 We have two systems we have first A x equal to b I am calling that system 1 and C x equal to b calling that system 2 okay then what we have observed yesterday is the following if each equation of 2 which equation of system 2 is a linear combination of the equations of system 1 which equation of system 2 is a linear combination of the equations of system 1 then each solution of 1 is a solution of 2 and conversely, okay let me make it more precise. This statement for the you can one can interchange the roles of systems 1 and 2 and say that if any equation of system 1 is a linear combination of the equation of system 2 then any solution of system 2 is a solution of 1, okay. In this case the solution set of these two systems is the same, okay the solution set for these two systems is the same that is the first theorem. Let me give this definition before I state this theorem systems 1 and 2 are said to be equivalent if the above holds the statement that I made just now, okay I will not write down all the details here. Systems 1 and 2 are said to be equivalent if the if this statement and the corresponding statement for solutions of 2 being solutions of 1, okay together with that statement these two statements if these two hold then we say that systems 1 and 2 are equivalent, okay. Then we have the following theorem equivalent systems have the same solution set equivalent systems have the same solution set, okay. See there are two apparently different notion that we have discussed till now one is the row equivalence of matrices that is one is doing elementary row operations on a matrix the other one is linear combinations of equations of two systems, okay. These are two different notions that we have discussed these two are related that is what we will discuss today. These two notions are related and let us see how these two are related. So recall the definition of row equivalent matrices A is row equivalent to B this is the notation for that A is row equivalent to B if B can be obtained from A by a sequence of by a finite sequence elementary row operations, okay and remember that we have seen that this is an equivalence relation. So let us now combine these two notions and approve the following theorem let A be row equivalent to C let A be row equivalent to C then the homogeneous systems A x equal to 0 and C x equal to 0 have the same solution set if A is row equivalent to C then the homogeneous systems A x equal to 0 and C x equal to 0 homogeneous system by which I mean the right hand side requirement vector is a 0 vector these two systems have the same solution set, okay. So can you now see that these two notions are related one is solutions sets being the same the other one is doing elementary row operations on a matrix to get another matrix row equivalent matrix, okay. Let us see how the proof goes A is row equivalent to C so there is a finite sequence of elementary row operations that one does on A to get C, okay. So let us say I have A going to A 1 going to A 2, etc going to A k I will call this as the matrix C so this is my finite sequence of elementary row operations that I have performed on A to get the matrix C I must show that the systems A x equal to 0 and C x equal to 0 have the same set of solutions, okay. One does not have to consider all the terms of the sequence it is enough if we prove the statement for one reduction can you see why sufficient to show that if A is row equivalent to C upon a single operation if A is row equivalent to C upon a single elementary row operation if I am able to show that the solution sets are the same I do not have to consider each term of the finite sequence I am claiming that it is enough to show that it is enough to show the following suppose A is obtained C is obtained from A by a single elementary row operation I show that the systems A x equal to 0 and C x equal to 0 have the same solution set I hope this is clear instead of writing down the proof let me just tell you orally you can fill up the details you derive C from A by a single elementary row operation look at each of the 3 row operations that we have written down E 1 of A is the ijth term of E 1 of A is I am looking at the first operation alpha A ij for alpha not 0 A sj if i is equal to s and it is A ij if i is not equal to s this is the first operation the second operation is replacing the s row by the tth row s row by s row plus alpha times the tth row replace the s row by alpha times the tth row so this is an operation performed only on the s row all the other entries are the same all the other rows remain the same finally interchanging of any 2 rows s and t so that is A tj if i is equal to s it is A sj if i is equal to t it is A ij if so the other rows are left as they are what is to be observed is that each of these operations can you see that it is a linear combination that is being performed on the rows away each of these operations if amounts to performing linear combination on the rows away okay so consider A x equal to 0 and C x equal to 0 each equation in C x equal to 0 is a linear combination of certain equations of A x equal to 0 because of the fact that an elementary row operation on A is a linear combination of the rows away and so the solutions of A x equal to 0 satisfy the system C x equal to 0 okay that is a first observation the second part I must show that every solution of C x equal to 0 satisfies A x equal to 0 but we had seen that each of these elementary row operations has an inverse operation and each of the inverse operation this is an elementary row operation of the same type and so one could go from C x equal to 0 to A x equal to 0 that is any solution of C x equal to 0 is a solution of A x equal to 0 and hence the systems are equivalent they have the same set of solutions okay so write down the details but I have told you essentially what are the steps involved in the proof okay so this is the connection between elementary row operations and linear combination of equations between two systems okay okay let us look at one or two numerical examples okay I want to look at the problem of deriving solutions of a homogeneous system how it is done by using elementary row operations okay okay let us look at the first problem let us say we need to solve the system A x equal to 0 where the coefficient matrix A has let us say three rows and four columns okay so this is my matrix A I am now seeking solutions of the system A x equal to 0 I will do it by using the elementary row operations what I know is that by the theorem that we have seen just now what I know is that if I get a matrix C upon doing elementary row operations with a particular structure of C in mind then the solution set of A x equal to 0 it is the same as a solution set of C x equal to 0 C must be simple in order for me to write down the solutions probably immediately okay so let us do the elementary row operations with a certain structure of C in mind and then we will formalize why this structure of C we need in the form of what are called as row reduced echelon matrix okay so let us now proceed we will let us do one elementary row operation at a time first I will interchange row 1 and row 2 okay I will denote that by R 1 double arrow R 2 interchange row 1 row 2 it will be clear in the next step as to why we are doing this then A is equivalent to so I will use this symbol A is row equivalent to the matrix the first row is 1 1 minus 1 0 second row is 3 minus 1 2 3 the last row remains the same the last row remains the same then the second operation I would like to make these 2 entries 0 okay why I would like to make these entries 0 that will be made clear a little later so I will do these 2 operations now row 2 I am replacing that by minus 3 times row 1 plus row 2 okay this is one operation I will do a similar operation for row 3 also row 3 the entry is 1 so I will replace row 3 by minus 1 times row 1 plus row 3 then the row reduced matrix row equivalent to A is the first row remains as it is 1 1 minus 1 0 the second row is minus 3 times this plus this 0 minus 3 minus 1 minus 4 3 plus 2 5 this remains as 3 that is a second row okay please check the calculations minus 3 times this is 0 minus 3 times 1 that is minus 3 minus 1 minus 4 minus 3 times this is 3 plus 2 is 5 this is 0 so this will remain as it is the next operation is minus first row plus the third row that is 0 0 2 and 1 so this is what I get after performing 3 elementary row operations the next step would be we could proceed taking several different steps but I would like to proceed in this example as follows I will interchange row 2 and row 3 I will interchange row 2 and row 3 to get the following row equivalent matrix row 1 remains the same 1 1 minus 1 0 0 0 2 1 0 minus 4 5 3 this is my latest row equivalent matrix the next step is clear I divide the second row by the constant 2 okay so row 2 will be replaced by 1 by 2 times row 2 and I will keep row 3 as it is then I get the following row equivalent matrix see the objective right now may not be to do a computationally efficient apply a computationally efficient procedure I am trying to arrive at a particular structure of C okay then a is row equivalent to 1 1 minus 1 0 second row is 0 0 1 1 by 2 third row remains as it is 0 minus 4 5 3 the next step will be to divide the third row by minus 4 okay and so third row is minus 1 by 4 times the third row then a is equivalent row equivalent to 1 1 minus 1 0 0 0 1 half 0 1 minus 5 by 4 minus 3 by 4 actually I could stop here to write down the solutions or do 1 more elementary row operation okay let us say I stop here and write down the solution set what are the equations corresponding to A x equal to 0 now A has been read this is the matrix C I want to look at C x equal to 0 what are the 3 equations that give me C x equal to 0 the first equation gives me x 1 plus x 2 minus x 3 plus 0 times x 4 equal to 0 so probably I will remove that second equation gives me see remember A is 3 by 4 matrix so the number of unknowns is 4 the second equation gives me x 3 plus x 1 by 2 x 4 that is equal to 0 the third equation gives me x 2 minus 5 by 4 x 3 minus 3 by 4 x 4 equal to 0 okay I am multiplying C on the right by the column matrix x 1 x 2 x 3 x 4 the vector of unknowns so I get these 3 equations so what is clear is that if I fix x 4 the solution set can be determined that is x 1 x 2 x 3 all the 3 can be determined in terms of x 4 okay so let us say I fix x 4 let us call x 4 as alpha for some alpha arbitrary then x 3 can be determined x 3 can be determined from the second equation x 3 is minus 1 by 2 alpha x 2 can be determined from the last equation x 2 is 5 by 4 x 3 plus 3 by 4 x 4 so this is minus 5 by 8 plus 6 by 8 1 by 8 alpha is it okay minus 5 by 8 plus 6 by 8 alpha that is 1 by 8 alpha finally x 3 can be determined from the first equation x 3 is x 1 plus x 2 x 1 I want x 1 x 1 is x 3 minus x 2 so that is minus 1 by 2 alpha minus 1 by 8 alpha so that is minus 5 by 8 alpha minus 4 minus 1 minus 5 by 8 alpha so that gives me the solution set for this system so I am sure you will now agree that this system C x equal to 0 is much easier to handle than the original system A x equal to 0 okay but of course you need to do this you take this effort of reducing A to a rho reduced to a rho equivalent matrix so let me write down the solution set for this example the solution set S is given by x 1 is minus 5 by 8 alpha x 2 is 1 by 8 alpha x 3 is minus 1 by 2 alpha and x 4 is alpha where alpha is an arbitrary real number alpha is an arbitrary real number so what is first clear is that there are infinitely many solutions there are infinitely many solutions and please observe that the number of equations is less than strictly less than the number of unknowns okay this will be a precursor to what we are going to prove a little later that is if you have a rectangular system of homogeneous equations where the number of equations is strictly less than the number of unknowns it always has a non-trivial solution okay we are going to prove this is an example which already sort of gives a trailer I can take alpha outside and write this as set of all alpha times minus 5 by 8 1 by 8 minus 1 by 2 and 1 alpha belongs to R this gives me the set of all solutions of the system homogeneous system A x equal to 0 let us look at another example okay a little simpler than this okay there are 4 equations in 2 unknowns the objective is to determine completely the set of all solutions of the homogeneous equation A x equal to 0 okay so let us do again elementary row operations on this A, A is equivalent to 1st row is kept as it is the 2nd row is okay so this time I am not writing down what are the operations okay maybe I will do that here row 2 is minus row 1 plus row 2 row 3 is row 1 plus row 3 and row 4 is kept as it is because the entry is already 0 so I get minus this plus this 0 minus 1 this plus this 0 3 0 1 the next step will be to keep the 1st row as it is the 2nd row I perform this operation multiply by minus 1 okay so I do not think I need to write that now multiply by minus 1 I get 0 1 3rd and 4th are kept as they are the next step will be so I write that here by the side I will keep the 2nd row as it is and then do the operations based on the 2nd row so I will keep the 2nd row as it is 1st row will be minus 2 times the 2nd row plus the 1st row the new 1st row is minus 2 times the 2nd row plus the 1st row so that gives me 1 and 0 3rd row is minus 3 times the 2nd row plus the 3rd row 0 0 4th row is minus times minus 1 times the 2nd row plus the 4th row 0 0 and I should stop here okay that is clear by looking at the entries of the matrix which we call C so can you tell me what is the solution set there are 2 unknowns 4 equations so what this says is that the last 3 equations are redundant they are unnecessary the solution set is given by the 1st 2 equations 1st equation gives me x 1 is 0 2nd equation gives me x 2 is 0 there are only 2 unknowns so the solution set for this problem I will again call that as s is just 0 0 so this system has only one solution and that is a 0 solution okay so this system has only the trivial solution okay so these 2 numerical examples have been given in order to consolidate what we have learnt till now that is the idea of performing elementary row operations on a matrix and also see how it is related to solutions of homogeneous equations so we are right now concerned with solutions of homogeneous equations non-homogeneous equations will come a little later we need the notion of a row reduced echolon matrix for that and the other difference between a homogeneous system and non-homogeneous system is that a homogeneous system always has a solution a non-homogeneous system may not have a solution okay if a homogeneous system always has a solution follows from the fact that 0 is a solution a non-homogeneous equation system in general may not have a solution okay so that needs a different treatment so we will come to that later let us now look at the particular structure of the final matrix C that we are arriving at okay this has a specific structure let us formalize that is we will discuss what is called as a row reduced echolon matrix and then formalize what we have done till now okay so the next topic is row reduced echolon matrices and what does one do with row reduced echolon matrices you will see that it gives the solution set completely for a system homogeneous or non-homogeneous okay so I would like to discuss the notion of a row reduced echolon matrix a row reduced echolon matrix a matrix R I will assume that it is of order m by n m rows n columns a matrix R is said to be a row reduced echolon matrix if it satisfies the following conditions the first condition is the first non-zero entry of each row the first non-zero entry of each row is 1 that is the first property the first non-zero entry of each row is 1 we will call this as a leading non-zero entry this will be called the leading non-zero entry that is the first non-zero entry will be called the leading non-zero entry so we require that the leading non-zero entry of each row each non-zero row is 1 okay so one could include this here also the first non-zero entry of each non-zero row the first non-zero entry of each non-zero row of R that must be 1 so the leading non-zero entry of each non-zero is 1 we will use this terminology that is the first condition for a row reduced echolon matrix the second condition each column of R each column of R containing containing the leading non-zero entry each column of each column of R containing a leading non-zero entry containing the leading non-zero entry of some row if I have a column which has a leading non-zero entry corresponding to some row then all the other entries must be 0 let me say has all the other entries 0 each column has the other entries 0 what is non-zero the only non-zero entry is the one that corresponds to the leading non-zero entry of a particular that is the second condition the third condition every 0 row of R appears below every non-zero row of R every 0 row of R appears below every non-zero row of R okay just to clarify it if it is not been made clear to earlier a row is called a 0 row if all its entries are 0 it is called a non-zero row if it has at least one non-zero entry okay okay so condition 3 says that these 0 rows are stacked at the bottom the 0 rows are stacked at the bottom something like what has happened in the second example the final condition that must be satisfied by a row reduced echelon matrix is condition 4 let i equal to 1, 2, 3, etc R denote the non-zero rows of R R has non-zero rows at the top 0 rows at the bottom let us say that there are R non-zero rows now each non-zero has a leading non-zero entry appearing in a certain column okay each non-zero row has the leading non-zero entry appearing in a certain column let us call these as columns 1, 2, 3, R C 1, C 2, C 3, C R let C 1, C 2, etc C R denote okay C standing for the column denote the columns in which the leading non-zero entries of the rows 1, 2, 3, etc R appear okay there are R rows each row has a leading non-zero entry I look at the column in which these entries appear column C 1, column C 2, etc column C R okay then what is the condition that must be satisfied for the matrix R to be a row reduced echelon matrix this condition must be satisfied C 1 strictly less than C 2 less than C 3, etc less than C R okay these are the 4 conditions that a row reduced echelon matrix must satisfy. Let me conclude with 2 or 3 examples let us look at the following okay to serve as a means to consolidate so let us look at the first example let us say A is 0 0 1 0 1, 0 0 0 1, 2, 0 0 0 0 0 this is not a row reduced echelon matrix because the leading non-zero entry of the second row is not 1 okay so this is not row reduced echelon R R E this is not a row reduced echelon matrix the leading non-zero entry of each row must be 1 let us look at another example I will call this A 1 this is A 2 0 0 1 0 1 0 0 0 okay 1 0 1 0 0 0 0 very similar to the previous example 2 entries have been changed this has the first property that the leading non-zero entry is 1 but it does not have the second property. The third column has the leading non-zero entry of the first row okay we must have this entry 0 in order for this to be a row reduced echelon matrix that is not the case so this is not a row reduced echelon matrix example 3 I will call it A 3 simplest row reduced echelon matrix okay this is a row reduced echelon matrix a non-trivial example one could look at example 2 if you want A 4 that is 1 0 0 1 0 0 0 0 is a row reduced echelon matrix okay we will discuss further properties of a row reduced echelon matrices and how they help in solving a non-homogeneous system if it has a solution in the next lecture you have any questions C 1 corresponds to let us say the second column here C 1 corresponds to C 1 is 1 C 2 is 2 C 3 is 3 here C 1 is 1 C 2 is 2 C 1 is the column in which the leading non-zero entry of that particular row appears so C 1 C 2 etc they are numbers okay other questions in this definition we require that C 1 strictly less than C 2 strictly less than C 3 etc strictly less than C R not equal to C equal number 1 is C 1 equal to 1 in the first example no for the fourth example they mean C 1 equal to 1 fourth example is this one yeah C 1 is 1 but it is not the case for first example if you explain the first example it would be better see for the fourth example C 1 is 1 C 2 is 2 the column number where it up yeah so what is the problem with the first example what is the context can I explain the first example this is not a row reduced echelon matrix because the leading non-zero entry of the second row is 2 not 1 this is not a row reduced echelon matrix because the leading non-zero entry of the first row appears in the third column the third column the other entries must be 0 that is your condition 2 that is not the case the third column the leading non-zero entry is 1 the other entries must be 0 that is not the case so for this reason it is not row reduced echelon matrix okay any other question so C 1 C 2 etc C R are the column numbers are the column numbers okay yes 0 0 3 it is a row reduced echelon function it is not the leading non-zero entry of the second row is 2 it is not 1 so it is not row reduced echelon matrix and the only echelon matrix contain only 1 all the leading non-zero entry must be 1 about other entries they are not necessarily 0 maybe I will give another examples next time other end see it does not mean all the other entries are 0 so this is possible this is also a row reduced echelon matrix okay so let me stop here.