1.2 Solving a System of Linear Equations

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1 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 1. Solving a System of Linear Equations 1..1 Simple Systems - Basic De nitions As noticed above, the general form of a linear system of m equations in n variables is of the form >< > a 11 x 1 + a 1 x + + a 1n x n = b 1 a 1 x 1 + a x + + a n x n = b a m1 x 1 + a m x + + a mn x n = b m Some of the coe cients a ij can be zero. In fact, the more coe cients are zero, the easier the system will be to solve. The easiest system to solve is of the form a 11 x 1 = b 1 >< a x = b a x = b > a mn x n = b m Its corresponding augmented matrix is a b 1 0 a 0 b 0 0 a 0 b a mn b m To solve, we simply divide by the coe cient in front of each variable. This is illustrated below. < x 1 = 10 Example 0 Solve x = 9 x = The solution is x 1 = 10 =, x = 9 =, x =. When all the coe cients in the system above are 1, the system in the above example is said to be in reduced row-echelon form. The precise de nition is given below. De nition 1 Consider a system of linear equations. 1. The system is said to be in reduced row-echelon form if it satis es the four properties below

2 CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS AND MATRICES (a) If a row does not consist entirely of zeros, then the rst nonzero number is a 1. This number is called the leading 1. (b) The rows consisting entirely of zeros, if there are any, are at the bottom of the system. (c) If two consecutive rows contain a leading 1, then the leading 1 of the higher row is further to the left than the leading 1 of the lower row. (d) Each column that contains a leading 1 contains 0 everywhere else.. The system is said to be in row-echelon form if the properties a - c are satis ed. Thus, a system in reduced row-echelon form is necessarily in row-echelon form. Remark The same de nitions apply to the augmented matrix of a system. Remark Since we can switch back and forth between a system of linear equations and its corresponding augmented matrix, we can either work on a system, or on its corresponding augmented matrix. We will illustrate both. Example The matrices below are in reduced row-echelon form Example The matrices below are in row-echelon form A system in row-echelon form is also easy to solve. We illustrate how with an example.

3 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 9 Example Consider the system < x 1 +x x = 10 x +x = x = Its corresponding augmented matrix is Such a system is also fairly easy to solve. The method used is called backsubstitution. Since we know x =, we can replace x by its value in the second equation to obtain x + = or x =. Now that we know x and x, we can replace in the rst equation to obtain x 1 + = 10 or x 1 =. It is called back substitution because we work backward. We start by nding the value of the last variable, then the next to the last and so on, until we nd the value of the rst variable. At this point, we are done. Remark In order to be able to do back-substitution, it is not necessary that a ii = 1 in the i th equation. De nition Two systems are said to be equivalent if they have the same solution set. Similarly, two augmented matrices are said to be equivalent if they are the augmented matrices of two systems having the same solution set. You remember for elementary mathematics, to solve a given equation in one variable say x, we transform it into an equation of the form x = S where S is then the solution we are after. To transform it, we use certain transformations which do not change the solution set of the equation, in other words, we obtain equivalent equations until we have one of the form x = S. You will recall that the transformations that can be applied when solving an equation are 1. Add the same number on both sides of the equation.. Multiply both sides by the same nonzero number. We follow a similar procedure to solve systems of equations. There are actually two procedures we can use. They are outlined below, then explored thoroughly in the next two subsections. The rst technique, is called Gaussian elimination. Algorithm 9 (Gaussian Elimination) To solve a system of linear equations, follow the steps below 1. Transform its augmented matrix into an equivalent augmented matrix in row-echelon form.

4 10 CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS AND MATRICES. Use back-substitution to nish solving. The second technique is called Gauss-Jordan elimination Algorithm 0 (Gauss-Jordan Elimination) To solve a system of linear equations, follow the steps below 1. Transform its augmented matrix into an equivalent augmented matrix in reduced row-echelon form.. The solution is given by the last column of the augmented matrix in reduced row-echelon form. With either technique, since the systems are equivalent, they will have the same solution set. Usually, getting from a system to an equivalent system in row-echelon form or reduced row-echelon form requires several steps, each step producing an equivalent system. The question is which transformations can we apply to a system so that it is transformed into a system in row echelon form or reduced row-echelon form. The key is that the new system should be equivalent to the original one, so they have the same solutions. It would not help to get a system which is simpler to solve, but which does not have the same solutions as the original one. There are three transformations which can be applied, to a system that will produce an equivalent system. For simplicity, let us label the equations of a system E 1, E,, E m. We will use the same names for the rows of the corresponding augmented matrix. Proposition 1 Each of the following operations on a system of linear equations produces an equivalent system 1. Equation E i can be multiplied by a nonzero constant, with the result used in place of E i. This operation is denoted (E i )! (E i ). Equation E j can be multiplied by any constant and added to equation E i, with the result used in place of E i. This operation is denoted (E j + E i )! (E i ). Equations E i and E j can be transposed in order. This operation is denoted (E i ) $ (E j ) Remark The same operations can be applied to the augmented matrix of a system. Simply replace the term equation by row in the above proposition. Remark These operations are called elementary row operations. Let us look at some examples to illustrate how these transformations are done.

5 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 11 Example Consider If we perform (E 1 )! (E ), we obtain Example Consider If we perform 1 E! (E ), we obtain This is useful when we want to obtain a leading 1 on a given row. Example Consider If we perform ( E 1 + E )! (E ), we obtain This is useful when we want to make an entry equal to 0. In this case, we made the rst entry of the second row equal to 0. We now look at each method to solve a system in greater detail. 1.. Gaussian Elimination Gaussian elimination was brie y outlined above. It has two main steps. First, we get a system (or its augmented matrix) in row-echelon form. Then, we use back substitution to nish solving. We have already explained back substitution. The rst step is always done in a very orderly fashion. In fact, it is very easy to implement on a computer. It always proceeds as follows (the algorithm below works for both a system and its augmented matrix. The augmented matrix part appears in parentheses)

6 1 CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS AND MATRICES Remark Proceed from the rst equation (row) to the last. Look at the rst equation (row), make sure no equation (row) below it has a leading entry further to the left. If one does, switch the two. Eliminate all the entries below the leading entry of the rst equation (row). Repeat the procedure for equation (row), then, In general, we look at the i th equation (row), make sure the coe cient for x i is not 0. If it is, you will have to nd an equation in which it is not, and interchange the two equations. Then, eliminate x i from all the equations below the i th equation. We do this for i = 1; ; ; n 1 if the system has n equations. Recall that Gaussian elimination is a systematic procedure which transforms a system into an equivalent system in row-echelon form. We illustrate it with examples. Example Consider the system E 1 x 1 +x +x = >< E x 1 +x x +x = 1 E x 1 x x +x = > E x 1 +x +x x = (1.) Its corresponding augmented matrix is We follow the algorithm described above. We begin with the rst row. Its leading entry is in column 1. The goal is to set to 0 all the other entries in column 1 and rows -. In order to achieve this, we perform (E E 1 )! (E ), (E E 1 )! (E ) and (E + E 1 )! (E ). The resulting augmented matrix is Next, we work on the second row. First, we make its leading entry 1 by performing ( E 1 )! (E 1 ). The resulting augmented matrix is

7 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 1 Then we make the entries below the leading entry of the second row equal to 0 by performing (E + E )! (E ) and (E E )! (E ). The resulting augmented matrix is We actually were lucky. In the process, we did part of the next step. Looking at row, we already set the entry below it to 0. We just have 1 to set the leading entry of row to 1 by performing E! E. The resulting augmented matrix is We are on the last row, the only thing to do here is to set its leading entry 1 to 1 by performing 1 E! (E ). The resulting augmented matrix is Now, we see that the matrix is in row-echelon form. The corresponding system is > < > E 1 x 1 +x +x = E x +x +x = E x + 1 x = 1 E x = 1 (1.) The system in (1) has the same solutions as the original system (1). We nish solving the system in (1) using back-substitution. From E, we get x = 1 We can now use E to nd x as follows x + 1 x = 1 x = 1 (1 x ) x = 0 since x = 1

8 1 CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS AND MATRICES We now continue with E. Finally, using E 1 yields x x x = x = (x + x ) x = x 1 + x + x = x 1 = x x + x 1 = 1 The solutions of the system in (1) and therefore of the system in (1) are x 1 = 1; x = ; x = 0; and x = 1 This is an example in which the procedure produced a unique solution, we had a consistent system. Following the same procedure, how will we detect if we have a system with no solutions, or one with an in nite number of solutions? The next two examples illustrate this. To also illustrate that the same procedure can be applied to system as well as augmented matrices, we do the next example using the system. < Example 9 Solve the system x 1 + x + x = x 1 + x + x = x 1 + x + x = To eliminate x 1 from E and E, we perform (E E 1 )! (E ) and (E E 1 )! (E ). The resulting system is < x 1 + x + x = x = x = The next step would be to eliminate x from E. This was done in the previous step. The only thing left is to set the leading coe cient of the second equation to 1 by performing ( E )! (E ). The corresponding system is < x 1 + x + x = x = x = This system has an in nite number of solutions given by x = and x 1 +x = or x 1 = x. The free variable is x. In parametric notation, the solution set is < x 1 = t x = t x =

9 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 1 < Example 0 Solve the system x 1 + x + x = x 1 + x + x = x 1 + x + x = To eliminate x 1 from E and E, we perform (E E 1 )! (E ) and (E E 1 )! (E ). The resulting system is < x 1 + x + x = x = x = The next step would be to eliminate x from E. This was done in the previous step. The only thing left is to set the leading coe cient of the second equation to 1 by performing ( E )! (E ). The corresponding system is x 1 + x + x = x = x = This system has no solutions because the last two equations are in contradiction. Remark 1 It is important to notice that the Gaussian elimination procedure is always done in a very orderly fashion. In fact, it is very easy to implement on a computer. It always proceeds as follows Look at the rst equation, make sure the coe cient for x 1 is not 0. If it is, you will have to nd an equation in which it is not, and interchange the two equations. Then, eliminate x 1 from all the equations below the rst equation. Next, look at the second equation, make sure the coe cient for x is not 0. If it is, you will have to nd an equation in which it is not, and interchange the two equations. Then, eliminate x from all the equations below the second equation. The procedure continues the same way. In general, we look at the i th equation, make sure the coe cient for x i is not 0. If it is, you will have to nd an equation in which it is not, and interchange the two equations. Then, eliminate x i from all the equations below the i th equation. We do this for i = 1; ; ; n 1 if the system has n equations. 1.. Gauss-Jordan Elimination Here, we will work with augmented matrices. Recall that Gauss-Jordan elimination is a systematic procedure which transforms a system into an equivalent system in reduced row-echelon form. To obtain a reduced row-echelon form, we rst obtain a row-echelon form, then we go further. We need to make the columns containing the leading entries equal to 0, except for the leading entry of course. We illustrate it with an example.

10 1 CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS AND MATRICES Example Solve the system x 1 +x +x = >< x 1 +x x +x = 1 x 1 x x +x = > x 1 +x +x x = Its corresponding augmented matrix is This is the system we did above. We will take it from its row-echelon form Now, we start from the bottom and work our way up. There is nothing to do on the last row. For row, we make the th entry 0 by performing E 1 E! (E ). The resulting augmented matrix is For row, we make the rd and th entry 0. Let s do them one at a time. First, we perform (E E )! (E ). We obtain Next, we perform (E E ) to obtain

11 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 1 Finally row 1. We make entries and equal to 0 by performing rst (E 1 E )! (E 1 ), this gives us then (E 1 E )! (E 1 ), this gives us The solution is now read from the last column. Thus, we see that x 1 = 1, x =, x = 0, and x = Homogeneous Linear Systems We nish this section by looking at a special type of system, homogeneous systems. These system as we will see, always have at least one solution. De nition (Homogeneous System) A system of linear equations is said to be homogeneous if the constant terms are always 0. In other words, a homogeneous system is of the form a 11 x 1 + a 1 x + + a 1n x n = 0 >< a 1 x 1 + a x + + a n x n = 0 > a m1 x 1 + a m x + + a mn x n = 0 De nition The solution x i = 0 for i = 1,,, n is called the trivial solution. Other solutions are called nontrivial solutions. Remark A homogeneous system always has at least the trivial solution. So, a homogeneous system is always consistent. Remark Another important fact about homogeneous systems is that the elementary transformations will not alter the last column of their augmented matrix. Thus, when we transform a homogeneous system in row-echelon form or reduced row-echelon form, we still have a homogeneous system. It turns out that a homogeneous system of linear equations either has only the trivial solution, or has in nitely many solutions. We state this result without proof. We will prove it later in the course. If the homogeneous system has more equations than unknowns, then it will have in nitely many solutions.

12 1 CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS AND MATRICES Example Solve < x 1 +x +x = 0 x 1 +x x +x = 0 x 1 x x +x = 0 The corresponding augmented matrix is To set the 1st entries in rows and, we perform (E (E E 1 )! E. We obtain E 1 )! E and Set the leading entry of row to 1 by performing ( E )! E. We obtain Set the second entry in row to 0 by performing (E + E )! E. We obtain Set the leading entry in row to 1 by performing 1 E! E. We obtain The corresponding system is < x 1 +x +x = 0 x +x +x = 0 x + 1 x = 0 We can use back substitution. Letting x be the free variable and setting x = t, we get from equation that x = t. From equation, we get 1 x = x x = 1 t t = t

13 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 19 Therefore x 1 = x x = t t = t Thus, the solution in para,etric form is x 1 = >< x = x = > x = t 1.. Concept Review t t 1 t Know what a system in row-echelon form and reduced row-echelon form is. Be able to solve a system in row-echelon form, using back-substitution. Know the transformations which produce equivalent systems. Be able to perform Gaussian elimination to transform a system of linear equations into an equivalent system in row-echelon form. Be able to perform Gauss-Jordan elimination to transform a system of linear equations into an equivalent system in reduced row-echelon form. Be able to solve a system of linear equation using either Gaussian elimination or Gauss-Jordan elimination. 1.. Problems On pages 19-1, do # 1,,,,,,, 1, 1, 1, 1, 19, 0,,,,.