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Solving Systems using Reduced Row Echelon Form
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As we saw in The Matrix and Solving Systems using Matrices section, the reduced row echelon form method can be used to solve
systems.
With this method, we put the coefficients and constants in one matrix (called an augmented matrix, or in coefficient form) and then, with
a series of row operations, change it into what we call reduced echelon form, or reduced row echelon form. This is when the leading entry in
every nonzero row is , and each leading
is the only nonzero entry for that column, such as
. Then the numbers in the last
column are the answers!. This is also called Gaussian Elimination, or Row Reduction.
To get the matrix in the correct form, we can 1) swap rows, 2) multiply rows by a non-zero constant, or 3) replace a row with the
product of another row times a constant added to the row to be replaced. Here is an example:
Solve the following system of equations:
.
1. Find the first non-zero entry in the first column; this is the pivot. Move this to the first row. This is just 5, so the augmented matrix is
still
.
2. Make the pivot (the 5) 1 by multiplying this row by
Now we have
3. Now we want the elements below the
:
.
.
(first column) to be . To get them to be , add multiples of the first (pivot) row to each of the
rows below, replacing those rows. Note that we took the opposite (negative) of the first element in that row (like
multiplier of the pivot row:
) to get the
.
Now we have
.
4. Now the second row is the pivot row. To get the second column, second row a , divide this row by
(multiply by reciprocal):
.
Now we have
.
5. Repeat the procedure from Step 3 above, using the first and third rows, to make their second column entries :
.
.
Now we have
6. Now multiply the third row by
to have the
in the third column for another pivot row:
.
Now we have
, which is in row echelon form.
7. Repeat step 3 above again to get ’s in the third column (first two rows) to get reduced row echelon form:
.
We get
Yeah! Our answer is
.
, and
!
Note if at the end we get all ’s in the last row, we have an infinite number of solutions, and if we get all ’s except for the last row, fourth
column, we have no solution. We saw these examples here in the Matrix and Solving Systems with Matrices section.
For Practice: Use the Mathway widget below to try a problem. Click on Submit (the blue arrow to the right of the problem) and click on
Find Reduced Row Echelon Form to see the answer.
You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by
topic.
If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at
any time (to get any type of math problem solved!).
On to Introduction to Linear Programming – you are ready!
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