Elementary Operations for Matrices

Elementary operations for matrices play a crucial role in finding the inverse or solving linear systems. They may also be used for other calculations. On this page, we will discuss these type of operations. Before we define an elementary operation, recall that to an nxm matrix A, we can associate n rows and m columns. For example, consider the matrix

\begin{displaymath}A = \left(\begin{array}{rrrr}
0&1&-1&3\\
0&2&3&1\\
-1&0&2&-3\\
\end{array}\right).\end{displaymath}

Its rows are

\begin{displaymath}\left(\begin{array}{rrrr}
0&1&-1&3\\
\end{array}\right), \; ...
...),\; \left(\begin{array}{rrrr}
-1&0&2&-3\\
\end{array}\right).\end{displaymath}

Its columns are

\begin{displaymath}\left(\begin{array}{rrrr}
0\\
0\\
-1\\
\end{array}\right),...
...; \left(\begin{array}{rrrr}
3\\
1\\
-3\\
\end{array}\right).\end{displaymath}

Let us consider the matrix transpose of A

\begin{displaymath}A^{T} = \left(\begin{array}{rrrr}
0&0&-1\\
1&2&0\\
-1&3&2\\
3&1&-3\\
\end{array}\right).\end{displaymath}

Its rows are

\begin{displaymath}\left(\begin{array}{rrrr}
0&0&-1\\
\end{array}\right), \; \l...
...ght),\; \left(\begin{array}{rrrr}
3&1&-3\\
\end{array}\right).\end{displaymath}

As we can see, the transpose of the columns of A are the rows of AT. So the transpose operation interchanges the rows and the columns of a matrix. Therefore many techniques which are developed for rows may be easily translated to columns via the transpose operation. Thus, we will only discuss elementary row operations, but the reader may easily adapt these to columns.

Elementary Row Operations.

1.
Interchange two rows.
2.
Multiply a row with a nonzero number.
3.
Add a row to another one multiplied by a number.

Definition. Two matrices are row equivalent if and only if one may be obtained from the other one via elementary row operations.

Example. Show that the two matrices

\begin{displaymath}A= \left(\begin{array}{rrrr}
1&-1&0\\
2&1&1\\
\end{array}\r...
... \left(\begin{array}{rrrr}
3&0&1\\
0&3&1\\
\end{array}\right)\end{displaymath}

are row equivalent.

Answer. We start with A. If we keep the second row and add the first to the second, we get

\begin{displaymath}\left(\begin{array}{rrrr}
3&0&1\\
2&1&1\\
\end{array}\right).\end{displaymath}

We keep the first row. Then we subtract the first row from the second one multiplied by 3. We get

\begin{displaymath}\left(\begin{array}{rrrr}
3&0&1\\
3&3&2\\
\end{array}\right).\end{displaymath}

We keep the first row and subtract the first row from the second one. We get

\begin{displaymath}\left(\begin{array}{rrrr}
3&0&1\\
0&3&1\\
\end{array}\right)\end{displaymath}

which is the matrix B. Therefore A and B are row equivalent.

One powerful use of elementary operations consists in finding solutions to linear systems and the inverse of a matrix. This happens via Echelon Form and Gauss-Jordan Elimination. In order to appreciate these two techniques, we need to discuss when a matrix is row elementary equivalent to a triangular matrix. Let us illustrate this with an example.

Example. Consider the matrix

\begin{displaymath}\left(\begin{array}{rrrr}
0&0&1&3\\
2&4&0&-8\\
1&2&1&-1\\
\end{array}\right).\end{displaymath}

First we will transform the first column via elementary row operations into one with the top number equal to 1 and the bottom ones equal 0. Indeed, if we interchange the first row with the last one, we get

\begin{displaymath}\left(\begin{array}{rrrr}
1&2&1&-1\\
2&4&0&-8\\
0&0&1&3\\
\end{array}\right).\end{displaymath}

Next, we keep the first and last rows. And we subtract the first one multiplied by 2 from the second one. We get

\begin{displaymath}\left(\begin{array}{rrrr}
1&2&1&-1\\
0&0&-2&-6\\
0&0&1&3\\
\end{array}\right).\end{displaymath}

We are almost there. Looking at this matrix, we see that we can still take care of the 1 (from the last row) under the -2. Indeed, if we keep the first two rows and add the second one to the last one multiplied by 2, we get

\begin{displaymath}\left(\begin{array}{rrrr}
1&2&1&-1\\
0&0&-2&-6\\
0&0&0&0\\
\end{array}\right).\end{displaymath}

We can't do more. Indeed, we stop the process whenever we have a matrix which satisfies the following conditions
1.
any row consisting of zeros is below any row that contains at least one nonzero number;
2.
the first (from left to right) nonzero entry of any row is to the left of the first nonzero entry of any lower row.

Now if we make sure that the first nonzero entry of every row is 1, we get a matrix in row echelon form. For example, the matrix above is not in echelon form. But if we divide the second row by -2, we get

\begin{displaymath}\left(\begin{array}{rrrr}
1&2&1&-1\\
0&0&1&3\\
0&0&0&0\\
\end{array}\right).\end{displaymath}

This matrix is in echelon form.

An application of this, namely to solve linear systems via Gaussian elimination may be found on another page.

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Author: M.A. Khamsi

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