Solve a System with More Variables Than Equations

Transcription

1 Question 4: Do all systems of linear equations have unique solutions? Earlier we eamined two systems where the numbers of variables was equal to the number of equations. Often there are fewer equations than variables or more equations than variables. Even in systems where the number of variables initially equals the number of equations, one of the rows may become all zeros, meaning that the equation was unnecessary. In all of these cases, we can still use the strategy from Question to solve the problem. Eample 6 Solve a System with More Variables Than Equations Solve the system of equations by transforming its augmented matri to reduced row echelon form. Solution This system does not use, y, and z like earlier eamples. However, if we let the first column in the augmented matri correspond to, the second column to, and the third column to, we can write the augmented matri for this system as To put this matri into reduced row echelon form, we must transform it using row operations to 0?? 0?? Since there is no third row, we have no hope of solving for a solution where,, and are each a unique value. Instead, we ll be able to solve for and in terms of. 9

2 The original augmented matri already has a in the first column and row, so there is no need to put a pivot there. To put a 0 below the pivot, multiply the first row by -, add it to the second row, and put the result in place of the second row: : 40 : The previous row operation not only put a 0 in the first column, but also placed the pivot in the second row and column (very lucky!). To put this matri into reduced row echelon form, we need to put a zero above the pivot in the second column. Multiply the second row by -, add it to the first row, and place the result in the first row: : : This matri is in reduced row echelon form, but we can t read the solution off as we have in earlier eamples. If we convert this back to a system of equations, we get Notice that each of these equations contains and it is easy to solve for in the first equation and and we get in the second equation. If we solve for

3 Although we don t have specific numbers for each variable, we do have a recipe for finding values. If we choose any value for, we can find the corresponding values for and. For instance, if we choose 00, then If 0, then Since can be any number, there are an infinite number of solutions to the system. However, not just any combination of numbers works. Once a value for is chosen, the equations above must be used to calculate corresponding values for and. This can be summarized by writing can be any real number In this case, we have more variables than equations so we would epect to be able to solve for only some of the variables eplicitly. In each row we can solve for a variable eplicitly as long as the row is not entirely zeros. Any variables beyond the number of nonzero rows, in this case one, will be values that we can pick. These variables are called parameters. Parameters are variables whose values are arbitrary and can be picked to be anything that is reasonable for the system of equations.

4 Eample 7 Solve a System with More Equations Than Variables Solve the system of equations by transforming its augmented matri to reduced row echelon form. Solution The augmented matri for this system is To put this matri into reduced row echelon form, we need to place pivots in the first and second columns and zeros in the rest of these columns. The entry in the first column and row is a so the pivot is already in place. To put zeros below the pivot, : 4 4 : : : In the second column and row, we need to transform the to a by multiplying the row by :

5 Now place zeros in the rest of the column, : 0 : 0 4 : 0 : If we convert this matri back to a system of equations, we find that 4 and. Notice that the last row of the matri is all zeros and does not contribute to the solution. This means that there are two variables and two nonzero rows so no parameters are needed in the solution. We can check the solution by substituting 4, into each equation: First equation: 4? TUE Second equation: 4 8 6? 6 6 TUE Third equation: 4 5? 5 5 TUE

6 Since, 4, satisfies each equation in the system, it is the solution to the original system of equations. Eample 8 Solve a System with More Equations Than Variables Solve by transforming its augmented matri to reduced row echelon form. Solution Like the last eample, we ll let the first column in the augmented matri correspond to and the second column to. The augmented matri for this system is To create a pivot in the first row and column, we have two possibilities. We could multiply the first row by 0 or interchange the first and third row. Since interchanging rows does not introduce and fractions into the matri, we ll do that to yield Now we ll use row operations to fill the rest of the first column with zeros: 4

7 : : : : To create a pivot in the second row and column, multiply the second row by : Now use row operations to place zeros above and below the pivot in the second column: : 0 6 : 8 0 : 0 6 : The first two rows in the reduced row echelon form suggest a solution, but the third is problematic. If we write this row as an equation, we get

8 The left side is simply 0. Since 0 cannot equal 7, this implies that there are no solutions to this system. This system is an inconsistent system. These eamples illustrate what may happen when there are more equations than variables or when there are more variables than equations. It is possible for the system to have a unique solution as in Eample 5. The system may be a system with infinitely many solutions like Eample 6 or have no solutions like the inconsistent system in Eample 8. 6