Using Matrices to Solve Systems of Equations
This tutorial: Part C: More Gauss-Jordan; Inconsistent and Underdetermined Systems
- Online Pivot & Gauss-Jordan Utility
- Excel On-Line Pivot & Gauss-Jordan Utility (downloadable Excel workbook)
- Online Matrix Algebra Tool
- Pivot Program for the TI-82 and TI-83
Good! You now know how to pivot and reduce a matrix in order to solve many systems of linear equations. However, the preceding tutorial never made it absolutely clear exactly what one means by "reduced". This is our first ask.
Recall that the leading entry of a row is the first non-zero entry in that row.
Row-Reduced Echelon Form
A matrix is in row-reduced echelon form (or reduced for short) if:
Example for you
Decide whether each of the following matrices is row-reduced or not.
Q How does one reduce a matrix to row-echelon form?
A Recall the procedure you used in Part B to solve systems of linear equations. (If you don't, go back to that part of the tutorial by pressing "Previous Topic".) You used pivoting to clear the columns containing the leading entries, and then, as a final step, you converted the leading entries into 1s. Thus, what you were left with, apart from a possible rearrangement of the rows, was a row-reduced echelon matrix! Thus, at this stage you already know how to reduce any matrix to row-reduced echelon form!
Q Wait a second! What is this about "rearranging" the rows? How can that be done with row operations?
A One can arrange the rows of a matrix in any order by repeatedly swapping pairs of rows (swapping two rows is one of the possible row operations listed in Part A of this tutorial). This may be needed to accomplish property (3) of row-reduced echelon form.
Each of the following matrices is one step away from being row-reduced. Complete the row reduction and show the result.
How to Swap Two Rows: To swap, say, Rows 1 and 2, type swap(R1,R2) or swap(R2,R1) next to any one row, and leave the others blank.
Q So what is so interesting about row-reduced echelon form?
A Recall at the very beginning of Part B of this tutorial that we used the row-reduced form of a matrix to read off the solution by simply converting it back into equations; for instance, the reduced matrix
Q What about those strange-looking row-reduced matrices that do not have 1s running down the diagonal, like
- Row 1: 0x + 1y + 0z + 0u = 4; that is, y = 4
Row 2: 0x + 0y + 0z + 1u = -5; that is, u = -5
- (x, y, z, u) = (100, 4, -600, -5)
y = 4
u = −5
Q OK then, how about
- Row 1: y + 3z = 4 or, solving for y, y = 4 - 3z
Row 2: u = -5
|General Solution||A Particular Solution|
How to Obtain the General Solution of a System of Linear Equations
1. Reduce the associated augmented matrix to row-reduces echelon form.
Examples for you:
The reduced form of the augmented matrix of a certain system of linear equations is
The general solution is:
The general solution is: A particular solution of the system represented above is:
Spotting an Inconsistent System.
Up to this point we never mentioned what happens if the leading entry of some row in your row-reduced augmented matrix is in the last (answer) column, as in, say
The moral of the story: If, during the process of row-reduction, you get any row of the form [0 0 0 0 ... 0 k] where k is nonzero, then you can stop working because the original system is inconsistent.
You now have several options:
- Try some of the questions in the true/false quiz (warning: it covers the whole of chapter 2) by going to "Everything for Finite Math"
- Try some of the questions in the Review Exercises by pressing the link on the side
- Try some of the exercises in Section 2.2 of or .
Copyright © 2009 Stefan Waner