Row transformations

Elementary transformations on rows of matrices

Now let's introduce various elementary transformations on rows of matrices, which we have already seen used in the topic about matrix representation of linear systems. Each system equation corresponds to a row of a matrix. Therefore, transformations of rows correspond to transformations of equations. To start with, we are going to define which transformations we consider elementary. For illustration we will use the following matrix:

$A = \left( \begin{array}{ccc} 2 & 6 & 3\\ 4 & 2 & 7\\ 8 & 3 & 1\\ \end{array} \right)$

The elementary operations are:

Interchanging any two rows of the matrix. Interchanging the second and the third rows of matrix $A$ will result in this:

$\left( \begin{array}{ccc} 2 & 6 & 3\\ \color{red}8 & \color{red}3 & \color{red}1\\ \color{red}4 & \color{red}2 & \color{red}7\\ \end{array} \right)$

Multiplying or dividing each element of the row by a fixed non-zero number. After multiplying the first row of matrix $A$ by 2 the resulting matrix will look like this:

$\left( \begin{array}{ccc} \color{red}4 & \color{red}12 & \color{red}6\\ 4 & 2 & 7\\ 8 & 3 & 1\\ \end{array} \right)$

Subtracting or adding one row from another. Subtraction of the second row from the first turns matrix $A$ into this:

$\left( \begin{array}{ccc} \color{red}-2 & \color{red}4 & \color{red}-4\\ 4 & 2 & 7\\ 8 & 3 & 1\\ \end{array} \right)$

Equivalent matrices

We can say that matrix $B$ is equivalent to matrix $A$ and vice versa if and only if one matrix can be represented as the other matrix by applying elementary operations to it. Such matrices are called equivalent. Equivalent matrices are designated this way: A $\sim$ B. Note that the order of matrices doesn't matter for equivalency (that is, A $\sim$ B is the same as B $\sim$ A), since we can transform B back into A by using the combination of inverse elementary operations.

The equivalence property is very important. Using elementary operations, we can transform matrices to the upper triangular form. In this form, systems of linear equations become easy to solve.

Now we will see how a chain of transformations that consists of a number of elementary operations preserves the equivalence.

For example, consider the following matrix:

$\left( \begin{array}{ccc} 2 & 7 & 1\\ 3 & 9 & -3\\ 2 & 6 & -1\\ \end{array} \right)$

Multiply the second row by $\frac{1}{3}$:

$\begin{pmatrix} 2 & 7 & 1\\ 3 & 9 & -3\\ 2 & 6 & -1\\ \end{pmatrix} \rightarrow \begin{pmatrix}2 & 7 & 1\\ \color{red}1 & \color{red}3 & \color{red}-1\\ 2 & 6 & -1\\ \end{pmatrix}$

Subtract the third row from the first row:

$\begin{pmatrix} 2 & 7 & 1\\ 1 & 3 & -1\\ 2 & 6 & -1\\ \end{pmatrix} \rightarrow \begin{pmatrix}\color{red}0 & \color{red}1 & \color{red}2\\ 1 & 3 & -1\\ 2 & 6 & -1\\ \end{pmatrix}$

Then exchange the first and second rows, and then subtract the first row multiplied by 2 from the third row:

$\begin{pmatrix} 0 & 1 & 2\\ 1 & 3 & -1\\ 2 & 6 & -1\\ \end{pmatrix} \rightarrow \begin{pmatrix}\color{green}1 & \color{green}3 & \color{green}-1\\ \color{green}0 & \color{green}1 & \color{green}2\\ \color{red}0 & \color{red}0 & \color{red}1\\ \end{pmatrix}$

Therefore, the following matrices are equivalent:

$\begin{pmatrix}2 & 7 & 1\\ 3 & 9 & -3\\ 2 & 6 & -1 \end{pmatrix} \sim \begin{pmatrix} 1 & 3 & -1\\ 0 & 1 & 2\\ 0 & 0 & 1\\ \end{pmatrix}$

This is a very typical example because we have reduced the matrix to its equivalent upper triangular matrix. In the future, this will help us in solving linear systems.

In order to determine if two matrices are equivalent, we would need to try to transform them to the same matrix -- for example, into an identity matrix. If we succeed, then the matrices are equivalent.

Note that it is possible to carry out similar operations with columns, but we won't consider this. In terms of linear equations, operations with columns correspond to variable changes. We cannot say that systems of linear equations have the same solutions if their matrices can be obtained from each other by column operations.

Non-equivalent matrices

But not all matrices are equivalent. For example, consider two following matrices:

$\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} \qquad \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 1 & 1 & 0\\ \end{pmatrix}$

The first matrix is an identity matrix, we will leave it as it is. Now the second matrix can be transformed a bit: subtract from the third row both the first and the second rows. We will get this as a result:

$\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ \color{red}0 & \color{red}0 & \color{red}0\\ \end{pmatrix}$

You may notice that no transformation would be able to create a nonzero in the third column. Therefore, it's impossible to transform the second matrix into the first, and thus they are not equivalent.

Conclusion

In this topic, we have discussed various elementary transformations on matrices as well as a crucial property of equivalence. Now, let's do a quick recap:

• There are three elementary transformations on matrices: Interchanging, multiplication, or its inverse — division, and addition, or its inverse — subtraction.

• Matrix $A$ is equivalent to matrix $B$ if and only if it can be represented as the latter using a number of elementary transformations on it.

• If there exists no combination of elementary transformations, such that if applied to arbitrary matrix $A$ to represent it as matrix $B$ means that they are non-equivalent.