Identifying Dependent and Inconsistent Systems

Recall that a solution to a linear system is an assignment of numbers to the variables such that all the equations are simultaneously satisfied. A solution of a system of equations in three variables is an ordered triple $(x, y, z)$, and describes a point where three planes intersect in space. 

There are three possible solution scenarios for systems of three equations in three variables:

• Independent systems have a single solution. Solving the system by elimination results in a single ordered triple $(x, y, z)$. Graphically, the ordered triple defines a point that is the intersection of three planes in space.
• Dependent systems have an infinite number of solutions. Graphically, the solutions fall on a line or plane that is the intersection of three planes in space.
• Inconsistent systems have no solution. Graphically, a system with no solution is represented by three planes with no point in common.

Dependent Systems of Equations with Three Variables

We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions (see below for a graphical representation). Or two of the equations could be the same and intersect the third on a line (see the example problem for a graphical representation).

Dependent systems

An example of three different equations that intersect on a line.

For example, consider this system of equations:

$\left\{\begin{matrix} \begin {aligned} 2x + y - 3z &= 0 \\ 4x + 2y - 6z &= 0 \\ x - y + z &= 0 \end {aligned} \end{matrix} \right.$

First, multiply the first equation by $-2$ and add it to the second equation:

$\begin {aligned} -2(2x + y - 3z) + (4x + 2y - 6z) &= 0 + 0 \\ (-4x + 4x) + (-2y + 2y) + (6z - 6z) &= 0 \\ 0 &= 0 \end {aligned}$

We do not need to proceed any further. The result we get is an identity, $0 = 0$, which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying the third equation by $−2$, and adding it to the first equation. We would then perform the same steps as above and find the same result, $0 = 0$.

If we were to graph each of the three equations, we would have the three planes pictured below. Notice that two of the planes are the same, and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.

Dependent system

Two equations represent the same plane, and these intersect the third plane on a line.

Inconsistent Systems of Equations with Three Variables

Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as $3 = 7$, or some other contradiction.

Inconsistent systems

All three figures represent three-by-three systems with no solution. (a) The three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. (c) All three planes are parallel, so there is no point of intersection.

For example, consider the system of equations

$\left\{\begin{matrix} \begin {aligned} x - 3y + z &= 4\\ -x + 2y - 5z &= 3 \\ 5x - 13y + 13z &= 8 \end {aligned} \end{matrix} \right.$

Using the elimination method for solving a system of equation in three variables, notice that we can add the first and second equations to cancel $x$:

$\begin {aligned}(x - 3y + z) + (-x + 2y - 5z) &= 4+3 \\ (x - x) + (-3y + 2y) + (z-5z) &= 7 \\ -y - 4z &= 7 \end {aligned}$

Next, multiply the first equation by $-5$,  and add it to the third equation: 

$\begin {aligned} -5(x - 3y + z) + (5x - 13y + 13z) &= -5(4) + 8 \\ (-5x + 5x) + (15y - 13y) + (-5z + 13z) &= -20 + 8 \\ 2y + 8z &= -12 \end {aligned}$

Now, notice that we have a system of equations in two variables:

$\left\{\begin{matrix} \begin {aligned} -y - 4z &= 7 \\ 2y + 8z &= -12 \end {aligned} \end {matrix} \right.$

We can solve this by multiplying the top equation by 2, and adding it to the bottom equation:

$\begin {aligned} 2(-y-4z) + (2y + 8z) &= 2(7) -12 \\ (-2y + 2y) + (-8z + 8z) &= 14 - 12 \\ 0 &= 2 \end {aligned}$

The final equation $0 = 2$ is a contradiction, so we conclude that the system of equations in inconsistent, and therefore, has no solution.