Solutions of Linear Systems - Gauss Jordan Method
Linear Algebra Project
Objectives:
To investigate solutions of systems of linear equations by
(a) Matrix Method
(b) Step-by-step Gauss-Jordan Elimination
(c) Gauss-Jordan Elimination in one-step
These steps will be carried out using Maple commands.
(d) To discuss the solution set of a system by graphing whenever possible.
Solved Example 1:
Solve: 4x + 3y = 2
7x + 5y = 3
Solution:
(a) Matrix Method
First activate the linalg and plot packages of Maple.
| > | with(linalg): with(plots): |
Warning, new definition for norm
Warning, new definition for trace
Matrix Method: Define the coefficient matrix: A:= matrix(2,2,[4,3,7,5]).
Define the constant matrix (or vector): c:= vector([2, 3]);
| > | A:= matrix(2,2,[4,3,7,5]); |
| > | c:= vector([2,3]); |
Use the command "linsolve" to solve the system.
| > | linsolve(A,c); |
(b) Gaussian Elimination - Step by Step:
First define the augmented matrix.
| > | Ac:= augment(A,c); |
Now we use the three elementary row operations to get an echelon
| > | A1:= mulrow(Ac, 1, 1/4); |
| > | A2:= addrow(A1,1,2,-7); |
| > | A3:= addrow(A2,2,1,3); |
| > | A4:=mulrow(A3,2,-4); |
The solution is x = -1, y = 2.
(c) Gaussian Elimination - in one step.
| > | Acr:=rref(Ac); |
This operation reduces the matrix to r ow r educed e chelon f orm (rref) and basically solves the problem.
rref is also called "reduced normal form."
(d) Graph of the system
We use the command "implicitplot".
| > | implicitplot({4*x + 3*y = 2,7*x + 5*y = 3}, x =-1.5..-.5, y = 1..3); |
Solved Example 2:
Solve the linear system
x + 2y + 3z = 0
x+ 3y + 8z = 2
x + 2y + 2z = 0
by Gauss-Jordan method.
Solution:
First define the augmented matrix then use RREF command.
| > | A:= matrix(3,3,[1,2,3,1,3,8,1,2,2]); |
| > | d:= vector([0, 2, 0]); |
| > | Ad:= augment(A,d); |
| > | Adr:= rref(Ad); |
Solution: x = -4, y = 2, z = 0.
Graph of the system
Geometrically, each linear equation in three variables is a plane. To find the solution of the system is to find the point where the three planes intersect each other.
We use the "implicitplot3d" command to graph an equation in three variables.
| > | implicitplot3d({x + 2*y + 3*z = 0,x+ 3*y + 8*z = 2,x + 2*y + 2*z = 0},x=-5..-3,y=1..3, z= -1..1, title=`Graph of the System`); |
We can get some idea about the location where the three planes meet by rotating the box with y-axis backward and forward, x-axis left and right, z-axis up and down, and noting that the planes intersect on the lines through x = -4, y = 2 and z = 0.
We zoom repeatedly to get the exact solution. We can find the exact location by rotating the axes.
| > | implicitplot3d({x + 2*y + 3*z = 0,x+ 3*y + 8*z = 2,x + 2*y + 2*z = 0},x=-4.0001..-3.9999 ,y=1.9999..2.0001, z= -0.0001..0.0001, title=`Graph of the System`); |
Solved Example 3:
Find all solutions of the homogeneous system of equations:
x1 + 2x2 + 2x4 + 3x5 = 0
x3 + 3x4 + 2x5 = 0
x3 + 4x4 - x5 = 0
x5 = 0
Solution:
The number of variables is more than the number of equations. This has infinitely many solutions.
| > | A:= matrix(4, 5,[1,2,0,2,3,0,0,1,3,2,0,0,1,4,-1,0,0,0,0,1]); |
| > | c:= vector([0,0,0,0]); |
| > | Ac := augment(A,c); |
| > | Acr:= rref(Ac); |
x1 + 2x2 = 0 or x1 = -2x2
x3 = 0
x4 = 0
x5 = 0
Solution: We choose x2 = lambda (
), the greek letter, as a variable. Then the solution is
(-2
, 0, 0, 0)
Since
can have any value the solutions are infinite.
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ASSIGNMENT
In Problems 1 and 2 solve the system using
(a) 'linsolve' command.
(b) Step-by-step Gauss-Jordan Elimination.
(c) Gauss-Jordan Elimination in one-step.
(d) Graphing .
Problem 1:
-4x + 3y = -15
3x + 5y = 33
Problem 2:
2x + z = 1
-x + y - 3z = 4
2x + y - 4 z = 0
Solve Problems 3 and 4 by one-step elimination.
Problem 3:
- 2b + 3c = 1
3a + 6b - 3c = -2
6a + 6b + 3c = 5
Problem 4:
2w - 3x + 4y - z = 0
7w + x - 8y + 9z = 0
2w + 8x + y - z = 0
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MSEIP Grant #P120AA010031: "Four Colleges: Calculus + Enhancements", 2001-2004