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<< 7.1. Systems of Linear Equations in Two Variables

7.2. Systems of Linear Equations in Three Variables

In the previous section we explained how to solve systems of two linear equations in two variables, using the matrix technique. In this section we use the same technique to solve systems of three linear equations in three variables. The steps are the same as before, namely applying one of the following three elementary row operations to the matrix representing the system of equations:

- Multiplying (or dividing) a row by a nonzero constant.
- Adding or subtracting a multiple of a row to another row.
- Swapping a pair of rows.

We will start with the example we gave in Section P.9.

**Example 1:** Solve the following system of three linear equations in three variables:

2x + 5y - z = 4

x - 3y + 2z = 3

3x - 2y + z = 8

**Solution:** We start by writing our system as a matrix as follows:

Next we swap the first and second rows, resulting in the following matrix:

Now we subtract twice the first row from the second row, yielding the matrix

Now we subtract three times the first row from the third row, yielding

Subtracting the third row from the second now yields

Next we divide the second row by 4, yielding

Subtracting seven times the second row to the first row yields

Dividing the third row by -5 yields

Adding three times the second row to the first row yields

Finally, subtracting twice the third row from the first row yields

Thus the solution is: x = 51 / 20; y = -1/4; z = -3 / 20.

As with systems of two linear equations in two variables, systems of three linear equations in three variables may also be inconsistent or dependent as the following examples show:

**Example 2:** Solve the following system of three linear equations in three variables:

2x - y - z = 1

-x + 2y - z = 1

-x - y + 2z = 1

**Solution: **We first represent the system by the following matrix:

Swapping the first and second rows yields the matrix

Negating the first row yields

Subtracting twice the first row from the second row yields

Adding the first row to the third row now yields

Adding the second row to the third row yields

Since the third row is equivalent to the equation 0 = 3, we see that this system is inconsistent.

**Example 3:** Solve the following system of three linear equations in three variables:

2x - y - z = 0

-x + 2y - z = 0

-x - y + 2z = 0

**Solution: **We first represent the system by the following matrix:

Swapping the first and second rows yields the matrix

Negating the first row yields

Subtracting twice the first row from the second row yields

Adding the first row to the third row now yields

Adding the second row to the third row yields

At this point we see that our system is dependent. Dividing the second row by 3 yields

Finally, subtracting the second equation from the first yields

This is as far as we can go with this matrix. Converting it back into a system of equations yields

x - z = 0

y - z = 0

0 = 0

The first equation tells us x = z and the second tells us y = z. Together, these equations tell us x = y = z, or in other words, that all the variables have the same (unknown) value. The third equation adds no useful information.

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