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# Row Operations and Row Echelon Forms

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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?

#### Guidance

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 form Row 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:

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

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:

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

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 mathematicians.

Example A

Put the following matrix into reduced row echelon form.

$\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}$

Solution:  In each step of the solution,  only one of the three row operations will be used.  Specific shorthand will be introduced.

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

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

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

Row reducing a $2 \times 2$  matrix to become the identity matrix is an exercise that illustrates the fact that the rows were linearly independent.

Example B

Put the following matrix into reduced row echelon form.

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

Solution:

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

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

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

Again, row reducing a $3 \times 3$  matrix to become the identity matrix is just an exercise that illustrates the fact that the rows were linearly independent.

Example C

In a single $3 \times 3$  matrix, describe the general approach of Gauss-Jordan elimination.  In other words, which locations would you try to focus on first?

Solution:  One approach is to try to get a one in the A position.  Then get a zero in position B and position C by multiplying by a multiple of row 1.  Then try to get a zero in position D.

$\begin{bmatrix}A & I & G\\B & H & F\\C & D & E\end{bmatrix}$

Every matrix may have a different strategy and as long as you use the three row operations, you will be on the right track.  One thing to be very careful of is to try to avoid fractions within your matrix.  Scale the row to eliminate the fraction.

Concept Problem Revisited

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:

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

#### Vocabulary

Row operations are swapping rows, adding a multiple of one row to another or scaling a row by multiplying through by a scalar.

Row echelon form is a matrix that 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.

Reduced row echelon form is the same as row echelon form with one additional stipulation: that every other entry in a column with a leading one must be zero.

#### Guided Practice

1. Reduce the following matrix to reduced row echelon form.

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

2. Reduce the following matrix to row echelon form.

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

3. Reduce the following matrix to reduced row echelon form.

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

1.

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

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

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

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

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

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

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.

#### Practice

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.

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

5. Reduce matrix  $A$ to row echelon form.

6. Reduce matrix  $A$ to reduced row echelon form.  Are the rows of matrix  $A$ linearly independent?

Use the following matrix for 7-8.

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

7. Reduce matrix  $B$ to row echelon form.

8. Reduce matrix  $B$ to reduced row echelon form.  Are the rows of matrix  $B$ linearly independent?

Use the following matrix for 9-10.

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

9. Reduce matrix  $C$ to row echelon form.

10. Reduce matrix  $C$ to reduced row echelon form.  Are the rows of matrix  $C$ linearly independent?

Use the following matrix for 11-12.

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

11. Reduce matrix  $D$ to row echelon form.

12. Reduce matrix  $D$ to reduced row echelon form.  Are the rows of matrix  $D$ linearly independent?

Use the following matrix for 13-14.

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

13. Reduce matrix  $E$ to row echelon form.

14. Reduce matrix  $E$ to reduced row echelon form.  Are the rows of matrix  $E$ linearly independent?

Use the following matrix for 15-16.

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

15. Reduce matrix  $F$ to row echelon form.

16. Reduce matrix  $F$ to reduced row echelon form.  Are the rows of matrix  $F$ linearly independent?

### Vocabulary Language: English

Gauss-Jordan elimination

Gauss-Jordan elimination

Putting a matrix into reduced row echelon form is a result of performing Gauss-Jordan elimination.
linearly independent

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

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

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

Row operations include swapping rows, adding a multiple of one row to another, or scaling a row by multiplying through by a scalar.