Section 6.1 – Matrix Solutions to Linear Systems
If a matrix has M rows and N columns, then it is called an M × N matrix.
Matrix elements are denoted:
aij
ie, element a24 is in row 2, column 4
Solving Linear Systems Using Matrices:
Example: Write the augmented matrix for each of the following systems:
2 x  5 y  10
x  3 y  1
a) 
x  y  z  1

b)  2 x  3 z  4
 y  4 z  10

c) Write the system that corresponds to this augmented matrix:
0
1

0
2 3
4
1
0
3
6
7 
9 
A matrix is in Row-Echelon form when ones are on the main diagonal, and zeros are in the
bottom left “corner”. Systems in Row-Echelon form can be solved using back substitution.
Example: Solve the following:
1
0

0
2 4
1
2
0
1
5
3
3
1
A matrix is in Reduced Row-Echelon form when ones are on the main diagonal, and zeros are in
the bottom left and top right corners.
Example:
1
0

 0
0
1
0
0
0
1
6 
4 
13 
Performing Matrix Row Operations (Gaussian Elimination):
The following are the legal operations:
Example:
1. Swap two rows
2. Multiply a row by a non-zero constant
3. Add a multiple of one row to another row.
Perform the following operations on:
1
5

0
2
4
6
4
1
1
3
0 
1 
a) –5R1 + R2
b) 1/14R2
c) 6R2 + R3
2
In Gaussian Elimination, your goal is to get one of the following matrices:
1 0 0 0 # 
1 0 0 # 
0 1 0 0 # 
1 0 # 
0 1 0 # 


0 1 # 


0 0 1 0 # 


0 0 1 # 


0 0 0 1 # 
Create zeros and ones in the following order:
1
2

4
3
#
# 
1
2

3
5
4
6
9
8
7
#
# 
# 
1
2

3

4
6
10 16
5
7
8
11 15
9 14
12 13
#
# 
#

#
* You create a ONE by multiplying your row by the reciprocal.
* You create a ZERO by multiplying the pivot row by the opposite and adding to your row.
Example: Solve the following system using Gaussian Elimination:
2 x  4 y  8

x  3 y  9
3
Example: Solve the following system using Gaussian Elimination:
 x 7 y  23

3x  2 y  23
Example: Solve the following system using Gaussian Elimination:
 x  2 y  10 z  6

2 x  y  4 z  3
3x  4 z  7

4
Example: A furniture company produces 3 types of desks: children’s, office, and deluxe. Each desk
is manufactured in 3 stages: cutting, construction, and finishing. The time requirements for each
model and manufacturing stage are given in the table below. The furniture company has available a
maximum of 100 hours for cutting, 100 hours for construction, and 65 hours for finishing. If all
available time must be used, how many of each type of desk should be produced each week?
Children’s
Finishing
1 hr
Construction
2 hr
Cutting
2 hr
Office
1 hr
1 hr
3 hr
Deluxe
2 hr
3 hr
2 hr
TOTAL
5
Section 6.2 – Inconsistent and Dependent Systems and Their Applications
Non-Unique solutions:
If you end up with a matrix that looks like:
5 
1 0 0
0 1 0
 1

0 0 0 12 
If you end up with a matrix that looks like:
5
1 0 0
0 1 0
2 

0 0 0 0 
then there is no solution.
then there are infinitely many solutions.
(your solutions will be equations)
A system of equations with no solution is called inconsistent.
A system of equations with at least one solution is called consistent.
Example: Use your calculator to put the matrix in reduced-row echelon form then state the
solution(s) if any.
 x  2 y  z  5

2 x  3 y  z  0
3x  4 y  z  1

Example: Use your calculator to put the matrix in reduced-row echelon form then state the
solution(s) if any.
3x  4 y  z  6

2 x  y  z  1
4 x  7 y  3z  13

6