math111_logoUniqueness

1. Structure of solutions

Given a system, we first check the consistency. If the system is not consistent, then we cannot say anymore about the solutions. If the system is consistent, then we may further ask specific questions about the solutions.

In the subsequent discussion, we always assume the systems are consistent.

Example Suppose the augmented matrix of a system Ax = b has the following row echelon form.

[ 5 2 1 -9 0 -3 ]
0 0 -7 1 4 1
0 0 0 3 5 -2

The last equation is 3x4 + 5x5 = -2. Because the pivot entry 3 (the coefficient of x4) is nonzero, we can move terms other than x4 to the right side and then divide 3. The result is an expression of x4 in terms of x5: x4 = - 2/3 - 5/3 x5.

The middle equation is -7x3 + x4 + 4x5 = 1. Because the pivot entry -7 (the coefficient of x3) is nonzero, we can move terms other than x3 to the right side and then divide -7. The result is an expression of x3 in terms of x4 and x5. By further substituting the earlier expression x4 = - 2/3 - 5/3 x5, we see that x3 is expressed in terms of x5.

Similarly, in the top equation, because 5 (the coefficient of x1) is a pivot, we can move terms other than x1 to the right side and then divide the coefficient 5. The result is an expression of x1 in terms of x2, x3, x4, x5. By further substituting the earlier expressions of x3 and x4 in terms of x5, we see that x1 is expressed in terms of x2 and x5.

The key observation here is that the pivot coefficient can be divided because it is nonzero. As a result, the corresponding "pivot variable" can be expressed in terms of "later variables". The back substitution process then further replace the "later variables" with "nonpivot variables".

Example In this example, we studied a system with the augmented matrix having the following row echelon form.

[ * # # # # ]
0 0 * # #
0 0 0 * #
0 0 0 0 0

By back substitution and successively dividing the coefficients of x4, x3, x1, we may write the general solution as expressions of x1, x3, x4 in terms of x2, with x2 arbitrary. Alternatively, by further row operations, we have the reduced row echelon form.

[ 1 a 0 0 b1 ]
0 0 1 0 b2
0 0 0 1 b3
0 0 0 0 0

Then the general solution of the system is x1 = b1 - ax2, x3 = b2, x4 = b3, with x2 arbitrary.

Further examples can be found here. It is easy to conclude from the examples that, for a consistent system, the general solution can be presented as expressions of some nonfree variables (which cannot be arbitrarily chosen) in terms of some free variables (which can be arbitrarily chosen). Moreover, under the correspondence

variables of system ⇔ columns of coefficient matrix,

we have

free variable ⇔ nonpivot column
nonfree variable ⇔ pivot column

Example In an earlier example, we considered a system

x1 + 3x2 + 2x3   + x5 = 0
- x1 - x2 - x3 + x4   = 1
  4x2 + 2x3 + 4x4 + 3x5 = 2
x1 + 3x2 + 2x3 - 2x4   = 0

The row echelon form of the augmented matrix is

[ 1 3 2 0 1 0 ]
0 2 1 -1 0 1
0 0 0 2 1 0
0 0 0 0 0 0

which implies the system is consistent. Moreover, the solution is not unique because we may express nonfree variables x1, x2, x4 in terms of free variables x3, x5. Later in the same example, we took away the freedom x3, x5 (i.e., assuming both are zero) and considered the system

x1 + 3x2   = 0
- x1 - x2 + x4 = 1
  4x2 + 4x4 = 2
x1 + 3x2 - 2x4 = 0

The row echelon form of the augmented matrix is

[ 1 3 0 0 ]
0 2 -1 1
0 0 2 0
0 0 0 0

The system is still consistent. Since all the columns of the coefficient matrix (i.e., the first three columns) are pivot, all the variables are nonfree, so that the solution is unique. The uniqueness is to be expected because all the freedoms have been taken away.


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