . (b) Continue using row operations, we have U = 1 2 1 4 1 0 2 1 3 - 1 0 0 0 0 0 ; 1 0 0 1 2 0 2 1 3 - 1 0 0 0 0 0 ; 1 0 0 1 2 0 1 1 / 2 3 / 2 - 1 / 2 0 0 0 0 0 (c) Since U is obtained from R by row operations, they have the same null-space. (d) No. For example, 1 2 1 lies in C ( A ), but any elements in C ( U ) has its third coordinates zero. REMARK: In general, null-space N ( A ) is invariant under invertible row oper- ations. In contrast, column vector space C ( A ) is invariant under invertible column operations. (Non-invertible operations in general may not preserve the spaces.) Problem 2: If you do column elimination steps (instead of row eliminations) on a matrix A to get some other matrix U (like in problem 6 of pset 1), does N ( A ) = N ( U )? Come up with a counter-example if false, or give an explanation why this should always hold if true. Solution (10 points) 2
No. We can give a counter-example as follows. A = ± 1 1 2 2 ² , U = ± 1 0 2 0 ² . Then, the null-space N ( A ) of A is spanned by ± 1 - 1 ² ; in contrast, the null-space N ( U ) of U is spanned by ± 0 1 ² . They are very diﬀerent. For invariance under row or column operations, please see the remark in previous problem. Problem 3: Suppose that column 3 of a 4 × 6 matrix is all zero. Then x 3 must be a variable. Give one special solution for this matrix. Solution (5 points) The variable x 3 is a free variable. A special solution for this variable can be taken to be 0 0 1 0 0 0 . Problem 4:

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pset3-s09-soln - 18.06 Problem Set 3 Solution Due Wednesday...

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