 Solving Systems of Linear Equations - Maple Programming Help

Home : Support : Online Help : Math Apps : Algebra and Geometry : Matrices : MathApps/SolvingSystemsOfLinearEquations

Solving Systems of Linear Equations

Main Concept

A system of linear equations is a collection of n equations involving the same n variables, where each equation equates a linear combination of the variables to a constant. A solution to this system of equations is a set of values, one assigned to each of the variables, such that all of the equations are simultaneously satisfied.

There are several ways to determine the solution to a system of equations, and the following three methods will be explored here:

 • Substitution and elimination
 • Gaussian elimination
 • Gauss-Jordan elimination

Trying to solve a linear system of equations gives us one of the following three results:

No solution Infinitely many solutions Unique solution Consider the equations of the planes shown in the plot to the left:

Consider the following system of equations:

x + y + z = 3

2x + 3y + 7z = 0

x + 3y - 2z = 17

In the following examples, this system of equations is solved using each of the methods listed above. Use the buttons to navigate through each solution method.

Method 1: Substitution and Elimination

Substitution refers to isolating for the value of one variable in one equation and then substituting this value for that same variable in the other equations to solve for the other remaining variables. Elimination refers to adding or subtracting multiples of one equation from another in order to get rid of one of the variables. Together, they can be used to effectively solve a system of equations:   Note: While it does not require much more effort to solve this simple system using substitution and elimination versus using Gaussian elimination or Gauss-Jordan elimination, as systems acquire more terms, Gaussian elimination or Gauss-Jordan elimination become more efficient.

Method 2: Gaussian Elimination

Gaussian elimination aims to orient the matrix formed by the system of equations such that only zeros appear below the pivot points of each row:   After performing Gaussian elimination, you are left with the following system of equations:

You can then solve for z:

$z=-\frac{26}{13}$

$z=-2$

and substitute it into the second equation:

$y+5\left(-2\right)=-6$

$y=-6+10$

$y=4$

Finally, you can substitute the y and z into the first equation to solve for x:

$x+4-2=3$

$x=3-2$

$x=1$

Therefore, the solution to the equation is:

or:

$\left(1,4,-2\right)$

Method 3: Gauss-Jordan Elimination

Gauss-Jordan elimination aims to diagonalize the matrix such that 1 appears at each pivot point, and 0 appears in all the other positions of the matrix:   Unlike simple Gaussian elimination, further simplification of the matrix is not needed to determine the solution of the system of equations. Rather, it is evident that the solution to this particular system of equations is:

or:

$\left(1,4,-2\right)$



 More MathApps