Systems of Linear Equations

Transcription

1 Systems of Linear Equations Linear Equation Definition Any equation that is equivalent to the following format a a ann b (.) where,,, n are unknown variables and a, a,, an, b are known numbers (the so called coefficients) is called linear equation in,,, n variables. The notion of equivalence of two equations means that they have precisely the same solutions. Solution is any sequence of numbers that in place of variables makes (.) true. Eample. is a linear equation and the equation is equivalent linear equation. y 4 y 4 Sequence 5, y is one solution of the equation above (either equation since they are equivalent) and so is /, y. As a matter of fact the equation above have infinitely many solutions. Systems of Linear Equations Our objective is to solve the system of two (or three) linear equations with two or three (or more) unknown variables simultaneously. System means that we have two or three (or more) equations simultaneously that ought to be true at the same time. Therefore the notion of solution is a sequence of numbers (in place of variables) that makes them all true simultaneously. Eample. y y is a system of two equations in two variables we need to find both. In this case solution is, y

2 One method of solving such a system is to sketch the graph of each equation (line) and to determine the intersection point of the lines, since this is the point that has coordinates making both equations true. We call this method solving system of equations graphically. One could also solve the system by the so called substitution method, Using substitution in the following manner y y

3 y ( Substitution equation) y y y ( Substitution) y y 4 y y. The principle is simple replace one variable using one equation in all other equations. This reduces the number of variables and the number of equations to consider Eample. Solve We can solve this using graphing y y

4 Now verify (check) the solution. Eample 4. Solve y 6 y. Graphical solution Check

5 Special Cases Eample 5. Solve y y. The system has no solution which we can see from the graph where the lines do not intersect, meaning there is no solution to the system. The system without a solution is usually called inconsistent. If you use substitution method you would reach a false statement (will be done in class). Eample 6. The equations in the following system are dependent meaning redundant since they represent the same line (coincident lines). Thus the solution is any point on that line. We shall define this notion of independence more precisely in case of any system later. 5

6 6y9 y. Using substitution method you would reach identity statement (will be done in class). Gauss-Jordan Elimination Method When we are dealing with systems of more than equations (in more than two unknowns) then the so called elimination method works to solve the system (i.e., to find all the solutions of the system). Eample 7. Solve y z y z yz. You can avoid to work with fractions by simply multiply any equation as needed (with common denominator), 6

7 E 4y z 4 E E y z y z. Note For the purpose of tracking our work we also labeled the equations. We shall rename and reorder the equations so that the simplest is the first F y z F 4y z 4 F y z. Now we proceed with elimination by eliminating the first variable from all other equation after the first equation, F F 6y z F F y z. F y z Now we can proceed to solve the system again by reversing the second and third equation and proceed elimination G G y z yz G 6y z. G G G y z G yz z 4. and now it is easy to see that z 4. and back substitution of this value for z in the second equation yields y. Using both known values for y and z back substituting into the first equation we can find that. The format of the system as the last one is called row-echelon format. We shall define that later more precisely. You should always verify the solutions, y, z 4 in the original system. The last triangular shape of the system with slight modification as we shall multiply the second equation with / above, 7

8 G G y z G G z 4 y z has leading entries in each row normalized. We would need that in case we want to proceed with elimination method as we shall see later. We shall soon define this notion precisely. It is obvious that this format is very easy to solve by back-substitution. Notice another important thing, we have been doing the following operations with equations that did not change the solution of the system (i.e., every system above is equivalent to the other). Interchange two equations. Multiply any equation with non-zero number. Replace any (i-th ) equation by the sum of itself and another equation (j-th) The operations,, and as above are called elementary row operations. These three steps are equivalent to. Interchange two equations. Multiply any equation with non-zero number. Add non-zero multiple of one equation (j-th) to another equation (i-th) [Getting # from # and # is obvious while getting # from # and # would be as in following multiply j-th with a non-zero number k, add that equation to i-th equation and then divide j-th equation with k.] You can also interchange variables meaning put on third place and z on the first place in all equations for eample. This of course would not be operation with a row but rather with column. We shall define this more precisely in the net lecture. The goal of procedure above would be to re-write the system such as a a a b n n a a a b n n (.) a a a b n n mn n m into row-echelon (triangular) format so that is easy to solve it by back-substitution. 8

9 Eample 8 (tetbook). [Details in class only.] Solve the following system by Gauss-Jordan Elimination This method should reach the following row-echelon format 0 and it is also easy to continue to proceed with elementary row operations to reach the following system 0 the so called reduced row-echelon ( diagonalized ) system from which is entirely obvious what the solution is. This format we shall later call reduced row-echelon format. Note The systems as in the eamples 7 and 8 do have graphical (i.e., geometrical ) solution as well as the equations are actually equations of planes in -D space (three variables as dimensions) so we would be looking for intersection of three planes as the solution. What could happen, discuss! Obviously without holographics this would be very impractical. Homework Check online. 9