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1. Type 1: Interchange rows (columns) i and j.
 
1. Type 1: Interchange rows (columns) i and j.
  
<math> r_i \leftarrow \rightarrow r_j (c_i \rightarrow c_j) </math>
+
<math> r_i <<>> r_j (c_i \rightarrow c_j) </math>
  
 
'''Example'''
 
'''Example'''

Revision as of 21:38, 14 December 2012

Echelon form of a matrix

A m X n matrix is in row echelon form if it satisfies the following properties 1, 2, and 3. Furthermore A m X n matrix is in reduced row echelon form if it satisfies all of the following properties:

1. If there are any zero rows, they must be at the bottom of the matrix.

2. The first nonzero entry from the left of a nonzero row is a 1, which is also called the leading one of that row.

3. The leading one for each nonzero row appears to the right and below any leading ones in the previous rows.

4. For a column with a leading one, the other entries in that column are zero.

A similar definition can be made for reduced column echelon form and column echelon form.

Example

The following matrices are in row echelon form because they follow properties 1, 2, and 3 but not property 4.

$ A=\left[\begin{array}{cccccc}1&4&0&3&-5&-7\\0&1&0&4&2&10\\0&0&0&1&-4&2\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}\right] $

$ B=\left[\begin{array}{ccccc}1&2&3&4&5\\0&0&1&3&5\\0&0&0&1&-4\\0&0&0&0&0\end{array}\right] $

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

The following matrices are in reduced row echelon form because they follow properties 1, 2, 3, and 4.

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

$ E=\left[\begin{array}{ccccc}1&7&0&0&3\\0&0&1&-2&6\\0&0&0&0&0\end{array}\right] $

$ F=\left[\begin{array}{cccccc}1&0&0&0&-1&2\\0&1&0&0&3&5\\0&0&0&1&-4&2\\0&0&0&0&0&0\\0&0&0&0&0&0\end{array}\right] $

The following matrices are not in reduced row echelon form.

$ G=\left[\begin{array}{cccc}1&3&0&5\\0&0&0&0\\0&0&1&-7\end{array}\right] $

$ H=\left[\begin{array}{cccc}1&0&7&-4\\0&-4&3&6\\0&0&1&2\end{array}\right] $

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

$ J=\left[\begin{array}{cccc}1&6&5&2\\0&1&4&3\\0&0&1&-3\\0&0&0&0\end{array}\right] $

Why are these matrices not in reduced row echelon form?

1. Matrix G is not in reduced row echelon form because it violates property 1. Row 2 is a zero row and it is not at the bottom of the matrix.

2. Matrix H is not in reduced row echelon form because it violates properties 2 and 4. Row 2 has the first nonzero entry as a -4 when it needs to be a 1 which violates property 2. Column 3 has a leading one and the other entries in that column are not zero which violates property 4.

3. Matrix I is not in reduced row echelon form because it violates property 3. Row 3 has a leading one directly below the leading one in row 2.

4. Matrix J is not in reduced row echelon form because it violates property 4. Columns 2 and 3 have leading ones and the other entries in that column are not zero.

Getting a Matrix into Echelon Form

Every matrix can be put into row (column)echelon form or reduced row (column) echelon form. This can be done with the use of elementary row (column) operations. The following are the different types of operations that can be performed on a matrix.

1. Type I: Interchange any two rows (columns).

2. Type II: Multiply a row (column) by a nonzero number.

3. Type III: Add a multiple of one row (column) to another.

The notation for these elementary row and column operations are as follows.

1. Type 1: Interchange rows (columns) i and j.

$ r_i <<>> r_j (c_i \rightarrow c_j) $

Example

By using the elementary row operations we can transform the following matrix A into reduced row echelon form.

$ A=\left[\begin{array}{ccc}1&1&-1\\3&4&-1\\5&6&-3\\-2&-2&2\end{array}\right] $

Adding (-2) times row 1 of A to row 2, to obtain

$ A=\left[\begin{array}{ccc}1&1&-1\\0&1&2\\5&6&-3\\-2&-2&2\end{array}\right] $

Now adding (-5) times row 1 to row 3 of A, to obtain

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

Adding 2 times row 1 to row 4 of A, to obtain

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

Add (-1) times row 2 of A to row 3, to obtain

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

By adding (-1) times row 2 to row 3, we get

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

Therefore we get the reduced row echelon form of the given matrix to be:

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

We can confirm that this is in fact in reduced row echelon form by checking the properties that were mentioned earlier under the heading Echelon Form of a Matrix.

Solving Linear Systems

Using the echelon form of a matrix now we can more efficiently find the solution fo a linear system compared to the elimination method.

Example

Take the following augmented matrix which is in row echelon form.

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

The solution can be quickly found from the correspoding equations

$ x_1 + 2x_2 = 3 $

$ x_2 + x_3 = 2 $

$ x_3 =-1 $

From this we can find

$ x_3 =-1 $

$ x_2 = 2 - x_3 = 2 + 1 = 3 $

$ x_1 = 3 - 2x_2 = 3 -6 = -3 $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman