Discover Linear Algebra

Example 2.4.1. A system with one unique solution.

From Discovery 2.1, consider system

We form the augmented matrix for the system, and reduce.

Every variable column has a leading one, so there are no free variables and no parameters are required. We can solve for each variable as a specific number, so the system has one unique solution: \(x = 5\text{,}\) \(y = 2\text{,}\) and \(z = 3\text{.}\)

Example 2.4.2. A system with an infinite number of solutions.

From Discovery 2.2, consider system

We form the augmented matrix for the system, and reduce.

The second column does not contain a leading one, so variable \(y\) is free and we assign to it a parameter: \(y=t\text{.}\) We can then use the simplified system

to solve for \(x = 2 - 2t\) and \(z=-3\text{.}\) In parametric form, the general solution of the system can be expressed as

and every particular solution to the system can be obtained by choosing a value for \(t\text{.}\) For example, the particular solution associated to \(t=3\) is

and the particular solution associated to \(t=-\sqrt{2}\) is

Example 2.4.3. A system with no solution.

From Discovery 2.3, consider system

We form the augmented matrix for the system, and reduce.

Here we have a leading one in the “equals” column. If we turn that third row back into an equation, we have

but there are no possible values of \(x,y,z\) that satisfy this equation. Therefore, the system is inconsistent. (Of course, we could have seen that this would happen right from the second matrix, and could have stopped there. But we went all the way to RREF to have another example demonstrating the row reduction strategy. In practice, we should stop reducing as soon as we can see that the system will be inconsistent.)

Example 2.4.4. A homogeneous system.

From Discovery 2.4, consider system

For a homogeneous system, we only reduce the coefficient matrix, since elementary row operations will never change an “equals” columns that contains all zeros.

To solve, remember that this is just the coefficient matrix for the simplified system, so all columns correspond to a variable, and the “equals” column is still all zeros but does not appear. We have two free variables, corresponding to the lack of leading one in the second and fourth columns. So set parameters \(x_2=s\) and \(x_4=t\text{.}\) The first two rows turn into equations

from which we obtain the general solution in parametric form

Example 2.4.5. Patterns in the solutions to homogeneous/nonhomogeneous systems.

Consider the homogeneous system

As in the previous example, to solve we work with just the coefficient matrix

But notice that this is the same coefficient matrix as for the system in Discovery 2.2, and the same row reduction sequence we used to solve that system in Example 2.4.2 would reduce this coefficient matrix to

And from here we also take the same steps as in Example 2.4.2 to solve this system. Assign parameter \(t\) to free variable \(y\text{,}\) and use the simplified homogeneous system

to solve for \(x = -2t\) and \(z=0\text{.}\) Let's compare the parametric forms of the solutions to the original nonhomogeneous system from Discovery 2.2 and the corresponding homogeneous system solved here.