Recall the criteria for echelon form from Lecture 3:
When a matrix is in echelon form, the leading entry of each row is called a pivot. Even though matrices can have different echelon forms, the pivots will always be in the same locations. A column that contains a pivot is called a pivot column.
Our goal in this lecture is to learn the step-by-step process for using row operations to transform a matrix from its original form to an equivalent echelon form.
The row operations are:
The process of applying a sequence of row operations to a matrix until it is in echelon form is called row reduction. The process we will follow for row reduction goes like this:
As we will see, this outline gives us a lot of flexibility with how we perform the operations, and different sequences of steps can result in different echelon forms. However, while the same matrix can correspond to different echelon forms, all echelon forms of a given matrix will have the same pivot locations!
Example 1. Follow the outline above to find an equivalent echelon form for this matrix: \begin{bmatrix} 0 & 1 & 2 & 0 & -1 \\ -2 & 2 & 0 & 3 & 2 \\ 2 & 8 & 20 & 0 & 6 \end{bmatrix}
Notice that the first column is not all zeroes, so the first column is a pivot column and the pivot is at the top of the column. Unfortunately, there is a zero in that position, so we must perform a swap. We can choose to either swap Row 1 and Row 2, or to swap Row 1 and Row 3. For this example, we will swap Row 1 and Row 2: \[ \text{Swap Row 1 and Row 2}: \begin{bmatrix} -2 & 2 & 0 & 3 & 2 \\ 0 & 1 & 2 & 0 & -1 \\ 2 & 8 & 20 & 0 & 6 \end{bmatrix} \]
Now we will choose to use the optional Step 2 to scale the first row to change the pivot into a 1. We need to multiply \( -2 \) by \( -\frac 1 2 \) to get 1, so that is the scale we will use: \[ \text{Scale Row 1 by \( -\frac 1 2 \)}: \begin{bmatrix} 1 & -1 & 0 & -\frac 3 2 & -1 \\ 0 & 1 & 2 & 0 & -1 \\ 2 & 8 & 20 & 0 & 6 \end{bmatrix} \]
Next, in Step 3, we need to use replacement operations to get zeroes below our pivot. We already have a zero in the second row, but we have a 2 in the third row. The question we ask ourselves is "what number must we multiply Row 1 by so that, when we add the result to Row 3, the number in Row 3 is changed to a zero?" In this case, our pivot being 1 makes this easy to figure out: \[ \text{Replace Row 3 by Row 3 + \( (-2) \cdot \) Row 1}: \begin{bmatrix} 1 & -1 & 0 & -\frac 3 2 & -1 \\ 0 & 1 & 2 & 0 & -1 \\ 0 & 10 & 20 & 3 & 8 \end{bmatrix} \]
We have achieved the result we want in the first column: all zeroes below our pivot. We now look for the next pivot location. It should be in Row 2 since our previous pivot was in Row 1. Is it possible to get a nonzero entry in Row 2, Column 2? Yes, because there is already a nonzero number there! This is our next pivot.
Step 2 is unnecessary since the pivot is already a 1. Moving to Step 3, we want to use a replacement operation to change the 10 in Row 3 to a zero: \[ \text{Replace Row 3 by Row 3 + \( (-10) \cdot \) Row 2}: \begin{bmatrix} 1 & -1 & 0 & -\frac 3 2 & -1 \\ 0 & 1 & 2 & 0 & -1 \\ 0 & 0 & 0 & 3 & 18 \end{bmatrix} \]
Now we have achieved our desired result in the second column. Where is our next pivot? It should be in Row 3 since the previous pivot was in Row 2. Is the next pivot in Row 3, Column 3? It is possible to get a nonzero number into that position, either by replacement or swapping operations. However, these operations would "destroy" the work we have done in the previous columns. So, in fact, the next pivot cannot be in Row 3, Column 3. Instead, the pivot is in Row 3, Column 4. Since there are no rows below Row 3, the matrix is in echelon form and we are done. \( \Box \)
Example 2. Use row operations to find an echelon form for this matrix: \begin{bmatrix} 1 & 4 & 8 \\ 2 & 8 & -1 \\ -3 & 0 & 0 \\ 1 & -4 & 16 \end{bmatrix}
We already have a nonzero entry in Row 1, Column 1, and it's already the number 1. So, we immediately start performing replacement operations to eliminate the entries below that pivot: \[ \text{Replace Row 2 by Row 2 + \( (-2) \cdot \) Row 1}: \begin{bmatrix} 1 & 4 & 8 \\ 0 & 0 & -17 \\ -3 & 0 & 0 \\ 1 & -4 & 16 \end{bmatrix} \] \[ \text{Replace Row 3 by Row 3 + \( (3) \cdot \) Row 1}: \begin{bmatrix} 1 & 4 & 8 \\ 0 & 0 & -17 \\ 0 & 12 & 24 \\ 1 & -4 & 16 \end{bmatrix} \] \[ \text{Replace Row 4 by Row 4 + \( (-1) \cdot \) Row 1}: \begin{bmatrix} 1 & 4 & 8 \\ 0 & 0 & -17 \\ 0 & 12 & 24 \\ 0 & -8 & 8 \end{bmatrix} \]
The next pivot is in Row 2, Column 2. We can swap Row 2 with our choice of Row 3 or Row 4. Remember that we don't want to swap Row 2 with Row 1, and we also don't want to use replacement with Row 1 to get a nonzero number into this pivot position. We want to maintain the zeroes in column 1 that we just worked so hard to get. \[ \text{Swap Row 2 and Row 4}: \begin{bmatrix} 1 & 4 & 8 \\ 0 & -8 & 8 \\ 0 & 12 & 24 \\ 0 & 0 & -17 \end{bmatrix} \]
Now that we have a nonzero entry in our pivot position, we use replacement to eliminate all entries below that position. We can use a scaling operation to turn the \( -8 \) into a 1, but we don't have to: \[ \text{Replace Row 3 by Row 3 + \( \frac {12} 8 \cdot \) Row 2}: \begin{bmatrix} 1 & 4 & 8 \\ 0 & -8 & 8 \\ 0 & 0 & 36 \\ 0 & 0 & -17 \end{bmatrix} \]
We have our third and final pivot in Row 3, Column 3. We could scale it to be a 1, but we can complete the row-reduction with just one more operation: \[ \text{Replace Row 4 by Row 4 + \( \frac {17} {36} \cdot \) Row 3}: \begin{bmatrix} 1 & 4 & 8 \\ 0 & -8 & 8 \\ 0 & 0 & 36 \\ 0 & 0 & 0 \end{bmatrix} \]
This matrix is in echelon form, so the row-reduction is complete. \( \Box \)
Recall that a matrix is in reduced echelon form if it satisfies these conditions:
As we will see, sometimes we need reduced echelon form to fully analyze a matrix. In this case, we need to update the row-reduction process to put the matrix in its unique reduced echelon form:
There are only two changes from the previous outline: Step 2 is no longer optional, and in Step 3 we must get zeroes above and below each pivot.
Example 3. Use row operations to find the reduced echelon form for this matrix: \[ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -3 & -6 & -9 \\ 6 & 7 & 8 & 9 \end{bmatrix} \]
As before, there are different choices we can make throughout the row-reduction process. However, the resulting reduced echelon form is unique. This means that, no matter what choices we make along the way, we should always end up with the same reduced echelon form for a given matrix. \[ \text{Replace Row 2 by Row 2 + \( (-4) \cdot \) Row 1}: \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -3 & -6 & -9 \\ 6 & 7 & 8 & 9 \end{bmatrix} \] \[ \text{Replace Row 3 by Row 3 + \( (-6) \cdot \) Row 1}: \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -3 & -6 & -9 \\ 0 & -5 & -10 & -15 \end{bmatrix} \] \[ \text{Scale Row 2 by \( -\frac 1 3 \)}: \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & -5 & -10 & -15 \end{bmatrix} \]
Remember that we now have to get zeroes above and below our pivot in Column 2: \[ \text{Replace Row 1 by Row 1 + \( (-2) \cdot \) Row 2}: \begin{bmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & -5 & -10 & -15 \end{bmatrix} \] \[ \text{Replace Row 3 by Row 3 + \( (5) \cdot \) Row 2}: \begin{bmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
The matrix is now in reduced echelon form, and the process is complete. \( \Box \)
There are many electronic tools that can be used to row-reduce matrices automatically. Here are a few resources that you may find useful.
Our focus in this lecture was on the step-by-step process that we use to obtain an echelon form or the reduced echelon form for a given matrix. In the next lecture, we'll examine why we want the echelon form (or reduced echelon form) and what this tells us about the underlying system of linear equations.
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