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Linear Combinations


Suppose that the vector (1,3) represents the contents of a ``Regular'' can of mixed nuts (1 lb cashews and 3 lb peanuts) while (1,1) represents a ``Deluxe'' can (1 lb cashews and 1 lb peanuts). Can you obtain a mixture of 2 lb cashews and 3 lb peanuts by combining the two mixtures in appropriate amounts? The answer is to use tex2html_wrap_inline1266 cans of Regular and tex2html_wrap_inline1270 cans of Deluxe in order to satisfy the equality


This is none other than a system of two linear equations:


The solution of these equations (obtained by the Gauss-Jordan procedure) is tex2html_wrap_inline1524 , tex2html_wrap_inline1526 . So the desired combination is to mix 1/2 can of Regular nuts with 3/2 cans of Deluxe nuts. Thus if some recipe calls for the mixture (2,3), you can substitute 1/2 can of Regular mix and 3/2 can of Deluxe mix.

Suppose now that you want to obtain 1 lb cashews and no peanuts. This poses the equations,


The solution is tex2html_wrap_inline1530 . Thus you can obtain a pound of pure cashews by buying 3/2 cans of Deluxe mix and removing enough nuts to form 1/2 can Regular mix, which can be sold. In this case it is physically possible to use a negative amount of some ingredient, but in other cases it may be impossible, as when one is mixing paint.

A vector of the form tex2html_wrap_inline1532 is called a linear combination of the vectors tex2html_wrap_inline1534 and tex2html_wrap_inline1536 . In particular we just saw that the vector tex2html_wrap_inline1538 is a linear combination of the vectors tex2html_wrap_inline1534 and tex2html_wrap_inline1536 . And so is tex2html_wrap_inline1544 .

The question arises: can one obtain any mixture whatever by taking the appropriate combination of Regular and Deluxe cans? Is any vector tex2html_wrap_inline1546 a linear combination of tex2html_wrap_inline1534 and tex2html_wrap_inline1536 ?

To answer the question, solve the equations in general. You want a mixture of tex2html_wrap_inline1552 lb cashews and tex2html_wrap_inline1554 lb peanuts and set


or equivalently,


These equations have a unique solution, no matter what are the values of tex2html_wrap_inline1552 and tex2html_wrap_inline1554 , namely,


No matter what vector tex2html_wrap_inline1560 you want, you can get it as a linear combination of (1,3) and (1,1). The vectors (1,3) and (1,1) are said to be linearly independent.

Not all pairs of vectors can yield an arbitrary mixture tex2html_wrap_inline1560 . For instance, no linear combination of (1,1) and (2,2) yields (2,3). In other words, the equations


have no solution.

The reason is that (2,2) is already a multiple of (1,1), that is (2,2) is a linear combination of (1,1). The vectors (1,1) and (2,2) are said to be linearly dependent.

If (2,2) represents, for instance, a large Deluxe can of nuts and (1,1) a small one, it is clear that you cannot obtain any mixture you want by combining large and small Deluxe cans. In fact, once you have small cans, the large cans contribute nothing at all to the mixtures you can generate, since you can always substitute two small cans for a large one.

A more interesting example is (1,2,0), (1,0,1) and (2,2,1). The third vector is clearly a linear combination of the other two (it equals their sum). Altogether, these three vectors are said to be linearly dependent. For instance, these three vectors might represent mixtures of nuts as follows:


Then once you have brands A and B, brand C adds nothing whatever to the variety of mixtures you can concoct. This is because you can already obtain a can of brand C from brands A and B anyway. In other words, if a recipe calls for brand C, you can always substitute a mixture of 1 can brand A and 1 can brand B.

Suppose you want to obtain the mixture (1,2,1) by combining Brands A, B, and C. The equations are


If you try to solve these equations, you will find that there is no solution. So the vector (1,2,1) is not a linear combination of (1,2,0), (1,0,1) and (2,2,1).

The concepts of linear combination, linear dependence and linear independence introduced in the above examples can be defined more formally, for any number of n-component vectors. This is done as follows.




In linear programming, there are typically many more variables than equations. So, this case warrants looking at another example.


This is really a mixing problem. Rather than peanuts and cashews, the mixture will contain leather and labor. The items to be mixed are four activities: manufacturing a deluxe belt, manufacturing a regular belt, leaving a leftover leather strip in inventory, and leaving an hour of labor idle (or for other work). Just as each Regular can of mixed nuts contributes 1 pound of cashews and 3 pounds of peanuts to the mixture, each regular belt will consume 1 leather strip and 2 hours of labor. The aim is to combine the four activities in the right proportion so that 40 strips and 60 hours are accounted for:



tex2html_wrap_inline1266 = number of deluxe belts made
tex2html_wrap_inline1270 = number of regular belts made
tex2html_wrap_inline1650 = number of leather strips left over
tex2html_wrap_inline1652 = number of labor hours left over
In tableau form, the equations are:


Since there are only two equations, you can only solve for two variables. Let's solve for tex2html_wrap_inline1266 and tex2html_wrap_inline1270 , using Gauss-Jordan elimination. After the first iteration you get,


A second iteration yields the solution,


This tableau represents the equations




You can't say how many deluxe belts tex2html_wrap_inline1266 and regular belts tex2html_wrap_inline1270 the plant will make until you specify how much leather tex2html_wrap_inline1650 and labor tex2html_wrap_inline1652 will be left over. But (1.6) is a formula for computing how many belts of either type must be made, for any given tex2html_wrap_inline1650 and tex2html_wrap_inline1652 . So the equations have many solutions (infinitely many).

For example, if you want to have nothing left over ( tex2html_wrap_inline1670 ), you will make 20 of each. If you want to have 5 strips and 5 hours left over, you will make tex2html_wrap_inline1672 deluxe and tex2html_wrap_inline1674 regular belts.

The variables tex2html_wrap_inline1676 you solved for are called basic variables. The other variables are nonbasic. You have control of the nonbasic variables. Once you give them values, the values of the basic variables follow. A solution in which you make all the nonbasic variables zero is called a basic solution.

Can you have basic variables other than tex2html_wrap_inline1676 ? Sure. Any pair of variables can be basic, provided the corresponding columns in (1.3) are linearly independent (otherwise, you can't solve for the basic variables).

Equations (1.3), for instance, are already solved in (1.3) for basic variables tex2html_wrap_inline1680 . Here the two basic activities are having leftover leather and having leftover labor. The basic solution is tex2html_wrap_inline1682 . This means that if you decide to produce no belts ( tex2html_wrap_inline1684 ), you must have 40 leftover leather strips and 60 leftover labor hours.

The intermediate step (1.4) solves the equations with basic variables tex2html_wrap_inline1266 and tex2html_wrap_inline1652 . Here the basic solution is unrealizable. If you decide to participate only in the basic activities (making deluxe belts and having leftover labor), you must make 40 belts and have -20 leftover hours (i.e., use 20 more than you have), which you can't do within your current resources.





next up previous contents
Next: Inverse of a Square Up: Basic Linear Algebra Previous: Operations on Vectors and

Michael A. Trick
Mon Aug 24 13:24:14 EDT 1998