## Transcript

Now that we have covered the fundamentals of the complex numbers, we want to move on to study the vector space properties of the complex numbers. As with vector spaces over R, we will frequently find ourselves wanting to solve a system of linear equations over the complex numbers—that is to say, a system whose coefficients and variables are both complex numbers. Since we can add, subtract, multiply, and divide complex numbers just as we can real numbers, we find that the technique we used to solve systems of linear equations over R can also be used to solve systems of linear equations over C. That is to say, we will row reduce a matrix. The elementary row operations remain the same, except that when we now say “multiply a row by a constant”, we are now multiplying by a complex number, and when we add a multiple of one row to another, we are again multiplying by a complex number. The definitions of row echelon form and reduced row echelon form remain the same, and the notion of a bad row, where we have set 0 = a for any a not equal to 0, is also the same. When necessary, we still replace variables with parameters to write the general solution, again keeping in mind that our parameters will now be from C, not just R.

So, basically, it’s all the same, except it’s completely different. Well, since the complex numbers are still new to you, even a simple multiply-a-row-by-a-constant operation may prove difficult, so you are urged to take your time, and write out all your work as you become acquainted with row reductions in the complex numbers.

Here are some examples of solving systems of equations over the complex numbers, designed to give you an idea of the different situations you may encounter.

Our first example: solve the following system of linear equations. So to solve this system, we will row reduce its augmented matrix. We’ll write that here. Now, since our first row already has a 1 in the first column, we can establish this as a leading 1 by subtracting the correct multiples of row 1 from rows 2 and 3, to leave them with zeroes in the first columns. So our first row operations, which can be performed simultaneously as before, are to replace row 2 with (row 2 + (-2)(row 1)) and to replace row 3 with
(row 3 + (-2i)(row 1)). So note that I always prefer to add a negative multiple rather than subtract a positive multiple because I find it too easy to make a mistake when you end up combining negative signs and subtraction. It’s better to simply deal with the matter when you are multiplying by your constant.

So to replace row 2 with (row 2 + (-2)(row 1)), we first need to calculate (-2)(row 1). Multiplying by a real number is still a simple matter of distributing, so we quickly find that (-2)(row 1) will equal [-2, -2i, 6 – 2i | 2 + 2i]. And so we find that (R2 + (-2)(row 1)) could be calculated as follows.

Now, to replace row 3 with (row 3 + (-2i)(row 1)), we first need to calculate (-2i)(row 1). Multiplying by a complex number does not become a simple case of distribution because i-squared becomes -1, but with practice, it can become easy to calculate that (-2i)(R1) will be [-2i, -2(i-squared), 6i – 2(i-squared) | 2i + 2(i-squared)], which is [-2i, 2, 2 + 6i, -2 + 2i]. And so now we compute (R3 + (-2i)(R1)) as seen here.

Having done our calculations, we can now replace row 2 with (row 2 + (-2)(row 1)) and row 3 with (row 3 + (-2i)(row 1)) to get this new matrix. Now, this matrix is already in row echelon form, and from it, we can see that our system will have a unique solution. To find this solution, we continue to row reduce until we reach reduced row echelon form. The first step in that process will be to multiply rows 2 and 3 by constants to make the first non-zero entry a 1. So we need to multiply row 2 by 1/(1 + i), and we need to multiply row 3 by 1/(3i).

But what are these values? Remember that for any complex number z, we have that 1/z = (the conjugate of z)/((the modulus of z)-squared). So when we multiply by 1/(1 + i), we are multiplying by (1 – i)/(1-squared + 1-squared), which equals (1 – i)/2. And when we multiply by 1/(3i), we are multiplying by (-3i)/(0-squared + 3-squared), which is –i/3. So rows 2 and row 3 will become, well, first we’ll see that ((1 – i)/2)(row 2) will be [0, 1, 2((1 – i)/2) | (2 + 4i)((1 – i)/2)]. This becomes [0, 1, 1 – i | and then on the other side of the augmented line, we’ll get a (1 + 2i)(1 – i)], which becomes 1 – i + 2i – 2(i-squared), which we see is 3 + i. It’s a little quicker to calculate that (-i/3)(row 3) will be [0, 0, 1 | and then (-3 + 3i)(-i/3)], so again, on the augmented side, we’re seeing that this is i – i-squared, which is 1 + i. Going ahead and replacing R2 with ((1 – i)/2)(R2) and R3 with (-i/3)(R3), we get this new matrix.

Now we begin our back-substitution steps by creating zeroes above the leading 1 in row 3. That means we need to replace row 1 with (row 1 + (3 – i)(row 3)), and we need to replace row 2 with (row 2 + (-1 + i)(row 3)). Now, since the first two columns in row 3 are zeroes, these operations will not change the value in the first two columns of rows 1 and row 2. Additionally, our row operation has been specifically chosen to make the third columns of row 1 and 2 equal to 0. So the only calculation that actually needs to be performed is in our fourth column. So instead of looking at these calculations on the entire rows, I will simply focus on the calculation needed for the fourth column.

So these are, as we look at (row 1 + (3 – i)(row 3)) in our fourth column, we’ll get a (-1 – i) + (3 – i)(1 + i). Well, this is -1 – i + 3 + 3i – i – i-squared, which will be a (-1 + 3 + 1) + i(-1 + 3 – 1), which is 3 + i. Doing the same thing in row 2, we’re looking at (row 2 + (-1 + i)(row 3)) simply in the fourth column becomes a (3 + i) + (-1 + i)(1 + i), which is 3 + i – 1 – i + i + i-squared, which we can group as (3 – 1 – 1) + i(1 – 1 + 1), which equals 1 + i. And so, our new matrix becomes this one seen here.

The final step in the row reduction is to replace R1 with (R1 +
(-i)(R2)). Again, we have that the only calculations needed are in the fourth column, since the first and third columns of row 2 have a 0 in them, and thus, these columns of row 1 will be unchanged, and we have chosen our row operation to turn the entry in the second column of row 1 into a 0. In the fourth column, we will get that (row 1 + (-i)(row 2)) will be (3 + i) + (-i)(1 + i), which equals 3 + i – i – i-squared, which becomes a (3 + 1) + i(1 – 1), which equals 4. And so we see that our reduced row echelon form matrix is as follows. And at this point, we can read off the solution to our system as z1 = 4, z2 = 1 + i, z3 = 1 + i.

Let’s look at another example. Now we will look at solving the following system of linear equations. And again, to solve the system, we will row reduce its augmented matrix, seen here. Notice that I went ahead and distributed any subtractions when I created this matrix. For example, the –(1 + i)z4 in the first equation becomes a -1 – i.

Now, our first row operations are to replace R2 with (R2 + (-1 + i)(R1)) and to replace R3 with (R3 + (-i)(R1)). So first, we will need to calculate (-1 + i)(R1). Well, this will become [1(-i + 1), (1 + i)(-i + 1), (-2i)(-1 + i), (-1 – i)(-1 + i) | (2 + i)(-1 + i)]. Whew, that’s a lot of multiplication. Let’s calculate all of these entries individually. Now obviously, 1(-1 + i) will equal -1 + i. Next, we find that (1 + i)(-1 + i) will equal -1 + i – i + i-squared, which equals (-1 – 1) + i(1 – 1), which equals -2. In the third column, we’ll get that (-2i)(-1 + i) = 2i – 2(i-squared), which equals 2 + 2i. In the fourth column, we get that (-1 – i)(-1 + i) = 1 – i + i – i-squared, which equals (1 + 1) + i(-1 + 1), which equals 2. And lastly, in the fifth column, we get that (2 + i)(-1 + i) = -2 + 2i – i + i-squared, which equals (-2 – 1) + i(2 – 1), which equals -3 + i. And so, at long last, we’ve calculated that (-1 + i)(R1) equals [-1 + i, -2, 2 + 2i, 2 | -3 + i]. And now we can use this to compute that (R2 + (-1 + i)(R1)) equals [0, 0, i, -1 + i | -2 – i]. Well, that was our first row operation.

Next, to replace row 3 with (row 3 + (-i)(row 1)), we will need to calculate (-i)(row 1). This, thankfully, is easier, so (-i)(row 1) will become a [1(-i), a (1 + i)(-i), a (-2i)(-i), a (-1 – i)(-i) | and a (2 + i)(-i)]. This becomes [-i, -i – i-squared, 2(i-squared), i + i-squared | -2i – i-squared], and this is [-i, 1 – i, -2, -1 + i | 1 – 2i]. And we can use this to compute that (row 3 + (-i)(row 1)) will equal [0, 0, -2 + i, -3 – i | -5i].

So, once we replace row 2 with (row 2 + (-1 + i)(row 1)) and row 3 with (row 3 + (-i)(row 1)), we get the following matrix.

The next thing we want to do is establish a leading 1 in the third column, but do we want to do so in row 2 or row 3? Neither one may look good to start with, but row 2 is by far the easiest, since it happens that dividing by i is the same thing as multiplying by a –i. So we can multiply row 2 by –i and get that [0, 0, i(-i), (-1 + i)(-i) | (-2 – i)(-i)] becomes [0, 0, -(i-squared), i – i-squared | 2i + i-squared], which is [0, 0, 1, 1 + i | -1 + 2i]. and by replacing row 2 with this new row 2, we get the following matrix.

Next, we want to replace row 3 with (row 3 + (2 – i)(row 2)). To do this, we must first compute (2 – i)(row 2). Well, here we’ll get 0, 0, our 2 – i, and then it’s the next two columns that become interesting. In the first, in the fourth column, we get (1 + i)(2 – i), and in the fifth column, we get (-1 + 2i)(2 – i). These calculations are not too difficult to perform, so we’ll simply distribute it out, and you’ll see in the end that we end up with the row [0, 0, 2 – i, 3 + i | 5i]. And now we use this fact to compute that (row 3 + (2 – i)(row 2)) becomes the row [0, 0, 0, 0 | 0].

So, this is our new matrix. This matrix is in row echelon form, and since there are no bad rows, we know that it has a solution. Moreover, we can already tell that our general solution will have two parameters. But let’s do one last row reduction step to put our matrix into reduced row echelon form before we attempt to find the general solution. And this operation will be to replace row 1 with (row 1 + (2i)(row 2)). Now since the first two columns of row 2 contain 0, this calculation will not change the first two columns of row 1. Now moreover, our operation was chosen to change the entry in the third column of row 1 into a 0, so no calculation is needed there either. All that remains is to do the calculations for columns 4 and column 5.

In column 4, the calculation will be (-1 – i) + (2i)(1 + i). This equals -1 – i + 2i + 2(i-squared), which equals (-1 – 2) + i(-1 + 2), which is -3 + i. In column 5, our calculation will become (2 + i) + (2i)(-1 + 2i), which equals 2 + i – 2i + 4(i-squared), which is (2 – 4) + i(1 – 2), which equals -2 – i. So, if we replace row 1 with (row 1 + (2i)(row 2)), our matrix becomes this.

Now, this matrix is in reduced row echelon form, and it is equivalent to the following system. If we replace the variable z2 with the parameter s, and the variable z4 with the parameter t, we see that the general solution to our system becomes that our vector [z1; z2; z3; z4] will equal [(-2 – i) – (1 + i)s – (-3 + i)t; s; (-1 + 2i) – (1 + i)t; t], which we can write in standard form as [-2 – i; 0; -1 + 2i; 0] + s[-1 – i; 1; 0; 0] + t[3 – i; 0; -1 – i; 1]. And so we’ve solved our system of equations.

Let’s do another example. We’ll solve the following system of linear equations. To solve this system, we will row reduce its augmented matrix, seen here. Our first goal is to create a 1 in the first column. At first glance, it may not seem obvious where to do this, and you may even be tempted to do this in the second row, even if it does create some fractions. But if you remember again that dividing by i is the same as multiplying by –i, you see that it is easy to create a 1 in the first column of the first row by multiplying it by –i/2. This will get you the following matrix.

Now we need to replace row 2 with (row 2 + (-2)(row 1)) and row 3 with (row 3 + (-1 – i)(row 1)). Since it is easy to calculate (-2)(row 1), we can quickly compute that (row 2 + (-2)(row 1)) will be [2, -2 – 4i, 1 – 2i | 1 – i], then a [-2, 2 + 4i, 4i | 4 + 6i], and when combined, we get [0, 0, 1 + 2i | 5 + 5i].

Now to compute (R3 + (-1 – i)(R1)), we will first need to compute (-1 – i)(R1), which is seen here. Let’s go ahead and calculate all of these entries individually. Now obviously, 1(-1 – i) will still be -1 – i. Next, we find that (-1 – 2i)(-1 – i) will equal 1 + i + 2i + 2(i-squared), which is (1 – 2) + i(1 + 2), which is -1 + 3i. In the third column, we get the calculation (-2i)(-1 – i) = 2i + 2(i-squared), which is -2 + 2i. And lastly, in the fourth column, we find that (-2 – 3i)(-1 – i) will equal 2 + 2i + 3i + 3(i-squared), which groups as (2 – 3) + i(2 + 3), which is -1 + 5i. And so we have found that (-1 – i)(row 1) is the following, and this lets us compute (R3 + (-1 – i)(R1)) as follows, getting that this new row will be [0, 0, 1 + 3i | 2 + 3i]. So going ahead and replacing row 2 with (row 2 + (-2)(row 1)) and row 3 with (row 3 + (-1 – i)(row 1)), our matrix becomes this.

The next step is to create a leading 1 in the third column. So we can do this in row 2 by multiplying it by 1/(1 + 2i). Well, this will equal (1 – 2i)/(1-squared + 2-squared), which is (1 – 2i)/5. So we’ll get that ((1 – 2i)/5)(row 2) will be 0, 0, (1 + 2i)/(1 + 2i) (let’s go ahead and leave it like that for simplicity), but in our fourth column, we’ll see that it’s (5 + 5i)((1 – 2i)/5). So, left of the augmented line, we’ll get [0, 0, 1]. All the interesting things happen on the right side, where we now are looking at (1 + i)(1 – 2i), which is 1 – 2i + i – 2(i-squared). If we group that, it becomes (1 + 2) + i(-2 + 1), which is a 3 – i. And so we see that our new matrix is this one.

Now we need to replace row 3 with (row 3 + (-1 – 3i)(row 2)). So this will make the first, second, and third entries in row 3 all equal to 0, so the only calculation we need to perform is in the fourth column, where we’ll get that (2 + 3i) + ((-1 – 3i) times our entry from row 2, which is (3 – i)), which equals 2 + 3i – 3 + i – 9i + 3(i-squared). Grouping our terms, this becomes a (2 – 3 – 3) + i(3 + 1 – 9), which is -4 – 5i. And so replacing row 3 with (row 3 + (-1 – 3i)(R2)), our matrix becomes this. In this case, our last row is a bad row, and so we see that our system is inconsistent and has no solutions.