### Uniqueness

##### 2. Uniqueness of solutions

A consequence of our discussion on the structure of solutions is the criterion for the uniqueness of the solution. Combining the following facts

• uniqueness ⇔ no freedom (i.e., no free variables)
• free variable ⇔ nonpivot column
• nonfree variable ⇔ pivot column

we have

A consistent system of linear equations has a unique solution
⇔ All variables are nonfree
⇔ All columns of the coefficient matrix are pivot

Example In a previous example, we saw that the system

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

is consistent. Moreover, in the row echelon form of the coefficient matrix,

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

[col 3] and [col 5] are not pivot. Therefore the solution is not unique.

The following system is obtained by taking away x3 and x5.

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

The system was also studied before and was known to be consistent. In the row echelon form of the coefficient matrix,

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

all columns are pivot. Therefore the solution is unique.

We remark that the criterion for the uniqueness does not involve the right side. In other words, if both Ax = b and Ax = c are consistent, then the solution of Ax = b is unique if and only if the solution of Ax = c is unique. We also note that the homogeneous system Ax = 0 always has x = 0 as the trivial solution. The uniqueness for Ax = 0 simply means that the trivial solution is the only solution. In summary, we have the following important facts about uniqueness.

The uniqueness is independent of the right side. Moreover,

A consistent system Ax = b has a unique solution
⇔ The homogeneous system Ax = 0 has only the trivial solution

Example The following system

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

has the obvious solution x1 = x2 = x3 = x5 = 0, x4 = 1. In the previous example, we saw that another system with the same coefficient matrix has non-unique solutions. Therefore this system also has non-unique solutions. In other worlds, the system must have solutions other than the obvious one. Similarly, the homogeneous system

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

must have nontrivial (i.e., nonzero) solutions. On the other hand, the system

 x1 + 3x2 + 2x3 + x5 = 23 - x1 - x2 - x3 + x4 = -45 4x2 + 2x3 + 4x4 + 3x5 = 78 x1 + 3x2 + 2x3 - 2x4 = 12

does not have an obvious solution. Since it has the same coefficient matrix as before, we conclude there are two possibilities.

1. The system has no solution
2. The system has more than one solutions

Of course, to find out which case exactly happens, one has to do more computations (in order to determine the consistency).