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Existence and Uniqueness of Solutions

A system of linear equations can have 0,1,∞0,1, \infty number of solutions. The pivot positions in the echelon form of the corresponding augmented matrix determine which of these cases occurs.

Existence of Solutions

Theorem (Existence of Solutions).
A linear system is inconsistent if and only if row echelon form of its augmented matrix has a row of the form: [000⋯0b]\left[\begin{array}{ccccc|c} 0 & 0 & 0 & \cdots & 0 & b \end{array}\right] where b≠0b \neq 0.

Such a row corresponds to the equation 0=b0 = b, which makes the system inconsistent. This is also called a pivot in the augmented column.

When we perform row operations, we're essentially manipulating the equations. If the system is inconsistent, the equations contradict each other in a fundamental way.

For example, consider:

{x+y=2x+y=5\begin{cases} x + y = 2 \\ x + y = 5 \end{cases}

These two equations contradict each other (the same combination x+yx + y cannot equal both 2 and 5). When we row-reduce:

[112115]→r2→r2−r1[112003]\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 1 & 1 & 5 \end{array}\right] \xrightarrow{r_2 \to r_2 - r_1} \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 3 \end{array}\right]

The second row says 0=30 = 3, which is impossible. The system is inconsistent.

Determine whether the following system is consistent:

{x1+3x2−5x3=42x1−x2+3x3=1−x1+4x2−8x3=10\begin{cases} x_1 + 3x_2 - 5x_3 = 4 \\ 2x_1 - x_2 + 3x_3 = 1 \\ -x_1 + 4x_2 - 8x_3 = 10 \end{cases}

Row Reduction to Check Consistency

Step 0 of 3

After row reduction to echelon form, we find that there is a pivot in the augmented column. Thus, this system is inconsistent.

Free Variables and Basic Variables

Once we know a system is consistent, we need to understand either the system has a unique solution or infinite many solutions. This depends on whether there are free variables.

Consider the following system of linear equations:

{x1−5x3=1x1+x2−4x3=5\begin{cases} x_1 \quad \quad - 5x_3 = 1 \\ x_1 + x_2 - 4x_3 = 5 \end{cases}

Its row reduced echelon form is computed by

[10−5111−45]→r2→r2−r1[10−510114]\left[\begin{array}{ccc|c} 1 & 0 & -5 & 1 \\ 1 & 1 & -4 & 5 \\ \end{array}\right] \xrightarrow{r_2 \to r_2 - r_1} \left[\begin{array}{ccc|c} \boxed{1} & 0 & -5 & 1 \\ 0 & \boxed{1} & 1 & 4 \\ \end{array}\right]
  • Pivot columns: Columns 1 and 2 (containing boxed pivots)
  • Basic variables: x1x_1 and x2x_2 (must be expressed in terms of other variables)
  • Free variables: x3x_3 (can take any value)
  • Consistent system: Since the augmented column does not contain a pivot

We write the row reduced echelon form of the augmented matrix back to system of linear equations:

{x1−5x3=1x2+x3=4\begin{cases} x_1 \quad \quad - 5x_3 = 1 \\ \quad \quad x_2 + x_3 = 4 \end{cases}

Then, we express the basic variables in terms of free variables, and the general solution is:

{x1=1+5x3x2=4−x3x3 is free\begin{cases} x_1 = 1 + 5x_3 \\ x_2 = 4 - x_3 \\ x_3 \text{ is free} \\ \end{cases}

x3x_3 is free in the sense that we can pick any value for x3x_3, and the corresponding values for x1,x2x_1, x_2 will give us a solution to the system. For example, we can pick x3=0x_3 = 0, and we get a solution (x1,x2,x3)=(1,4,0)(x_1,x_2,x_3) = (1,4,0). We can also pick x3=1x_3 = 1, and we get a solution (x1,x2,x3)=(6,3,1)(x_1,x_2,x_3) = (6,3,1).

Since x3x_3 can be any real numbers, this system has infinitely many solutions.

Uniqueness of Solutions

Theorem (Uniqueness of Solutions):
For a consistent system, it has a unique solution if and only if there are no free variables.

Decision Flowchart: Determining Number of Solutions

The following flowchart summarizes how to determine the number of solutions for any linear system after row reduction: