Solution by Row Operations

The idea behind finding solutions by row operations is to transform a given system of equations into another with the same mathematical properties and hence the same solution. The aim is to transform a system in such a way as to produce a simpler system which is easier to solve. Three types of operations are permitted to transform a system:

1. Multiply an equation by a nonzero constant.

2. Add a multiple of one equation to another.

3. Interchange two equations.

In practice, not all of these operations may be required to simplify a given system, while in a complicated system it may be necessary to employ these methods repeatedly before a solution becomes apparent. Notice also that the first of these procedures includes multiplication by a reciprocal which, of course, amounts to division of an equation by a constant, while the second includes adding to one equation another equation which has been multiplied by —1 which amounts to subtraction.

Example 7.10 We will apply these rules to the following system of equations. The choice of operation at each stage is simply determined by inspection and by looking for patterns in the equations:

Solution

Step 1 Add the second and third equations, and replace the third equation with the result i

Step 2 Add the second equation to the first with the result

Step 3 Multiply the third equation by 2 and subtract from the second equation:

Step 4 Multiply the first equation by 4 and add to the second equation:

Step 5 Divide the lirst of these equations by 21:

Step 6 Subtract the second equation from the lirst:

Step 7 Divide the second equation by 4:

Step 8 Finally, subtract the third equation from three times the second equation:

We noted in die last section that systems of two linear equations may have no solution, exactly one solution, or infinitely many solutions. The same is true when there are m > 2 linear equations. The following provides an example of infinitely many solutions.

Example 7.11 Attempt to solve the system of equations:

Solution

Step 1 Add the first and third equations:

Step 2 Subtract the third equation from the second equation, and interchange the second and first equations:

Step 3 Add three times the second equation to the first equation, and divide the first equation by 3:

The source of the difficulty is now apparent, since we are unable to reduce the second and third equations further to obtain unique solutions. Both equations imply that y = z — I. Any values of y and - satisfying this relationship ate solutions and, of course, there are infinitely many such values. So, even though we have a unique solution for x ~ 3, the system as a whole has an infinity of possible solutions. In this case, wc say that the variables y and z are not independent. We also note that the dependence or the relationship between y and c: is itself a linear dependence, and this is necessarily so. Through these steps, we have discovered another important property of systems of linear equations.

Definition 7.1

Linearly dependent equations are equations that may be derived from each other by a series of linear operations.

Theorem 7.3

If two or more variables in a linear system are (linearly) dependent, then two or more of the equations must be linearly dependent.

Example 7.11 illustrates this proposition since the first equation may be obtained by multiplying the second equation by —2, subtracting the third equation, and then dividing the result by 7. Thus, the first equation is a linear combination of the second and third equations.

Equations that are not linearly dependent are linearly independent.The equations in example 7.11 are linearly dependent.

The following example illustrates a case in which there is no solution to a system of equations:

Example 7.12 Consider the following system:

—X

y — '

: = 0

0 0

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