Discover Linear Algebra

Example 4.4.7. A system as a matrix equation.

Let's again consider the system from Task b of Discovery 4.5. To solve, we row reduce the associated augmented matrix to RREF as usual.

Variable \(x_3\) is free, so assign a parameter \(x_3 = t\text{.}\) Then we can solve to obtain the general solution is parametric form,

Let's check a couple of particular solutions against the matrix equation \(A\uvec{x}=\uvec{b}\) that represents the system. Recall that for this system, \(\uvec{x}\) is the \(3\times 1\) column vector that contains the variables \(x_1,x_2,x_3\text{.}\) The particular solutions associated to parameter values \(t=0\) and \(t=3\) are

Let's collect the \(t=0\) solution values into the vector \(\uvec{x}\) and check \(A\uvec{x}\) versus \(\uvec{b}\text{:}\)

So the solution to the linear system we got by row reducing did indeed give us a vector solution \(\uvec{x}\) to the matrix equation \(A \uvec{x} = \uvec{b}\text{.}\) Let's similarly check the \(t=3\) solution, as in Task f of Discovery 4.5:

Again, our system solution gives us a solution to the matrix equation.

Example 4.4.8. Solutions in vector form: one unique solution.

The system from Discovery 2.1 can be expressed in the form \(A\uvec{x} = \uvec{b}\) for

We solved this system in Example 2.4.1 and determined that it had one unique solution, \(x = 5\text{,}\) \(y = 2\text{,}\) and \(z = 3\text{.}\) In vector form, we write this solution as

Example 4.4.9. Solutions in vector form: an infinite number of solutions.

The system from Discovery 2.2 can be expressed in the form \(A \uvec{x} = \uvec{b}\) for

We solved this system in Example 2.4.2, and determined that it had an infinite number of solutions. We expressed the general solution to the system using parametric equations

In vector form, we expand this solution as

Notice how the solution is the sum of a constant part

and a variable part

Further notice how the constant part is a particular solution to the system — it is the “initial” particular solution associated to the parameter value \(t=0\text{.}\)

Example 4.4.10. Solutions in vector form: a homogenous system.

The system from Discovery 2.4 is homogeneous, so it can be expressed in the form \(A\uvec{x} = \zerovec\) for

where \(\zerovec\) is the \(3 \times 1\) zero column vector. We solved this system in Example 2.4.4, and determined that it had an infinite number of solutions. We expressed the general solution to the system using parametric equations

In vector form, we expand this solution as

This time, the solution is a sum of two variables parts,

since there are two parameters. And there is no constant part to the general solution, because if we set both parameters to zero we obtain the trivial solution \(\uvec{x} = \zerovec\text{.}\) A homogeneous system will always work out this way. (So it would be more accurate to say that the general solution to the system from Discovery 2.4 has trivial constant part, instead of saying it has no constant part.)

Example 4.4.11. Solutions in vector form: patterns in homogeneous/nonhomogeneous systems.

In Example 2.4.5, we solved a homogenous system \(A\uvec{x} = \zerovec\) with

and found an infinite number of solutions, with general solution expressed parametrically as

In vector form, we express this as

This homogeneous system has the same coefficient matrix as in Example 4.4.9 above, so it is not surprising that their general solutions are related. In particular, notice that both systems have the same variable part, but that the nonhomogeneous system from Example 4.4.9 has a non-trivial constant part.