Describing solutions to linear systems in three unknowns

A system of three linear equations in three unknowns can have no solutions, one solution or infinitely many solutions. In this chapter, we explore the conditions for these three possibilities.

one equation

A single linear equation in three unknowns represents a plane in a $3$3-dimensional coordinate system. For example, in the equation $3x-y+2z=0$3x−y+2z=0, we could choose a fixed value for $z$z, say $z=t$z=t. Then, $y=3x-2t$y=3x−2t. But this is the equation of a line in the $x$x-$y$y plane at the level of $t$t. There must be such a line at every possible level of $t$t and we might imagine these infinitely many lines, side by side, forming a plane in $3$3-space.

Example 1

There are infinitely many sets of numbers $(x,y,z)$(x,y,z) that satisfy the equation $3x-y+2z=0$3x−y+2z=0 but there are also infinitely many sets of numbers that do not satisfy the equation. We can specify the solution set by choosing a parameter $t$t for $z$z and another parameter $s$s for $y$y. Then, we see that $x$x is constrained by the equation to be $x=\frac{1}{3}s-\frac{2}{3}t$x=13​s−23​t.

The solution set can be written in vector form as

The two vectors on the right define a plane and the solution set is the set of linear combinations of the two vectors as $s$s and $t$t range over the real numbers.

 

two equations

Two equations represent two planes. The planes might be parallel, in which case they have no points in common and there can be no solution that satisfies both equations. If the planes are not parallel, they intersect along a line and there are infinitely many solutions, all belonging to the line.

Example 2

The equations $3x-y+2z=0$3x−y+2z=0 and $3x-y+2z=6$3x−y+2z=6 represent parallel planes. Any attempt at a simultaneous solution leads to an absurdity like $6=0$6=0 and we must conclude that no solution exists.

On the other hand, a pair of equations like $3x-y+2z=0$3x−y+2z=0 and $x+y+z=0$x+y+z=0 can be solved to find a line of solutions. We might observe immediately that $(0,0,0)$(0,0,0) satisfies both equations. That is, both planes pass through the origin.

We could multiply the second equation by $-3$−3 and add it to the first equation. This gives $0x-4y-z=0$0x−4y−z=0.  We now have the equations $x+y+z=0$x+y+z=0 and $4y+z=0$4y+z=0 and these must have the same solution set as the original pair.

As before, we could choose to put $z=t$z=t. Then, from the $4y+z=0$4y+z=0 we find $y=-\frac{t}{4}$y=−t4​. Finally, from $x+y+z=0$x+y+z=0 we have $x=\frac{t}{4}-t=-\frac{3t}{4}$x=t4​−t=−3t4​. All three variables have been expressed in terms of the single parameter $t$t.

This time, the solution set in vector form is

The solution set is the set of scalar multiples of the vector on the right. The solution $(0,0,0)$(0,0,0) corresponds to $t=1$t=1.

 

three equations

Three linear equations in three unknowns represent three planes in coordinate space. There are four possibilities.

• All three may be parallel, in which case there is no intersection and no solution. The system of equations would be found to be inconsistent.
• The three planes could intersect pairwise in three different parallel lines, in which case there is no simultaneous solution. Again, the system would be inconsistent.
• All three could intersect in a single line so that there are infinitely many solutions. Solutions would be found as in Example 2.
• The three planes could intersect at a single point.

 

Example 3

Consider the system of three equations

If we subtract the first equation from the second and also subtract twice the first equation from the third, we obtain

Next, subtract $\frac{1}{2}$12​ the second equation from the third.

From the third of these equations, we find $z=5$z=5. Then, when this is substituted into the second equation, we get $y=3$y=3. And when these values of $z$z and $y$y are substituted into the first equation, we have $x=2$x=2.

So, the solution set is the single point

 

There is a more efficient way of setting out these calculations, explained at this link.