L1-5. Reducing Rows 11 Aug 2014

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1 L1-5. Reducing Rows 11 Aug 2014 Never send a human to do a machine's job. Agent Smith Primary concepts: Row-echelon form, reduced row-echelon form, Gauss-Jordan elimination method Online tetbook (Hefferon) Chapter 1, p Here is the process of solving a system of equation in matri form: Find a series of elimination matrices E that perform the same operations as equivalent equation operations. Each matri E must be the same size as the original matri A. In this eample, we will use two subscripts for each matri E jk to indicate that it is zeroing the entry at row j, column k of the matri it works on. As you work through this, keep track of all the elimination matrices we use. Let matri E 21 represent the same action as L1-4 s first equation operation (EO 1 ): the elimination of the 1 in the second row, first column of A. As in the equation operations, row 1 is used to eliminate 2 and 3 in column 1; row 2 is used to eliminate 2 in column 2, etc. The first elimination matri E appears on the left in blue in both the left and right hand sides of this equation. E A E b Note that the first and third rows of E 21 come straight from the identity matri; nothing happens to rows 1 and 3 in this step. Only the second row of E 21 is different; only the second row of the original A changes. Note also that the matrices E are always placed on the left side of whatever comes from the prior step: E multiplies A. The element of matri A in red (A 11 ) is known as the pivot for this operation. Be sure you can see that multiplication by matri E 21 on the left is eactly the same as the statement multiply the first row by negative one and add it to the second row to get the new second row, leaving everything else the same or in shorthand R1 + R2 R2. Your online tet cleverly uses the Greek letter to stand for row (imagine that).

2 What would go wrong if matri E 21 was applied to the right of matri A? Try it! Now finish matri E 31 (below) to eliminate the value in row 3, column 1 of the last left hand side above. Be sure to write the row operation in the R1 shorthand demonstrated above. Apply the same E 31 to the right hand side. E E A E E b ??? 2 6 5??? 6 Notice that we write the input to this step as E 21 A and E 21 b and that the new E matri appears on the left of the eisting matri product. We are still using the same pivot because we are still operating on an element of the first column of A. Multiply this step out to find the input to the net step. Now find matri E 32 to eliminate the entry at row 3, column 2 of the matri E 31 E 21 A and apply it to the right hand side. The element A 22 is the pivot for this step; nothing in this step should change the first column of the matri being reduced. Don t forget: All operations must also be done to the right hand side of the equation! If everything is peachy, you should have E E E A E E E b Note that you may have to multiply one or more rows by -1 to get the same signs all the way across. Proceed with back-substitution to obtain the vector : 3 = 4, etc. Because matri multiplication is associative (but NOT commutative!) we can combine all the E jk s into one matri E = E 32 E 31 E 21. Note that the final matri on the left hand side is upper triangular (only 0 s below the main diagonal). You will often see this entire process written as follows: A = b EA = Eb. Taking the symbol U to denote an upper triangular matri, U = EA; 2

3 hence U = c, where the vector c is just the product of E with the original vector b. In the U = c form, we re ready to solve. We love upper triangular matrices! Important Repetition: In the first two elimination steps, matrices E 21 and E 31, we depend upon the value in the 1,1 position of A being nonzero. At each subsequent step, this value (known as the pivot) remains nonzero. The pivots are always the values along the diagonal: the original value in the 1 st row, 1 st column of A, the subsequent value in the 2 nd row, 2 nd column of E 31 E 21 A and the final value in the 3 rd row, 3 rd column E 32 E 31 E 21 A. By the time we get to this last pivot, no further work needs to be done. The three pivots, corresponding to the three elimination steps, are shown in red. The column containing a pivot is called a pivot column. In this case, we started with a matri of order 3 and we found 3 nonzero pivots. We will soon find that when we have a nonzero pivot in each column, we are guaranteed a unique solution to the original problem. If we arrive at a step and find the pivot is 0, we can swap the pivot row with one of the rows below to obtain a nonzero pivot before the elimination step. It is silly to refer to nonzero pivots; pivots are nonzero by definition. Still going It is possible to continue working on the last EA in the eample above: We can turn this into an Identity matri using more row operations. Eliminate the nonzero upper triangular entries in the first row: Notice how identity rows in the E matri don t make any changes in each step. Now eliminate the final nonzero off-diagonal entry in 2,3: 3

4 There is the solution vector, revealed for all to see! What s more, we found a series of matrices E jk such that when the product of all the E jk s multiplies the original matri A, the result is I. In other words, the product of all the E jk s is A s inverse, written A -1. Did you remember to write down all the E jk s so you can see this inverse? Augment and multiply To help us remember to apply the matrices E to the right hand side column vector b, it is convenient to form the augmented matri A b: A b and the row operations need now proceed only on the augmented matri A b, which is still conformable for left multiplication by 33 elimination matrices The first step above is therefore E 21 A b and the result is a augmented matri with 0 in the second row, first column. At the end of the process, only the left 33 portion of the final augmented matri will be upper triangular. Complete the solution. Practice The augmented matrices shown below [A b] are partially reduced. Complete the reductions using row operations. Describe the solution to the original linear system, A = b. 4

5 a. b c. Find an equation involving g, h, k that make the augmented matri represent a consistent system of equations g h Ans. k + 2g + h = k d. Use an augmented matri to solve the system of simultaneous equations: ac0 bd 0 4a3d 2c7 a4b6c6 e. One serving (28 gm) of CrispieCrunchies supplies 110 calories, 3 gm of protein, 21 gm carbs and 3 gm fat. One serving of WinkyWheaties supplies 110 calories, 2 gm protein, 25 gm carbs, but only 0.4 gm fat. 1. Set up a matri A and vector such that A yields the total calories, protein, carbs and fat contained in 3 servings of Crunchies and 2 servings of Wheaties. Determine those values. 2. Determine the miture of the two cereals (if one such miture eists) that provides 110 calories, 2.25 gm protein, 24 gm carbs and 1 gm fat. We ve seen that getting a square matri into upper triangular form is very important and useful. However, rectangular matrices just don t get upper triangular. So there has to be a net-best thing! You should agree that the augmented matri A b is row-equivalent to If this started from A = b and was in fact an augmented matri, we would happily conclude that the original matri A was turned into a pleasant-looking upper triangular. 5

6 However, if the problem started as just a 34 rectangular matri, we ve got something that looks upper triangular, but isn t quite. Definition: A matri is in row-echelon form (abbreviated REF) if it meets all of the following conditions: 1. The leftmost nonzero entry of any row is equal to 1 (the leading one). 2. Given any two different rows with leading 1s, the lower row s 1 is to the right or the upper row s 1. Suppose that a 1 is located in row i, column j and another one is located in row s, column t. This requires that if t > j, s > j. In other words, the REF matri is doing its best to look upper triangular. 3. Any row where every entry is zero lies below all rows that contain a nonzero entry. In other words, rows of all zeros (if any) are at the bottom. Eamples: Which of the following are in REF? If not, what row operation(s) get it in REF? Remember, you can swap rows, but not columns. a b c Is there a unique REF form for any given matri A? Taking the REF matri down even further leads to reduced row-echelon form (RREF). The following criterion is added: 4. Each column containing a leading one contains 0 s at every other location in that column. Note that when dealing with an augmented matri, the augment is not included in these criteria. Eample is REF; is its RREF. How did we get RREF from REF? 6

7 An arbitrarily shaped RREF matri always contains a smaller identity matri at its left or upper left The augmented matri A b is RREF with an identity of order 2; the augmenting column is not considered part of the original matri. Why is RREF so important? Theorem (yes, an actual theorem): Given a matri A, there eists a unique matri B such that A and B are row-equivalent and B is RREF. This very powerful theorem allows us to be confident that once we have row-reduced a given matri, we are looking at a simple matri that has the same solution set as the original matri. Even better, the solution set may be easily determined from the RREF matri. Given that we only use row operations to go from A to B, we already have rowequivalence between A and B. Lucky for us, there are formal procedures to get from an m n matri A to an RREF B using row ops alone; the best-known is Gauss Jordan Elimination (yes, that Gauss; the other fellow is Wilhelm Jordan, a German geodesistmathematician). Bet you don t know what a geodesist is --- go look it up -- and see if you can find what Jordan did to earn a place in the mathematical heavens with Gauss. GJE for fun and profit This will look like a series of steps for a computer program: As you follow along, note that j is an inde variable that allows us to work the columns of A from left to right, one column at a time. The counter r keeps track of the number of nonzero rows as we go along. 1. Set j = 0 and r = 0. Programmers call this initializing the inde and counter variables. 2. Increase j by 1. If j now equals n + 1, STOP! If not, continue to step Eamine the entries of A in column j located in rows r + 1 through m. If all of these entries are zero, this column is done; return to step 2. If not, continue to step 4. 7

8 4. Increase r by 1. Remember, this is counting the number of nonzero rows and step 3 just found one. 5. If the entry at row r, column j is a 0, swap row r with the first row from row r+1 to row m that has a nonzero entry in column j. If not, continue to step If necessary, multiply row r by a scalar, converting the entry in row r, column j to Choose the first row from rows r + 1 through m with a nonzero entry in column j. Use row operations with row r to convert every other entry of column j (rows r+1 through m) to zero. 8. Return to Step 2 (it would make no sense to return to step 1 at this point). Eample A This could be either be a plain old 43 matri on its own or an augmented matri representing four equations in two unknowns in the A b form. However, note that the 3 rd column is the sum of the first two. Use the GJE steps above to convert A to REF. 1. Initialize: set j = 0, r = 0 2. Increase j by 1 (j is now equal to 1) 3. All entries in column 1 are nonzero. 4. Increase r by 1 (r is now equal to 1). 5. A 11 is nonzero, so no row swap is needed. 6. A 11 is already equal to 1 7. Use row operations to convert rows 2 through 4 in column 1 to 0 s A Return to step 2. We are now at j = 2 and r = Step 3 finds that column 2 has at least one nonzero entry in rows 2-4, step 4 increases r to 2. Step 5 must swap rows 2 and 3. 8

9 Step 6 turns A 22 into a 1 and the row op must also work on A Step 7 uses the pivot at A 22 to zero the 4 th row of column 2 (3 rd row is already 0, thank you very much) Now back to step 2, j is now 3 and r is still 2. Rows 3 and 4 of column 3 are already 0 s so step 3 bounces us back to step 2. This increases j to 4 while r is still 2. Step 2 now says STOP, you can t have j = 4! Note that we now have an upper triangular 33 sitting on top of a zero row; this is indeed REF. There were only two pivots, giving a rank of 2 for the original matri A. We can stop here if we want to use back-substitution to solve an A = b problem. To get to RREF, one more step (not included in the GJE algorithm above): What row operation was needed to do this last step? How would you modify the GJE algorithm to go all the way to RREF? 9

10 The final RREF matri has I of order 2 (in the little bo) sitting on top of the zero rows, with a column vector tacked on to its right. If this was originally an augmented matri from a problem of 4 equations in 2 unknowns, we would immediately have the solution 1 1. No back substitution necessary! Of course, when we noted that the 3 rd 2 1 column was the sum of the first two, this result was inevitable. Note that there were only two independent columns in the original matri; we only found two pivots. What does 1 this do to our ability to find a third component for, ie 2. Hmm, curiouser and 3 curiouser Practice Heffernon, p 8-9: 1.16b, 1.17a,c, e, 1.20 Download and play with the Mathematica demonstration, PlaneSolutionsandGaussianElimination.nbp FYI: Gauss had a personal motto: In Latin, Pauca sed matura or Few, but ripe. It is thought that this eplains his relatively small body of published work. See 10