Matrix Row Operations

As we venture further into the world of matrices, we will eventually need to learn how to carry out matrix row operations. These operations help us solve a number of problems with matrices, and they can be very useful. There are three basic matrix row operations that we need to cover: Switching rows, multiplying a row by a number, and adding rows. By combining these three techniques, we can rearrange our matrices in the same way that we rearrange algebraic equations. But how do these matrix row operations work, and when should we use them? Let''s find out:

When do we use matrix row operations?

We use matrix row operations primarily to solve systems of linear equations. As we may have learned, it is possible to solve systems of linear equations using matrices. A key part of this process is turning the left side of our matrix into an identity matrix.

We may recall that an identity matrix is the equivalent of the value "1" in the world of matrices. This means that if we can get one side of the matrix to look like an identity matrix, we can solve for variables x and y. If one part of the matrix represents the variable and the left side represents the value "1," this simply means we are multiplying our value by 1 to get our variable. In other words, the right side of the matrix is the correct variable value.

Recall that an identity matrix has elements of "1" along its main diagonal with all other elements as zeros. For example:

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Note that as long as an identity matrix follows the above pattern, it can have any number of elements. This means that we can turn any matrix into an identity matrix.

Switching rows

The first operation we''re going to learn is switching rows. As the name implies, this involves switching the rows around on our matrix:

$\left[\begin{array}{cccc}2& 3& -2& 6\\ 0& 0& 3& -6\\ 1& 0& 2& -3\end{array}\right]$

Let's carry out the following operations:

• Move row 1 to row 2
• Move row 2 to row 3
• Move row 3 to row 1

We are left with:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& -2& 6\\ 0& 0& 3& -6\end{array}\right]$

Why did we do this? Remember, we are in the process of creating an identity matrix. This means that we need a value of "1" in the top left corner, which we have achieved with this matrix row operation.

Multiplying a row by a number

Multiplying a row by a number is quite straightforward:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& -2& 6\\ 0& 0& 3& -6\end{array}\right]$

What happens if we multiply row 3 by $\frac{1}{3}$ ?

We get:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& -2& 6\\ 0& 0& 3& -6\end{array}\right]$

Now we have a 1 in row 3, column 3.

Adding rows

Another operation is to add two rows together and replace an entire row with the sum.

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& -2& 6\\ 0& 0& 1& -2\end{array}\right]$

What happens if we add rows 2 and 3 together, and then replace row 2 with the result?

Let's add those two rows together first:

$\left[\begin{array}{cccc}2& 3& -2& 6\end{array}\right]+\left[\begin{array}{cccc}0& 0& 1& -2\end{array}\right]=\left[\begin{array}{cccc}2& 3& -1& 4\end{array}\right]$

Now let's replace row 2 with the result:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& -1& 4\\ 0& 0& 1& -2\end{array}\right]$

Adding multiples of rows

While there are only three matrix row operations, it is possible to combine two of these operations into one. We call this "adding rows of multiples." As the name suggests, this is a combination of both adding rows and multiplying rows by a number.

Consider this matrix -- the same matrix we add after multiplying row 3 by $\frac{1}{3}$ :

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& -2& 6\\ 0& 0& 1& -2\end{array}\right]$

What if we combined the last two steps to make things go a little faster? In other words, instead of starting by adding row 2 to row 3, we can add $R_2+2R_3$ .

Starting with the brackets first, we multiply row 3 by 2 to get:

$\left[\begin{array}{cccc}2& 3& 0& 2\end{array}\right]$

Now we can replace row 2 with this result. We are left with:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 2& 3& 0& 2\\ 0& 0& 1& -2\end{array}\right]$

The benefit here is that we now have a 0 in row 2, column 3. We are one step closer to creating our identity matrix.

Now we can do exactly the same step once more, this time getting another zero in row 2, column 1. In this case, we multiply row 1 by (-2) and replace row 2 with this result:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 0& 3& -4& -8\\ 0& 0& 1& -2\end{array}\right]$

Now we have learned all of the various steps we need to find our identity matrix. Let's continue to finish creating this identity matrix:

Add $R_2+4R_3$ to get a 0 in row 2, column 3:

$\left[\begin{array}{cccc}1& 0& 2& -3\\ 0& 3& 0& 0\\ 0& 0& 1& -2\end{array}\right]$

$R_1\leftarrow R_1-2R_3$ to get a 0 in row 1, column 3:

$\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 3& 0& 0\\ 0& 0& 1& -2\end{array}\right]$

The final step is to multiply row 3 by $\frac{1}{3}$ :

$\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 0\\ 0& 0& 1& -2\end{array}\right]$

Now we have our identity matrix on the left side. So what this is really saying is:

The solutions for the variables $x_1,x_2,x_3$ are $\left(x_1,x_2,x_3\right)=\left(1,0,-2\right)$ .

Note that this strategy does not work when solving linear equations for parallel lines or equivalent expressions. This is because there are either infinite solutions or no solutions at all.

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