advertisement

Course 111: Algebra, 2nd Feb 2007 To be handed in at tutorials on Feb 5th and 6th. 1. Solve the following system of linear equations using both Gaussian elimination and Gauss-Jordan elimination. 3x1 − 4x2 = 10 −5x1 + 8x2 = −17 −3x1 + 12x2 = −12 What is the rank of the coefficient matrix describing this system? Solution The augmented matrix is 3 −4 10 −5 8 −17 −3 12 −12 Then using row operations to write this in reduced row-echelon form 3 −4 10 −5 8 −17 −3 12 −12 1/3R1 R2 +5R1 , R3 +3R1 3/4R2 R3 −8R2 −→ −→ −→ −→ 1 −4/3 10/3 8 −17 −5 −3 12 −12 1 −4/3 10/3 0 4/3 −1/3 0 8 2 1 −4/3 10/3 1 −1/4 0 0 8 2 1 −4/3 10/3 1 −1/4 0 0 0 0 This is the matrix in row-echelon form. Writing this as equations to solve by Gaussian elimination gives 4 10 10 4 x1 − x2 = ⇒ x1 = + x2 3 3 3 3 1 x2 = − 4 To continue to reduced row-echelon form we need an additional row operation: 1 −4/3 10/3 1 −1/4 0 0 0 0 R1 +4/3R2 −→ 1 0 3 0 1 −1/4 0 0 0 This is the reduced row-echelon form and now writing this as equations we have immediately that x1 = 3 x2 = − Agreeing with the solutions above. Rank is 2 1 4 2. Solve the following system of linear equations using both Gaussian elimination and Gauss-Jordan elimination. 7x1 + 2x2 − 2x3 − 4x4 + 3x5 = 8 −3x1 − 3x2 + 2x4 + x5 = −1 4x1 − x2 − 8x3 + 20x5 = 1 What is the rank of the coefficient matrix describing this system? Solution The augmented matrix is 7 2 −2 −4 3 8 2 1 −1 −3 −3 0 4 −1 −8 0 20 1 Then using row operations to write this in reduced row-echelon form 7 2 −2 −4 3 8 2 1 −1 −3 −3 0 4 −1 8 0 20 1 R1 +2R2 R2 +3R1 , R3 −4R1 R2 ↔R3 R3 +R2 1 R ,− 61 R3 15 2 −→ −→ −→ −→ −→ 1 −4 −2 0 5 6 −3 −3 0 2 1 −1 4 −1 −8 0 20 1 1 −4 −2 0 5 6 0 −15 −6 2 16 17 0 15 0 0 0 −23 1 −4 −2 0 5 6 0 0 0 −23 0 15 0 −15 −6 2 16 17 1 −4 −2 0 5 6 0 15 0 0 0 −23 0 0 −6 2 16 −6 1 −4 −2 0 5 6 0 0 0 −23/15 0 1 0 0 1 −1/3 −8/3 1 This is row-echelon form. You then need to solve: 1 8 x3 − x4 − x5 = 1 3 3 x2 = − x1 − 4x2 − 2x3 + 5x5 = 6 23 15 and you see that this can’t be solved uniquely for x1 , x2 , x3 , x4 and x5 . Reducing the matrix still further to RREF using Gauss-Jordan Elimination should give the same result ... 1 −4 −2 0 5 6 0 0 0 −23/15 0 1 0 0 1 −1/3 −8/3 1 R1 +2R3 R1 +4R2 −→ −→ 2 1 28 28 2 1 x1 − x4 − x5 = ⇒ x1 = + x4 + x5 3 3 15 15 3 3 23 x2 = − 15 1 8 1 8 x3 − x4 − x5 = 1 ⇒ x 3 = 1 + x4 − x5 3 3 3 3 Writing the equations this way you see there are two free variables. So, if we say x4 = t and x5 = s where s and t are any numbers then we can solve up to s and t, giving, 1 28 2 + t+ s 15 3 3 23 = 15 1 8 = 1+ t+ s 3 3 = t = s x2 x3 x4 x5 Rank is 3 1 0 0 −2/3 −1/3 28/15 0 0 −23/15 0 1 0 0 0 1 −1/3 −8/3 1 and this is reduced row-echelon form. Going back to write this as equations we get x1 = 1 −4 0 −2/3 −1/3 8 0 0 −23/15 0 1 0 0 0 1 −1/3 −8/3 1 3. Prove that if A is an n × n matrix (ie. A describes a system with as many equations as unknowns) then the reduced row-echelon form of the matrix will either contain at least one row of all zeroes or it will be the n × n identity matrix, In where 1 ... . . I = .. . . 0 ... 0 .. . {z } | 1 n cols, n rows Proof: Suppose B is the reduced row-echelon form of the matrix. If B has at least one row of all zeroes then we are done. Now suppose B does not have a row of all zeroes. This means that every row has a leading 1 in it since it is in RREF. The leading 1 of a row must be the right of the leading 1 of the row immediately above it. Since B is a square matrix (n × n) and does not have any rows of all zeroes we can figure out the position of each of the leading 1’s in B. First, suppose that the leading 1 in the first row is NOT b11 (i.e. b11 = 0). Then the next possible location of the leading 1 in the first row would then be b12 . If this is where the leading 1 is then B must have the following form. B= 0 0 .. . 1 b13 . . . b1n 0 b23 . . . b2n .. .. .. . . . 0 0 bn3 . . . bnn Now, assume the leading 1 of each of the lower rows is exactly one column to the right of the leading 1 above it. This however, leads us to a contradiction. Because the first leading 1 is in the second column by the time we reach the (n-1)st row the leading 1 will be in the nth column and this will in turn force the nth row to be a row of all zeroes, contradicting the initial assumption. Similarly, if the leading 1 is placed to the right by say 2 columns (or 3 or 4, etc.) the same problem results but with more than one row of all zeroes. Likewise the leading 1 in the first row in any of b13 , b14 , . . . , b1n leads to the same problem. So, to satisfy the assumption that we do not have any rows of all zeroes we must have that the leading 1 in the first row must be at b11 . Using a similar argument we can see that if the leading 1 on any of the lower rows moves to the right more than one column we will have a leading 1 in the nth column prior to hitting the nth row. This in turn forces at least the nth row to be a row of all zeroes contradicting the initial assumption. Therefore, the leading 1 in the first row is at b11 and the only way to not have a row of all zeroes at the bottom is to have the leading 1’s of a row be exactly one column to the right of the leading 1 of the row above it. This means that the leading 1 in the second row must be at b22 , the leading 1 in the third row must be at b33 , etc. And finally in the nth row leading 1 must be at bnn . Therefore the leading 1’s of B are on the diagonal and since B is in reduced row-echelon form all the entries above and below the leading 1’s are zeroes. This is exactly In . Therefore, if B does not have a row of all zeroes then we must have that B = In .