TBF 122 - General Mathematics - II Lecture – 2 : Gauss-Jordan Elimination Prof. Dr. Halil İbrahim Karakaş Başkent University

Row Operations on Matrices. To solve systems of linear equations by the method of elimination, we use the operations A, B, and C given in the following theorem. Theorem. Each of the following operations transforms a given system of linear equations to a system equivalent to the given system: A. Interchanging two equations. B. Multiplying an equation by a nonzero constant. C. Adding a constant multiple of one equation to another equation. Eac of the operations A, B and C , when applied to a system of linear equations , induces the respective row operation below on the augmented matrix of the system. Interchanging two rows of the matrix (corresponds to A). Multiplying a row by a nonzero constant(corresponds to B). Adding a constant multiple of one row to another row(corresponds to C). Recall that multiplying a row by a constant means multiplying each entry in that row by that constant. Adding a row to another row means adding each entry of the first row to the corresponding entry of the second row.

We will use the following notation for row operations when necessary: Interchanging two rows. Ri  Rj ( i-th row and j-th row are interchanged) Multiplying a row by a nonzero constant. sRiRi ( i-th row is multiplied by the number s0 ) Adding a constant multiple of one row to another row. sRi+Rj Rj ( s times the i-th row is added top the j-th row) Example. R1  R2 2R2  R2 2R1 +R3  R3

We have observed in our previous lecture that when we apply a row operation to the augmented matrix of a system of linear equations, we obtain the augmented matrix of a system of linear equations which is equivalent to the system in the beginning. From this observation we concluded that to solve a system of linear equations, one could apply suitable row operations to the augmented matrix of the system to obtain a suitable augmented matrix in such a way that the solution set of that suitable augmented matrix is easily determined. It is time to clarify what we mean by suitable augmented matrix here. The definition needed is below. A matrix A satisfying the following four conditions is called a reduced matrix: 1. Any row of A consisting entirely of zeros is below any row having at least one nonzero entry. 2. The first nonzero entry of a row is 1. 3. If a column contains the first nonzero entry 1 of a row, then all other entries in that column are zero. 4. A column that contains the first nonzero entry of a row lies on the right hand side of the column that contains the first nonzero entry of the row above.

Examples. is a redfuced matrix. is not reduced. is reduced. is not reduced.

Is it possible to transform every matrix to a reduced matrix Is it possible to transform every matrix to a reduced matrix? Recall the following Another example Theorem. Every matrix can be transformed to a reduced matrix by a finite number of row opera-tions.

Theorem. Every matrix can be transformed to a unique reduced matrix by a finite number of row operations. The unique reduced matrix obtained from a given matrix by a finite number of row operations is called the reduced form of that matrix. Some examples: reduced reduced reduced reduced reduced

Gauss - Jordan Elimination Gauss - Jordan Elimination. Every system of linear equations is equivalent to the system of linear equations corresponding to the reduced form of its augmented matrix. Since it is quite easy to determine the solution set of a system of linear equations whose augmented matrix is in reduced from, to solve a system of linear equations, we find the reduced form of its augmented matrix and determine the solution set by means of the reduced form. The method of solving a system of linear equations in this way is called Gauss-Jordan elimination. Example. The following table gives examples of systems of linear equations with reduced augmented matrices and their solution sets. Try to see how the solution sets are obtained. How the solution sets are obtained will be understood better as we proceed. Augmented matrix System Solution Set S={(3,2)} S={(3 , t) : tR} S={(-2+3t , 4+6t , t) : tR} S = 

The solution set of a system of linear equations can be determined from the reduced form of its augmented matrix. Tom express the results in this direction, the following definitions will be useful.. If all entries in a row of a matrix are zero, that row is called a zero row of the matrix. If at least one entry in a row is different from zero, then that row is called a nonzero row. The first nonzero entry(from left to right) of a nonzero row is called the leftmost entry of that row. With these terms defined, conditions in the definition of reduced matrix can be restated as folows 1. All zero rows are below nonzero rows. 2. The leftmost nonzero entry of each nonzero row is 1. 3. If a column contains the leftmost entry of a row, all other entries in that column are zero. 4. The column containing the leftmost entry of a row is right to the column containing the leftmost entry of the previous row.

In Gauss-Jordan elimination method, one finds the reduced form of the augmented matrix of the given system and determines the set of solutions by considering the following cases: 1. If the reduced form has a row of the form ( 0 0 . . . 0 | 1 ), then the system has no solution. 2. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is one more than the number of nonzero rows, then the system has a unique solution. 3. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is at least two more than the number of nonzero rows, then the system has infinitely many solutions. If the number of columns of the reduced matrix is r more than the number of nonzero rows, then a general solution depending on r-1 parameters can be written. Note. In the augmented matrix or its reduced form of a system of linear equations, the number of columns is one more than the number of variables; the number of rows is the number of equations.

In Gauss-Jordan elimination method, one finds the reduced form of the augmented matrix of the given system and determines the set of solutions by considering the following cases: 1. If the reduced form has a row of the form ( 0 0 . . . 0 | 1 ), then the system has no solution. 2. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is one more than the number of nonzero rows, then the system has a unique solution. 3. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is at least two more than the number of nonzero rows, then the system has infinitely many solutions. If the number of columns of the reduced matrix is r more than the number of nonzero rows, then a general solution depending on r-1 parameters can be written. Example. If the reduced form of the augmented matrix of a system of linear equations is then, the system under consideration has no solution, because the reduced form has a row ( 0 0 0 | 1 ) and this row corresponds to the equation 0 = 1.

In Gauss-Jordan elimination method, one finds the reduced form of the augmented matrix of the given system and determines the set of solutions by considering the following cases: 1. If the reduced form has a row of the form ( 0 0 . . . 0 | 1 ), then the system has no solution. 2. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is one more than the number of nonzero rows, then the system has a unique solution. 3. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is at least two more than the number of nonzero rows, then the system has infinitely many solutions. If the number of columns of the reduced matrix is r more than the number of nonzero rows, then a general solution depending on r-1 parameters can be written. Example. If the reduced form of the augmented matrix of a system of linear equations is the system has a unique solution, because it has no row of the form (0 0 0 | 1) and the number columns is one more than the number of nonzero rows ( 4 columns and 3 nonzero rows). The system of linear equations corresponding to the reduced form gives the solution set. Solution Set: S = {(3 , 4 , 1)}.

In Gauss-Jordan elimination method, one finds the reduced form of the augmented matrix of the given system and determines the set of solutions by considering the following cases: 1. If the reduced form has a row of the form ( 0 0 . . . 0 | 1 ), then the system has no solution. 2. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is one more than the number of nonzero rows, then the system has a unique solution. 3. If the reduced form has no row of the form ( 0 0 . . . 0 | 1 ) and if the number of columns of the reduced matrix is at least two more than the number of nonzero rows, then the system has infinitely many solutions. If the number of columns of the reduced matrix is r more than the number of nonzero rows, then a general solution depending on r-1 parameters can be written. If the reduced form of the augmented matrix of a system of linear equations satisfies the conditions in case 3 above, defining some new terms will be useful to describe the solution set. Recall that the last column of the augmented matrix is formed by the right hand side constants and each column before the last column corresponds to a variable. If the j-th column is not the last column, i-j entry of the augmented matrix gives the coefficient of the variable xj in the i-th equation of the system. If the reduced form of the augmented matrix has no row of the form (0 0 . . . 0 | 1), then each variable corresponding to a column containing the leftmost entry 1 of a nonzero row is called a dependent variable; the remaining variables are called independent variables.

In a reduced augmented matrix that does not have a row of the form (0 0 . . . 0 | 1), the number of columns is necessarily larger than the number of nonzero rows because the leftmost entry 1 of each nonzero row appears in a (different) column before the last column. In other words, the number of dependent variables is less than or equal to the number of variables. If the number of nonzero rows is equal to the number of columns other than the last column, then all the variables are dependent variables and the system has a unique solution. If the number of nonzero rows is less than the number of columns other than the last column, then there are independent variables; the solution set can be expressed in terms of as many parameters as independent variables. This is accomplished by expressing each dependent variable in terms of the independent variables and assigning new symbols (or parameters) like s, t, u for the independent variables. Example. If the reduced form of the augmented matrix of a system of linear equations is then the system has infinitely many solutions, because the matrix has no row of the form (0 0 0 | 1) and the number of columns is two more than the number of nonzero rows. x1 and x2 are dependent variables, x3 is independent. Setting x3=t , S = {(3-2t , 4+t , t) : t  ℝ}.

Example . Let us solve the following system of linear equations by Gauss-Jordan elimination. S = {(1 , 1 , 5)}.

Example . Let us solve the following system of linear equations by Gauss-Jordan elimination. S = {(2 , 0 , -1)}.

Example . Let us solve the following system of linear equations by Gauss-Jordan elimination. S = .

Example . Let us solve the following system of linear equations by Gauss-Jordan elimination. S = {(-3-t , 4+2t , t) : t  ℝ}.

Example . Let us solve the system of linear equations on the right by Gauss-Jordan elimination. S = {(7+2s+3t , -3-3s-2t , s , 2t , t) : s , t  ℝ}.

Sometimes one may need to solve a number of systems of linear equations having the same coefficients at the same time. Gauss-Jordan elimination can be applied to such systems at the same time. Here is an example: Solution sets are S = {(1 , 1 , 5)} , S= {(5 ,-3 , 4)} , S= {(3 , 1 , 11)} , respectively.

Another example: We see from the last (reduced matrix that the first and the third systems have no solution, the second system has infinitely many solutions depending on one parameter. The dependent variables for the second system are x1 and x3. x2 is independent and we have : S = {(1-t , t , 1) : t  ℝ}.

Problem(from ancient China) Problem(from ancient China). Rice produced in a farm will be packed by using three types of bags of small, medium and large sizes. After packing, 3 large bags, 2 small bags and 1 medium bag together weigh 40 kg; 2 large bags, 3 small bags and 1 medium bag together weigh 30 kg; 1 large bag, 2 small bags, 3 medium bags together weigh 28 kg. How much does each bag (full of rice) weigh? Solution. Let 1 (full) large bag weigh x kg , 1 small bag weigh y kg and 1 medium bag weigh z kg. The data of the problem gives the following system of linear equations. The augmented matrix of this system is Solution set: S = {(11 , 1 , 5)}. Each large bag weighs 11 kg, each small bag weighs 1 kg and each medium bag 5 kg.

Solution by Gauss-Jordan elimination is given below: Problem(2004 ÖSS Question). The sum of present ages of Aslı, Hakan and Tolga is 72. When Aslı will be Hakan’s present age, Tolga’s age will be twice Hakan’s age. Find Hakan’s present age. Solution. Denote the present ages of Aslı, Hakan and Tolga by x1 , x2 , x3, respectively. Then x1 + x2 + x3 = 72; x3+(x2 – x1) = 2(x2 + (x2 – x1) ). Thus we obtain the system of linear equations A quick solution can be given by adding (-1) times the second equation to the first equation. That yields 4x2 = 72 or x2 = 18. Hakan’s present age is 18 . Solution by Gauss-Jordan elimination is given below: Solution Set: S = {(54-t , 18 , t) : t  ℝ}. Note that ages of Aslı, Hakan and Tolga are integers and they are given in increasing order. The solution set shows that there are more than one possibility for ages of Aslı and Tolga. However Hakan’s present age is fixed. It is 18. Aslı is at least 1 and at most 17 years old at present. So 54- t1, t  53 and 54-t 17 , t  37. It follows that Solution Set : S= {(54-t , 18 , t) : 37  t  53}.

In our first lecture, a solution was given for the next problem without using matrices. Now we solve it by using Gaus – Jordan elimination. Problem. A part of 36 thousand TL is deposited in A-bank, a part of it in B-bank and the remaining part in C-bank. The total amount deposited in A-bank and B-bank is 6 thousand TL more than the amount deposited in C-bank; the total amount deposited in A-bank and C-bank is 3 thousand TL less than twice the amount deposited in B-bank. How much TL is deposited in each bank? Let the amount deposited in A-bank be x thousand TL, the amount deposited in B-bank be y thousand TL and the amount deposited in C-bank be z thousand TL. Then Thus the solution of the problem is reduced to the solution of the system The solution is on the next slide.

Augmented matrix: We find the reduced form of the augmented matrix: We see that the solution set is S = {(8, 13, 15)}. 8 thousand TL to A-bank, 13 thousand TL to B-bank and 15 thousand TL to C-bank.

b) Solve the sytem you have found. Problem. The traffic flow for a network of four one-way streets in a city is shown in the figure on the right. The number in the beginning of each street indicates the number of vehicles per hour that enter that street; the number at the end of each street indicates the number of vehicles per hour that leave that street. For a smooth traffic flow, the number of vehicles en-tering each intersection should always equal the number leaving. The variables x1, x2, x3 and x4 represent the flow of traffic between the four intersections in the network. 600 500 x4 400 800 700 x1 x3 900 North Str. South Str. West Str. East Str. x2 a) Find the system of linear equations to be satisfied for a smooth traffic flow. b) Solve the sytem you have found. c) What is the maximum number of vehicles that can travel per hour from East-South crossing to East-North crossing on East Street? d) If the traffic lights are adjusted in such a way that 200 vehicles travel per hour from East-South crossing to West-South crossing, determine the flow around the rest of the network.

its reduced form is obtainbed below: Solution. a) The number of vehicles entering East-South crossing is 1000, the number of vehicles leaving that crossing is x1+ x4. Comparing the number of vehicles entering and leaving each crossing we get the following system of linear equatios: 600 500 x4 400 800 700 x1 x3 900 North Str. South Str. West Str. East Str. x2 b) Augmented matrix: its reduced form is obtainbed below:

d) Take t = 200 in the above solution. Thus x1, x2 and x3 are dependent variables and x4 is independent variable. The system corresponding to the reduced augmented matrix and the solution set of the system are : S = {(1000-t, 200+t, 1300-t, t) : t  ℝ } 600 500 x4 400 800 700 x1 x3 900 North Str. South Str. West Str. East Str. x2 c) x1=1000 vehickels can travel per hour from East-South crossing to East-North crassing on East Street. The minimum number is d) Take t = 200 in the above solution. 800 vehicles per hour from East-South crossing to East –North crossing on East Str., 200 vehicles per hour from East-South crossing to South-West crossing on South Str. 1100 vehicles per hour from North-West crossing to South-West crossing on West Str., 400 vehicles per hour from West-North crossing to East-North crossing on North Str.