Discover Linear Algebra

Example 11.3.1. Turning a complex linear system into an augmented complex matrix.

Consider the system

As usual, we can represent this system as an augmented matrix, with one row per equation and one column per variable, with an extra augmented column for the constants on the right of the equals signs.

Example 11.3.2. Reducing a complex augmented matrix.

Here we will reduce the augmented matrix for the system from Example 11.3.1. There are always multiple ways to reduce a matrix, but, as usual, we are not interested in the most efficient way. Instead, we will demonstrate a systematic way to reduce while avoiding complicated calculations with complex numbers, especially division.

Rather than divide by a complex number to obtain a leading one in the top left, we start by using the complex conjugate to first convert the upper left entry to a real number. (For this particular matrix, we could achieve the same result by simply adding the second row to the first, but again we wish to demonstrate a systematic method in a simple example.)

Now we can easily obtain a leading one by dividing by a real number.

Note that these first two operations together are equivalent to dividing the first row by \(1 - \ci\text{.}\)

Now use the leading one to eliminate just the real part of the entry below it.

Then convert the leading entry in the second row to a real number, by multiplying by \(\ci\text{.}\)

Now we can easily eliminate the bottom left entry.

From here we proceed in a similar fashion to obtain and exploit the next leading one.

From the reduced matrix, we can see that the original complex linear system has a unique solution, \(x = 2 - \ci\) and \(y = 1 - 3 \ci\text{.}\)

Example 11.3.3. Solving a complex linear system.

To solve a complex linear system we proceed just as in the real case, by reducing, assigning parameters (if necessary), and solving for each variable either exactly or in terms of the assigned parameters. The only difference is that parameters should be allowed to take on every possible complex value, instead of just every possible real value.

Suppose we have reduced the coefficient matrix for a homogeneous complex linear system, and have arrived at the RREF

We have four variables, but only two leading ones, so we need two parameters. The variables \(x_2\) and \(x_4\) are free, so we should assign parameters to those and solve for the others. In parametric form, the general solution is

Each different choice of a pair of complex values for parameters \(s\) and \(t\) leads to a different specific solution. With a homogeneous system, we still always have the trivial solution, \(x_1 = x_2 = x_3 = x_4 = 0\text{,}\) in this case associated to choice \(s = 0\text{,}\) \(t = 0\text{.}\)

We also have two specific solutions attached to each parameter, one from choosing \(s = 1\text{,}\) \(t = 0\) to obtain solution

and another from choosing \(s = 0\text{,}\) \(t = 1\) to obtain solution

As mentioned at the beginning of this example, parameters \(s\) and \(t\) can take on any complex values. For example, choosing \(s = 7 \ci\) and \(t = 5 - 4 \ci\) yields particular solution