Using Matrices to Solve Systems of Equations

This tutorial: Part C: More Gauss-Jordan; Inconsistent and Underdetermined Systems
Some On-line Resources for This Topic:

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:

  • The leading entry in each row is a 1.
  • The column of every leading entry is clear; that is, the other entries in the column are all zero. (Note: Columns that do not contain leading entries do not need to be clear.)
  • The rows are arranged so that the leading entries go from left to right as you go down the rows. Further, rows of zeros (if any) are at the bottom.

    1. \mat2{\[\box{ 1 } , 0 , 0 , 3 \]! 0 , \box{ 1 } , 0 , 2 ! 0 , 0 , \box{ 1 }, 9 }       2. \mat2{\[\box{ 1 } , -9 , 0 , 3 \]! 0 , 0 , \box{ 1 } , 5 ! 0 , 0 , 0 , 0 }      3. \mat2{\[\box{ 1 } , 0 , 0 , 0 \]! 0 , \box{ 1 }, 0 , 0 ! 0 , 0 , \box{ 1 } , 0 ! 0 , 0 , 0 , \box{ 1 } }      4. \mat2{\[ 0 , \box{ 1 } , 3 , 0 , 0 \]! 0 , 0 , 0 , \box{ 1 }, 0 ! 0 , 0 , 0 , 0 , \box{ 1 } ! 0 , 0 , 0 , 0 , 0 }
Example for you

Decide whether each of the following matrices is row-reduced or not.

    Not reduced
    Not reduced
    Not reduced
    Not reduced

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

representing equations in the unknowns x, y, z translates back into equations as x = 3, y = 2, z = -3. This is a unique solution; it is the only solution that satisfies the system of linear equations.

Q What about those strange-looking row-reduced matrices that do not have 1s running down the diagonal, like

A Again, just translate the rows into equations (using the names x, y, z, u for the unknowns): Because these equations say absolutely nothing about x and z, we can give them any values we like and still satisfy the equations. For instance, is a solution of the above system because the values of the unknowns satisfy the two equations above. The general solution is as this includes all the possibilities for all the unknowns. Here, the system of linear equations has infinitely many solutions.

Q OK then, how about

A Translating the rows into equations gives: These equations tell us what y and u have to be (once we have z), but not what x and z have to be, so the solution is Choosing different values for x and z gives us different particular solutions. To summarize:
How to Obtain the General Solution of a System of Linear Equations

1. Reduce the associated augmented matrix to row-reduces echelon form.
2. Translate each row into an equation, and solve for the variable that occurs first (and corresponds to the leading entry for that row).
3. All variables not solved for are arbitrary.

Examples for you: The reduced form of the augmented matrix of a certain system of linear equations is
The general solution is:
A particular solution of the system represented above is:
    x = ,   y =
    Answers must be numbers; not formulas.
Another system has unknowns x, y, z, s, t and reduced augmented matrix
The general solution is:
    x =  
    y =
    z =     Enter arbitrary for all variables that are arbitrary.
    s =
    t =
A particular solution of the system represented above is:
    x = ,   y =   z = ,   s =   t =
    Answers must be numbers; not formulas.
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 next-to-last row translates into the equation
    0x + 0y + 0z + 0s = 1, or   0 = 1
which is a false statement. If we remember that the rows of the matrix tell us what equations any solution of the system needs to satisfy, we must conclude that there can be no solution, as any solution would need to satisfy a false equation! In other words, the original system of equations is inconsistent, and has no solutions.

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:

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Last Updated: October, 2009
Copyright © 2009 Stefan Waner