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# Row Operations and Row Echelon Forms

## Swapping rows, adding a multiple of one row to another, or scaling a row to get leading ones.

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Row Operations and Row Echelon Forms

Applying row operations to reduce a matrix is a procedural skill that takes lots of writing, rewriting and careful arithmetic.  The payoff for being able to transform a matrix into a simplified form will become clear later.  For now, what does the simplified form mean for a matrix?

### Row Operations and Row Echelon Forms

There are only three operations that are permitted to act on matrices.  They are the exact same operations that are permitted when solving a system of equations.

1. Add a multiple of one row to another row.
2. Scale a row by multiplying through by a non-zero constant.
3. Swap two rows.

Using these three operations, your job is to simplify matrices into row echelon formRow echelon form must meet three requirements.

1. The leading coefficient of each row must be a one.

2. All entries in a column below a leading one must be zero.

3. All rows that just contain zeros are at the bottom of the matrix.

Here are some examples of matrices in row echelon form:

\begin{align*}\begin{bmatrix} 1 & 14\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 4 \end{bmatrix}, \begin{bmatrix} 1 & 2 & 3 & 5 & 6\\ 0 & 0 & 1 & 4 & 7\\ 0 & 0 & 0 & 1 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\end{align*}

Reduced row echelon form also has one extra stipulation compared with row echelon form.

4. Every leading coefficient of 1 must be the only non-zero element in that column.

Here are some examples of matrices in reduced row echelon form:

\begin{align*}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 3\\ 0 & 1 & 4 \end{bmatrix}, \begin{bmatrix} 1 & 2 & 0 & 0 & 6\\ 0 & 0 & 1 & 0 & 7\\ 0 & 0 & 0 & 1 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\end{align*}

Putting a matrix into reduced row echelon form is a result of performing Gauss-Jordan elimination.  The process illustrated in this concept is named after those two mathematicians.

To put the a matrix into reduced row echelon form, use the row operations to change the matrix. Take the following matrix:

\begin{align*}\begin{bmatrix} 3 & 7\\ 2 & 5 \end{bmatrix}\end{align*}

In each step of reducing the matrix, only one of the three row operations will be used.  Specific shorthand will be introduced.

\begin{align*}\begin{bmatrix} 3 & 7\\ 2 & 5 \end{bmatrix} \begin{matrix} & \rightarrow & \\ \rightarrow &\cdot 3 & \rightarrow \end{matrix} \begin{bmatrix} 3 & 7\\ 6 & 15 \end{bmatrix} \begin{matrix} & \rightarrow & \\ \rightarrow & -2 \cdot I & \rightarrow \end{matrix} \begin{bmatrix} 3 & 7\\ 0 & 1 \end{bmatrix}\end{align*}

Note that the \begin{align*}\cdot 3\end{align*} in between the first two matrices indicates that the second row is scaled by a factor of 3.  The \begin{align*}-2 \cdot I\end{align*} between the next two matrices indicates that the second row has two times the first row subtracted from it.  The \begin{align*}I\end{align*} is a roman numeral referring to the row number.

\begin{align*}\begin{bmatrix} 3 & 7\\ 0 & 1 \end{bmatrix} \begin{matrix} \rightarrow & -7II & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 3 & 0\\ 0 & 1 \end{bmatrix} \begin{matrix} \rightarrow & \cdot \frac{1}{3} & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{align*}

Row reducing a \begin{align*}2 \times 2\end{align*} matrix to become the identity matrix illustrates the fact that the rows of the original matrix are linearly independent.

### Examples

#### Example 1

Earlier, you were asked what it means for a matrix to be simplified. There are two forms of a matrix that are most simplified.  The most important is reduced row echelon form that follows the four stipulations from the guidance section.  An example of a matrix in reduced row echelon form is:

\begin{align*}\begin{bmatrix} 1 & 0 & 0 & 2 & 43\\ 0 & 1 & 0 & 2 & 3\\ 0 & 0 & 1 & 98 & 5 \end{bmatrix}\end{align*}

#### Example 2

Put the following matrix into reduced row echelon form.

\begin{align*}\begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 1 & 2 & 4 \end{bmatrix}\end{align*}

\begin{align*}\begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 1 & 2 & 4 \end{bmatrix} \begin{matrix} & \rightarrow &\\ & \rightarrow &\\ \rightarrow & -\frac{I}{2} & \rightarrow \end{matrix} \begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 0 & 0 & 4 \end{bmatrix} \begin{matrix} & \rightarrow & \\ & \rightarrow & \\ \rightarrow & \div 4 & \rightarrow \end{matrix} \begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 0 & 0 & 1 \end{bmatrix} \begin{matrix} \rightarrow & \div 2 & \rightarrow\\ \rightarrow & \div 3 & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & \frac{1}{3}\\ 0 & 0 & 1 \end{bmatrix}\end{align*}

Note that in the preceding step, two operations were used.  This is acceptable when the operations do not interfere or interact with each other.

\begin{align*}\begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & \frac{1}{3}\\ 0 & 0 & 1 \end{bmatrix} \begin{matrix} & \rightarrow &\\ \rightarrow & -\frac{III}{3} & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{matrix} \rightarrow & -2II & \rightarrow\\ & \rightarrow &\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\end{align*}

Again, row reducing a \begin{align*}3 \times 3\end{align*} matrix to become the identity matrix is just an exercise that illustrates the fact that the rows were linearly independent.

#### Example 3

Reduce the following matrix to reduced row echelon form.

\begin{align*}\begin{bmatrix} 0 & 4 & 5\\ 2 & 6 & 8 \end{bmatrix}\end{align*}

\begin{align*}\begin{bmatrix} 0 & 4 & 5\\ 2 & 6 & 8 \end{bmatrix} \begin{matrix} \rightarrow & II & \rightarrow\\ \rightarrow & I & \rightarrow \end{matrix} \begin{bmatrix} 2 & 6 & 8\\ 0 & 4 & 5 \end{bmatrix} \begin{matrix}\rightarrow & \div 2 & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 3 & 4\\ 0 & 4 & 5 \end{bmatrix}\end{align*}

\begin{align*} \begin{matrix} & \rightarrow &\\ \rightarrow & -I & \rightarrow \end{matrix} \begin{bmatrix} 1 & 3 & 4\\ 0 & 1 & 1 \end{bmatrix} \begin{matrix} \rightarrow & -3II & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix}\end{align*}

#### Example 4

Reduce the following matrix to row echelon form.

\begin{align*}\begin{bmatrix} 3 & 6\\ 2 & 4\\ 5 & 17 \end{bmatrix}\end{align*}

\begin{align*}\begin{bmatrix} 3 & 6\\ 2 & 4\\ 5 & 17 \end{bmatrix} \begin{matrix} \rightarrow & \div 3 & \rightarrow\\ \rightarrow & \div 2 & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 2\\ 1 & 2\\ 5 & 17 \end{bmatrix} \begin{matrix} & \rightarrow &\\ \rightarrow & -I & \rightarrow\\ \rightarrow & -5I & \rightarrow \end{matrix} \begin{bmatrix} 1 & 2\\ 0 & 0\\ 0 & 7 \end{bmatrix} \begin{matrix} & \rightarrow & \\ \rightarrow & III & \rightarrow\\ \rightarrow & II & \rightarrow \end{matrix} \begin{bmatrix} 1 & 2\\ 0 & 7\\ 0 & 0 \end{bmatrix} \begin{matrix} & \rightarrow & \\ \rightarrow & \div 7 & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 1 & 2\\ 0 & 1\\ 0 & 0 \end{bmatrix}\end{align*}

#### Example 5

Reduce the following matrix to reduced row echelon form.

\begin{align*}\begin{bmatrix} 3 & 4 & 1 & 0\\ 5 & -1 & 0 & 1 \end{bmatrix}\end{align*}

\begin{align*}\begin{bmatrix} 3 & 4 & 1 & 0\\ 5 & -1 & 0 & 1 \end{bmatrix} & \begin{matrix} \rightarrow & \cdot 5 & \rightarrow\\ \rightarrow & \cdot 3 & \rightarrow \end{matrix} \begin{bmatrix} 15 & 20 & 5 & 0\\ &15 & -3 & 0 & 3 \end{bmatrix} \begin{matrix} & \rightarrow & \\ \rightarrow & -I & \rightarrow \end{matrix} \begin{bmatrix} 15 & 20 & 5 & 0\\ 0 & -23 & -5 & 3 \end{bmatrix} \\ &\begin{matrix} \rightarrow & \cdot 23 & \rightarrow\\ \rightarrow & \cdot 20 & \rightarrow \end{matrix} \begin{bmatrix} 345 & 460 & 115 & 0\\ 0 & -460 & -115 & 60 \end{bmatrix}\end{align*}

\begin{align*}\begin{bmatrix} 345 & 460 & 115 & 0\\ 0 & -460 & -115 & 60 \end{bmatrix} \begin{matrix} \rightarrow & +II & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} 345 & 0 & 0 & 60\\ 0 & -460 & -115 & 60 \end{bmatrix} \end{align*}

\begin{align*}\begin{matrix} \rightarrow & \div 345 & \rightarrow\\ \rightarrow & \div -460 & \rightarrow \end{matrix} \begin{bmatrix} 1 & 0 & 0 & \frac{60}{345}\\ 0 & 1 & \frac{115}{460} & -\frac{60}{460} \end{bmatrix}\end{align*}

Notice how fractions were avoided until the final step.  Adding and subtracting large numbers in a matrix is easier to handle than adding and subtracting small numbers because then you don’t need to find a common denominator.

### Review

1. Give an example of a matrix in row echelon form.

2. Give an example of a matrix in reduced row echelon form.

3. What are the three row operations you are allowed to perform when reducing a matrix?

4. If a square matrix reduces to the identity matrix, what does that mean about the rows of the original matrix?

Use the following matrix for 5-6.

\begin{align*}A=\begin{bmatrix} -3 & -4 & -12\\ 4 & 4 & 12\\ -11 & -12 & -35 \end{bmatrix}\end{align*}

5. Reduce matrix \begin{align*}A\end{align*} to row echelon form.

6. Reduce matrix \begin{align*}A\end{align*} to reduced row echelon form.  Are the rows of matrix \begin{align*}A\end{align*} linearly independent?

Use the following matrix for 7-8.

\begin{align*}B=\begin{bmatrix} 3 & -4 & 8\\ 9 & 0 & 1\\ 0 & 1 & -2 \end{bmatrix}\end{align*}

7. Reduce matrix \begin{align*}B\end{align*} to row echelon form.

8. Reduce matrix \begin{align*}B\end{align*} to reduced row echelon form.  Are the rows of matrix \begin{align*}B\end{align*} linearly independent?

Use the following matrix for 9-10.

\begin{align*}C=\begin{bmatrix} 0 & 0 & -1 & -1\\ 3 & 6 & -3 & 1\\ 6 & 12 & -7 & 0 \end{bmatrix}\end{align*}

9. Reduce matrix \begin{align*}C\end{align*} to row echelon form.

10. Reduce matrix \begin{align*}C\end{align*} to reduced row echelon form.  Are the rows of matrix \begin{align*}C\end{align*} linearly independent?

Use the following matrix for 11-12.

\begin{align*}D=\begin{bmatrix} 1 & 1\\ 3 & 4\\ 2 & 3 \end{bmatrix}\end{align*}

11. Reduce matrix \begin{align*}D\end{align*} to row echelon form.

12. Reduce matrix \begin{align*}D\end{align*} to reduced row echelon form.  Are the rows of matrix \begin{align*}D\end{align*} linearly independent?

Use the following matrix for 13-14.

\begin{align*}E=\begin{bmatrix} -5 & -6 & -12\\ -1 & -1 & -2\\ 2 & 2 & 4 \end{bmatrix}\end{align*}

13. Reduce matrix \begin{align*}E\end{align*} to row echelon form.

14. Reduce matrix \begin{align*}E\end{align*} to reduced row echelon form.  Are the rows of matrix \begin{align*}E\end{align*} linearly independent?

Use the following matrix for 15-16.

\begin{align*}F=\begin{bmatrix} -23 & 6 & 3\\ 2 & -\frac{1}{2} & 0\\ -8 & 2 & 1 \end{bmatrix}\end{align*}

15. Reduce matrix \begin{align*}F\end{align*} to row echelon form.

16. Reduce matrix \begin{align*}F\end{align*} to reduced row echelon form.  Are the rows of matrix \begin{align*}F\end{align*} linearly independent?

To see the Review answers, open this PDF file and look for section 8.5.

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### Vocabulary Language: English

TermDefinition
Gauss-Jordan elimination Putting a matrix into reduced row echelon form is a result of performing Gauss-Jordan elimination.
linearly independent The rows of a matrix are linearly independent if each row cannot be produced by a linear combination of the other rows. Matrices with linearly independent rows will reduce to the identity matrix.
reduced row echelon form A matrix is in reduced row echelon form if it has a leading one at the start of every non-zero row, zeros below every leading one, all rows containing only zeros at the bottom of the matrix, and only zeros to the right of leading ones in rows with leading ones.
row echelon form A matrix is in row echelon form if it has a leading one at the start of every non-zero row, zeros below every leading one and all rows containing only zeros at the bottom of the matrix.
row operations Row operations include swapping rows, adding a multiple of one row to another, or scaling a row by multiplying through by a scalar.