Solution Set 3, Fall '12

Transcription

1 Solution Set 3, 86 Fall '2 Do Problem 5 from 32 [ 3 5 Solution (a) A = Only one elimination step is needed to produce the 2 6 echelon form The pivot is the in row, column, and the entry to eliminate is the 2 just below the pivot, so the multiplier is 2 = 2 Therefore, subtract 2 times row from row 2 to get [ 3 5 U = [ 3 5 (b) B = Since the pivot and entry to eliminate are the same as in (a), perform the same elimination step: namely, subtract 2 times [ row from row 2 This is the same as multiplying B by the elimination matrix E 2 = : 2 So B = LU, where E 2 B = L = (E 2 ) = [ 2 [ 2 [ = [ [ 3 5, and U = 3 2 Reduce these matrices to their ordinary echelon forms U Which are the free columns and which are the pivot columns? (a) Solution First subtract row from row 2 to get 2 times row 2 from row 3 to get the ordinary echelon form U = Then subtract (b) Columns, 3, and 5 are pivot columns Columns 2 and 4 are free columns

2 Solution Subtract 2 times row from row 2, and subtract 3 times row from row 3, to 2 3 get 4 Then subtract 7 4 times row 2 from row 3 to get the ordinary echelon 7 form 2 3 U = 4 Columns and 3 are pivot columns Column 2 is a free column 3 Do Problem 25 from 32 Solution One possible solution is the matrix R = Let's demonstrate that this satises the conditions of the problem The column space really does contain, because R = = To determine the nullspace of R, rst note that R is already in reduced row-echelon form; columns, 2 and 3 are pivot columns, while column 4 is a free column To nd the special solution to Ax =, where x x = x 2 x 3, set the free variable x 4 to, and solve for the pivot variables x, x 2, x 3 by back x 4 substitution; it turns out that the special solution is consisting of all multiples of the special solution The nullspace of R is the line Other matrices R are possible In fact, the set of all solutions is precisely the set of 3 4 matrices that are expressible in the form AR, with A an invertible 3 3 matrix So, for example, R = would also work 4 Do Problem 2 from 33 Solution (a) The matrix is from each of rows 2 and 3, to get the row echelon form 2 For downward elimination, subtract row No upward

3 elimination is necessary Just divide row by 4 to get the reduced row echelon form R = (b) The matrix is row 2, and subtract 3 times row from row 3, to get For downward elimination, subtract 2 times row from subtract 2 times row 2 from row 3, to get the row echelon form upward elimination, add 2 times row 2 to row, to get divide row 2 by to get the reduced row echelon form 2 R = 2 3 (c) The matrix is of rows 2 and 3, to get the row echelon form , and then For Finally, For downward elimination, subtract row from each No upward elimination is necessary Just divide row by to get the reduced row echelon form R = 5 Do Problem 2 from 33 Solution For each matrix, we should calculate its ordinary echelon form, identify the pivot rows and pivot columns, and keep only the corresponding rows and columns of the original matrix [ 2 3 For A =, we subtract row from row 2 to nd the ordinary echelon form 2 4 [ 2 3 U = Both rows of U contain pivots, and the pivot columns are columns and 3 So A has[ rank 2, and we remove the free column 2 from A to reveal the invertible 3 submatrix S = (One could alternatively remove column from A) 4 3

4 [ 2 3 For B =, we subtract 2 times row from row 2 to nd the ordinary echelon [ 2 3 form U = Only row and column have a pivot So B has rank, and we remove row 2 and columns 2 and 3 to reveal the invertible submatrix S = [ (One could alternatively remove any one row and any two columns from A) For C =, if we were to follow the same procedure, we would exchange rows 2 and 3 to yield the ordinary echelon form U = In U, the pivot columns are columns 2 and 3, and the pivot rows are rows and 2 However, since we performed a row exchange, the corresponding rows of the original matrix C are actually rows and [ 3 So we remove row 2 and column from C to reveal the invertible submatrix S = In this case, it may have been easier to see this directly than to calculate the ordinary echelon form 6 Do Problem 34 from 34 Solution (a) According to the rst blue box on page 56, the number of special solutions to Ax = equals the number of columns minus the rank of A This equation has special solution, and A has 4 columns, so the rank of A must be 4 = 3 The complete solution to Ax = is the line consisting of all multiples of the special solution s (b) The last non-zero component of the special solution s = (2, 3,, ) occurs as the third component, so the free column of R must be column 3 The pivots, then, must occur in columns, 2, and 4, and we can already ll in this much of R: R = p q the entries p and q have yet to be determined Since s lies in the nullspace of A, it also lies in the nullspace of R In other words, 2 p Rs = q 3 = This tells us 2 + p = and 3 + q = Therefore, p = 2 and q = 3, and so 2 R = 3 (c) We noted in (a) above that A has rank 3 Since A also has 3 rows, A has full row rank Therefore, according to the blue box on page 59, Ax = b has a solution for every b in R 3 4 ;

5 7 Do Problem 2 from 35 Solution (a) The equation Ax = has only the solution x = because the columns of A, being a basis for R 5, are linearly independent (b) If b is in R 5 then Ax = b is solvable because the columns of A, being a basis for R 5, span R 5 8 Do Problem 24 from 35 Solution (a) False Consider the 2 matrix [ Its columns are dependent (indeed, the rst column is by itself ), but its row is independent, because it is not a row of all zeros (b) False Consider the matrix [ multiples of [ Its column space is the x-axis, ie, the line of all [, but its row space is the y-axis, ie, the line of all multiples of (c) True In every matrix, the rank equals both the dimension of the column space and the dimension of the row space (d) False Consider any matrix M of all zeros, for example the matrix [ The columns of M are dependent, so they don't form a basis for anything 9 You do not need to touch MATLAB, or even the computer, for this problem Here is what happens when one uses MATLAB's rank command: >> e=e-5; a=[+e ; ; rank(a) ans = (the rst command just means that e = 5 ) Show that the rank is not mathematically correct Why do you think MATLAB produces this answer? (No need to read MATLAB documentations a couple of sentences with a reasonable guess would suce) Is there a nearby matrix that really is rank? [ + ɛ Solution For any A = with ɛ, the rank of A is 2 To see this, exchange [ the two rows to get, and subtract ( + ɛ) times row from row 2 to get the row + ɛ echelon form [ ɛ Observe that both columns are pivot (as ɛ ), so A has rank 2 While attempting to calculate the rank MATLAB is performing some standard calculations (that ought to work for any input) and is rounding o intermediate results (computers usually deal with numbers up to xed precision) This is where the ɛ gets lost The matrix A is close to does have rank Computational problem See MATLAB le [, which 5