To describe the set of solutions to A of basic and nonbasic variables.
D E F I N I T I O N ■ After the Gauss–Jordan method has been applied to any linear system, a variable that appears with a coefficient of 1 in a single equation and a coefficient of 0 in all other equations is called a basic variable (BV). ■
Any variable that is not a basic variable is called a nonbasic variable (NBV). ■ Let BV be the set of basic variables for A
ables for A
following cases occurs.
Case 1 A
Ax b has no solution (recall Example 6). As an example of Case 1, suppose that when the Gauss–Jordan method is applied to the system Ax b, the following matrix is obtained:
1 0
0
A
In this case, A
Case 2 Suppose that Case 1 does not apply and NBV, the set of nonbasic variables, is empty. Then A
call that in solving
2x1 2x2 x3 9 2x1 x2 2x3 6 2x1 x2 2x3 5 the Gauss–Jordan method yielded
A
In this case, BV {x1, x2, x3} and NBV is empty. Then the unique solution to A (and Ax b) is x1 1, x2 2, x3 3.
Case 3 Suppose that Case 1 does not apply and NBV is nonempty. Then A
Ax b) will have an infinite number of solutions. To obtain these, first assign each non-basic variable an arbitrary value. Then solve for the value of each non-basic variable in terms of the nonbasic variables. For example, suppose
A
(15)
Because Case 1 does not apply, and BV {x1, x2, x3} and NBV {x4, x5}, we have an example of Case 3: A
see what these solutions look like, write down A
0x1 0x2 0x30x40x5 3 (15.1) 0x1 0x2 0x32x4 0x5 2 (15.2) 0x1 0x2 0x3 0x40x5 1 (15.3) 0x1 0x2 0x3 0x4 0x5 0 (15.4) Now assign the nonbasic variables (x4and x5) arbitrary values c and k, with x4 c and x5 k. From (15.1), we find that x1 3 c k. From (15.2), we find that x2 2
2c. From (15.3), we find that x3 1 k. Because (15.4) holds for all values of the variables, x1 3 c k, x2 2 2c, x3 1 k, x4 c, and x5 k will, for any values of c and k, be a solution to A
Our discussion of the Gauss–Jordan method is summarized in Figure 6. We have de-voted so much time to the Gauss–Jordan method because, in our study of linear pro-gramming, examples of Case 3 (linear systems with an infinite number of solutions) will occur repeatedly. Because the end result of the Gauss–Jordan method must always be one of Cases 1–3, we have shown that any linear system will have no solution, a unique so-lution, or an infinite number of solutions.
3 2 1 0 1 0 1 0 1 2 0 0 0 0 1 0 0 1 0 0 1 0 0 0
1 2 3 0 0 1 0 1 0 1 0 0
1 1
1 0 2 1 2 3 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0
Use the Gauss–Jordan method to determine whether each of the following linear systems has no solution, a unique tion, or an infinite number of solutions. Indicate the solu-tions (if any exist).
1 x1 x2 x3 x4 3 1 x1 x2 x3 x4 4 1 x1 2x2 x3 x4 8 2 x1 x2 x3 4 2 x1 2x2 x3 6 3 x1 x2 1 2 2x1 x2 3 2 3x1 2x2 4
4 2x1 x2 x3 x4 6 2 x1 x2 x3 x4 4 5 x1x2 x2 x4 5 2 x2x2x 2 2x4 5 2 x2x 2x3 0.5x4 1 x2 2x3 x4 3
6 x12x2 2x3 4 2 x1 2x2 x3 4 2 xx1 2x2 x3 0 7 x1 x2 2x3 2 2 x1 x2 2x3 3 2 x1 x2 x3 3 8 x1 x2 x3 x4 1 2 x1x2 2x3 x4 2 2 x1 x2 2x3x4 3 Group B
9 Suppose that a linear system Ax b has more variables than equations. Show that Ax b cannot have a unique solution.
2.4 Linear Independence and Linear Dependence
†In this section, we discuss the concepts of a linearly independent set of vectors, a lin-early dependent set of vectors, and the rank of a matrix. These concepts will be useful in our study of matrix inverses.
Before defining a linearly independent set of vectors, we need to define a linear com-bination of a set of vectors. Let V {v1, v2, . . . , vk} be a set of row vectors all of which have the same dimension.
†This section covers topics that may be omitted with no loss of continuity.
YES
YES NO
NO Does A b have a row [0 0 · · · 0 c] (c ≠ 0)?
Ax = b has no solution
Find BV and NBV
Is NBV empty?
Ax = b has a unique solution
Ax = b has an infinite number of solutions F I G U R E 6
Description of Gauss–Jordan Method for Solving Linear Equations
P R O B L E M S
Group A
D E F I N I T I O N ■ A linear combination of the vectors in V is any vector of the form c1v1 c2v2
ckvk, where c1, c2, . . . , ckare arbitrary scalars. ■ For example, if V {[1 2], [2 1]}, then
2v1 v2 2([1 2]) [2 1] [0 3]
v1 3v2 [1 2] 3([2 1]) [7 5]
0v1 3v2 [0 0] 3([2 1]) [6 3]
are linear combinations of vectors in V. The foregoing definition may also be applied to a set of column vectors.
Suppose we are given a set V {v1, v2, . . . , vk} of m-dimensional row vectors. Let 0 [0 0 0] be the m-dimensional 0 vector. To determine whether V is a linearly independent set of vectors, we try to find a linear combination of the vectors in V that adds up to 0. Clearly, 0v1 0v2 0vkis a linear combination of vectors in V that adds up to 0. We call the linear combination of vectors in V for which c1 c2 ck 0 the trivial linear combination of vectors in V. We may now define linearly inde-pendent and linearly deinde-pendent sets of vectors.
D E F I N I T I O N ■ A set V of m-dimensional vectors is linearly independent if the only linear combination of vectors in V that equals 0 is the trivial linear combination. ■ A set V of m-dimensional vectors is linearly dependent if there is a nontrivial linear combination of the vectors in V that adds up to 0. ■
The following examples should clarify these definitions.
Show that any set of vectors containing the 0 vector is a linearly dependent set.
Solution To illustrate, we show that if V {[0 0], [1 0], [0 1]}, then V is linearly dependent, because if, say, c1 0, then c1([0 0]) 0([1 0]) 0([0 1]) [0 0]. Thus, there is a nontrivial linear combination of vectors in V that adds up to 0.
Show that the set of vectors V {[1 0], [0 1]} is a linearly independent set of vectors.
Solution We try to find a nontrivial linear combination of the vectors in V that yields 0. This re-quires that we find scalars c1 and c2 (at least one of which is nonzero) satisfying c1([1 0]) c2([0 1]) [0 0]. Thus, c1and c2must satisfy [c1 c2] [0 0]. This implies c1 c2 0. The only linear combination of vectors in V that yields 0 is the trivial linear combination. Therefore, V is a linearly independent set of vectors.
Show that V {[1 2], [2 4]} is a linearly dependent set of vectors.
Solution Because 2([1 2]) 1([2 4]) [0 0], there is a nontrivial linear combination with c1 2 and c2 1 that yields 0. Thus, V is a linearly dependent set of vectors.
Intuitively, what does it mean for a set of vectors to be linearly dependent? To understand the concept of linear dependence, observe that a set of vectors V is linearly dependent (as
0 Vector Makes Set LD
E X A M P L E 8
LI Set of Vectors
E X A M P L E 9
LD Set of Vectors
E X A M P L E 1 0
long as 0 is not in V ) if and only if some vector in V can be written as a nontrivial linear combination of other vectors in V (see Problem 9 at the end of this section). For instance, in Example 10, [2 4] 2([1 2]). Thus, if a set of vectors V is linearly dependent, the vec-tors in V are, in some way, not all “different” vecvec-tors. By “different” we mean that the di-rection specified by any vector in V cannot be expressed by adding together multiples of other vectors in V. For example, in two dimensions it can be shown that two vectors are linearly dependent if and only if they lie on the same line (see Figure 7).