Elimination Method Streamlined There is a more streamlined version of elimination method where we do not have to write all of the steps in such an elaborate way. We use matrices. The system of n equations in m unknowns, which we shall call m n system a a a b n n a a a b n n (.) a a a b n n mn n m has augmented matri of the system as a a a n b a a an b am am amn bm (.) which we can use to make all the elementary row operations. The coefficient matri of the system (.) is similar only without the last column, a a a a a a a a a n n m m mn (.) and we shall need this notion later. Eample. We can repeat the same elimination method as eample 8 from the previous lecture using augmented matri only. The augmented matri of 0 8 8 5 5 0 is
0 0 8 8 5 0 5 0 first by dividing third row with 5 and interchanging second and third row 0 0 / 5 0 8 8 then subtracting the first row from the second row 0 0 0 8 8 [We keep on labeling the steps of the process on the right always referring only to the previous matri.] Now we shall divide the second row by and subtract it from the third row 0 0 0 0 6 6 and finally dividing the second row with and the third row with -6 0 0 / 0 0 / 6 and we arrived at row-echelon format. [Actually we had that already in the previous matri.] We can now rewrite the system of equations from this matri and solve it by back substitution or we can naturally continue to diagonalize the system as following 0 0 0 0
0 0 0 0 0 0 0 and we arrived at reduced row-echelon format where the solution is clearly visible from the last column. Notice the whole idea to reach diagonalize form we start elimination of coefficients downward from left to right and once we reach row-echelon format we repeat the same only upward from right to left. [We shall eplain this in more details in the class.] Now we can define what row-echelon and reduced row-echelon format of a matri means. We say that the matri (matri of the system, augmented matri of the system, or any matri whatsoever) is in row-echelon format if the following is true:. If a row contains only 0s then that row is in the bottom of the matri (there could be more of them).. In two successive rows not containing all 0s the lower row has leading entry strictly to the right of the upper row.. First non-zero entry in its row (we call that leading entry) has all other entries in same column below as 0s. [# actually follows from # and # but it is worth noticing it separately.] educed row-echelon format of a matri means just one more condition 4. Each leading (first non-zero) entry in any row is the only non-zero in its column and it has to be. When a system is written in such a format that its augmented matri is in reduced rowechelon format then we say it is in canonical form. Eample. All of the matrices below are in row-echelon format 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 5 0 0 4 4 0 0 0 and the following matrices are in reduced row-echelon format
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 The same system can have many augmented matrices in row-echelon formats, for eample the following two 0 0 0 0 0 4 0 0 0 0 0 0 0 6 0 0 0 [Can you see why?] but it can only have one reduced row-echelon format. This uniqueness will be more obvious later in our course. Eample. Solve the following system of, 4 using Gauss-Jordan Elimination method with matrices only. First we write augmented matri 4 0 First it is obvious that we need to interchange first and second row as second row already has as the first entry. 4 0 The net we shall eliminate all the entries below the in the first row. 4
0 5 5 0 0 6 8 40 The net step in to reduce second row so that it starts with. 0 6 / 5 0 6 8 40 Now we eliminate coefficient below in the second row. Since we intend to diagonalize matri we shall also eliminate all the entries in the same column (ecept that ). 0 0 6 0 0 4 6 Now we shall reduce the third row by diving it with -. 0 0 6 0 0 / We already have reduced row-echelon format but we shall continue to diagonalize by eliminating all the entries in the third column above in the third row. 0 0 0 0 4 0 0 Now the solution of the system is obvious,, 4,. You should always check the solution as this method is not very human-friendly and it can easily generate errors. We noted before that a system of equations does not have to be consistent (i.e., it might not have solutions) and also that it could be with more solutions. Eample 4. The following system (tetbook) is not consistent. 5
4 8 4 8 ` You can convince yourself that this system will have row-echelon format of augmented matri as which show the third equation as 0 5. Eample 5. The following system has many solutions. / / 0 4 8 0 0 0 5 4 8 9 If you follow the G-J-E method with augmented matri you will reach the following reduced row-echelon format, 0 0 5 0 4 8 0 0 0 0 which indicates that the third equation is redundant 0 0. But the first two equations are telling you all the solutions of the system: 0 5 4 8 because if we treat the third variable as a parameter variable (meaning it can be anything we want) then we get the following 0 5 4 8 and now for any choice of third variable we have one solution. For eample if we select we have solution 45, 6,. The solution format of the last system is 6
called parametric format where is a parameter of the solution. This is more commonly written as 0t5 4t8. Note: a system can have more than one parameter. One more definition we shall often use. The leading of any row in reduced row-echelon format is called pivot and the position of that entry is called pivot position. We have used elementary row operations within matrices just like we used that with equations. We say that #, #, # [or equivalently #, #, and # ] operations in contet of a matri are elementary row operations on a matri. If a matri A is reached from another matri B with finite number of these operations we shall say that these matrices are row equivalent and write A B. This relation is obviously a relation of equivalence. Theorem (on eistence and uniqueness of solution) System of linear equations is consistent when the last non-zero row of reduced rowechelon format of its augmented matri has leading before the last entry (meaning that the last entry cannot be a leading entry). When the system is consistent the solution can be unique or there could be many solutions when there are parameter variables. The proof of this theorem is obvious from the discussion we had above. Eample 7. (if time permits) Determine whether the following system is consistent and solve the system if it is. 4 4 9 4 4 Homework: Check online. 7