Connor Ahlbach Math 308 Notes

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1 Connor Ahlbach Math 308 Notes. Echelon Form and educed ow Echelon Form In this section, we address what we are trying to acheive by doing EOs. We are trying to turn a complicated linear system into s simpler one, but what do complicated and simpler actually mean. Also, how do we get from this simpler linear system to the set of solutions. In Example. with variables x, y, z, we did EOs The right set of equations is then x =, y = 6, z = 2, which just tells us the solution. If we can ever make the matrix of the linear system into ones on the main diagonal and zeros elsewhere - called the identity matrix for reasons we will see later, then the augmented part tell us the unique solution. However, we can not always get the matrix of a linear system to be the identity matrix. First, of all the matrix does not need to be square - it could be 2 3 for example, as in And, even if it is square, we may not be able to use EOs to get to the identity matrix. Consider , 0 0 whose second equation is 0 =. Thus this linear system has no solution. But why if there not some other EOs we could do to make the matrix into the identity matrix? Well, if we could, then this linear system would have a unique solution, which is false. If we can t always get the idneity matrix, what is the next best thing? The next best thing will be reduced row echelon form. We first introduce echelon form as an intermediate. Definition.. A linear system is in echelon form (EF) if the leading nonzero entries in each row of its augmented matrix proceed strictly from left to right as we go from top to bottom, and any rows of all zeros and are the bottom. Here are some examples of EFs: , , , where each can represent any number. Example : Why are the following linear systems are not in EF? Perform EOs to put them in EF

2 Solution: This is not in EF since the leading entries in each row all lie in first column This is not in EF since the leading entries in last row s leading entry is not left of leading entry in previous row, among other things Notice that the last example illustrates the utility of the swapping rows operation. Furthermore, by iteratively swapping two rows, we can acheive any rearragement of the rows we desire. We will discuss the usefulness of echelon form later. We now introduce reduced row echelon form (EF). Definition.2. A linear system is in reduced row echelon form (EF) if if satisfies the following: (i) It is an echelon form. (ii) All of the leading nonzero entries in each row are s. (iii) The rest of each column with a leading is all zeros. Here are some examples of EFs: , , , where each can represent any number. Example 2: Why are the following linear systems in EF not in EF? Perform EOs to put them in EF Solution:

3 This is not in EF since its leading entries are not s, and the columns with these leading entries have other nonzero entries This is not in EF since its leading entries are not s, and the columns with these leading entries have other nonzero entries So, why is EF the next best thing to the identity matrix? Just like with the identity matrix. We can go from a linear system in EF to describing its solution easily. But first, we need to discuss pivot and free columns and leading and free variables. Definition.3. In a linear system in EF, (including EF), the pivot columns are the coumns which contain a leading nonzero entry in one of the rows. The other non-pivot columns are free columns. If the linear system has a solution, we call the variables corresponding to pivot columns leading variables, and the variables corresponding to free columns free variables (for reasons that will be clear shortly). Since the last column does not have a variable associated to it, it does not give us a free or leading variable. 3 For example, in , columns, 2, are pivot columns, and columns 3, 4, 6, 7 are free columns, which makes x, x 2, x leading variables, and x 3, x 4, x 6 free variables. Given a linear system in EF, we can describe its set of solutions by first noting it has no solution if it has any instances of [ k ] for some k 0, or equivalently if the augmented part is a pivot column. Otherwise, it has solutions we can describe by making all of the free variables free, and then writing the leading variables in terms of the free variables using each equation. For example, consider , We can rewrite the equations in this linear system as x = x 3 3x 4 x 6 + 8, x 2 = 2x 3 4x 4 6x 6 + 9, x = 7x

4 4 Thus, we can describe the solution to this linear system as x 3, x 4, x 6 are free, and then x = x 3 3x 4 x 6 + 8, x 2 = 2x 3 4x 4 6x 6 + 9, x = 7x Example 3: Describe the set of solutions to the following linear systems: [ ] Here, x, x 4 are free variables, and then, by rewriting the equations, x 2 = 2x 4 + 3, x 3 = x Here, the third column is free. However, the last equation is 0 = 3, which has no solution, so this linear system has no solution. Strictly speaking, x 3 does not get to be a free variable since it does not get to be free when describing the solution. Here, x 3, x 4 are free variables, and then, by rewriting the equations, x = 2x 3 3x 4 + 4, x 2 = 3x 3 + 4x One can perform EOs to get from EF to EF without changing which of the columns are pivot or free by using the leading entries to kill (mathematical term meaning to make 0) all of the nonzero entries above it from right to left - see examples above. Going left to right also works, but right to left is computationally easier. Thus, we can tell if a linear system will have a solution, and a valid set of free variables to describe the solution set directly from EF. Lastly, EF has more significance than just allowing us to describe the set of solution. The EF of a linear system is unqiue! Alice and Bob can do 2 completely different set of EOs to a linear system get to EF, but they will always end up with the same EF. For this purpose, we also say two linear systems are row equivalent, denoted, if they can be obtained from one another by EOs - ecall we can undo EOs with EOs. For example, recalling our earlier examples, , Theorem.4. Every linear system is row equivalent to a unique linear system in EF. Proof. First, we need to argue that any linear system is row equivalent to some linear system in EF. To perform EOs to get a linear system to EF, first locate a nonzero entry in the first column, and swap that row up to the top. If there is no nonzero entry in the first column, move on to the next column. Then, use that nonzero entry to kill each other nonzero entry in the first column. Next, locate a leading nonzero entry in the second column, if there are any, swap it to the second row, and use it to kill all other nonzero leading entries in the second column. Continue to use

5 a nonzero leading entry in each column to kill all other nonzero leading entries in that column until you obtain EF. From EF, make all leading nonzero entries into s by dividing each of the equations by them. Proceed from right to left using the leading nonzero entries to kill all other entries in that column until you reach EF. In this way, one can obtain EF starting with any linear system. To prove uniqueness, it is unwise to try to examine all possible paths of EOs one can trace to get EF. Instead, we refer to something that is invariant under EOs, namely the set of solutions. Given any linear system, it has a set of solutons. We claim that is set of solutions uniquely determines the EF. We describe a one-to-one correspondence between possible EFs and a particular way of describing the set of solutions Consider the variables entering in reverse order: x n, x n,..., x 2, x, and as each variable enters, the linear system set of solutions tells it what it has to be given free variables that have entered already, or if it is free. Thus, for all k, x k can only be written in terms of the free variables among x n,..., x k+. For example, if x,..., x enter, the linear system could say: x is free, then x 4 = 2x +,, then x 3 is free, then x 2 = x 3 + 3x, then x = 7x 3 + x. So we have x 7x 3 x = 0, x 2 + x 3 3x =, x 4 = 2x +. This corresponds to the augmented matrix , which is in EF. Also, we can describe the solution set to any linear system in EF in this way, with x k written in terms of the free variables among x n,..., x k+ for all k, by solving for the leading variables. Since there is only one way to describe a set of solutions in this way, any two distinct EFs give riese to 2 different set of solutions, which means they cannot be row equivalent. Therefore, any linear sytem is row equivalent to a unique linear system in EF. The careful reader will notice that we did not cover the case where the linear system has no solution. To use the previous argument for this case, simply attach an augmented part of all zeros, ignoring previous augmented part, which forces there to be a solution. Then, by the same reasoning, this new linear system must have a unique EF. But removing the augmented part of all zeros does not affect if the linear system is in EF or not, so the unique EF of our old linear system must be new EF with this column of all zeros removed. Exercises : () educe the following linear systems to echelon form using EOs

6 (2) educe the following linear systems in echelon form to EF using EOs (3) Describe the set of solutions to the given linear system in EF. Specify which variables are free, and then write the rest of the variables in terms of them. Use x,..., x n for the variable names (d) [ 0 0 ] (4) Solve the following linear systems by reducing to EF by EOs and then describing the set of solutions as in previous problem

7 (d) Problems : () Specify which value(s) of the variables in each of the linear systems below give rise to no solutions, exactly one solutions, and infinitely many solutions h k 2 h k h 3 8 k 7