1.1 Systems of Linear Equations
Recall: a linear equation in x and y has
the form y = mx +b or ax + by = c. The
standard form is most useful here.
Recall: a system of linear equations in
two variables is a collection of equations:
ax  by  c
dx  cy  e
In reality, most situations depend on
several variables, so we write:
Definition: A linear equation in
x1 , x2 , x3 ...xn has the form
a1 x1  a2 x2  a3 x3      an xn  b
where the ai and b are constants.
Note: if an equation can be rearranged
algebraically to look like this, it is linear.


Example: x2  2 6  x1  x3 is linear
since it can be written
2 x1  x2  x3  2 6
Definition: A solution to a system of
equations in n variables is a list
x1  s1
x2  s 2
x3  s3

xn  s n
Or an ordered n-tupple
s1, s2 , s3 ,..., sn 
To solve a system of two linear
equations, you can use graphing,
substitution, or elimination.
Example:
x1  x2  7
2 x1  x2  8
We can solve graphically on the TI-84.
There are 3 possible outcomes when
solving a system of equations:
Higher Dimensions: Equations in R3 can
describe lines or planes.
►The solution set for a linear system is
the set of all possible solutions of the
linear system.
►Two linear systems with the same
solution set are called equivalent
systems.
Strategy for Solving a System
Replace one system with an equivalent
system that is easier to solve.
x1  2 x2  1
 x1  3x2  3
Add equation 1 to equation 2 to get
x1  2 x2  1
x2  2
Add 2 times equation 2 to equation 1:
x1  3
x2  2
Note: In two dimensions, each equation
represents a line.
In higher dimensions with more
variables, we introduce a way of
streamlining the the process.
Definition: A matrix is an array of
numbers.
An m x n matrix has m rows and
n columns.
We can use these to solve systems that
would be difficult or impossible to graph:
Example:
x1  2 x2
4
x2  x3  0
x1  3x2  2 x3  5
The coefficients of the xi can be put into
a matrix called the coefficient matrix:
0
1 2
0 1  1 


1 3  2
This is a 3 x 3 matrix where each entry is
a coefficient of a variable in the original
system.
Entries are identified by their row and
column number. For example, 2 is in the
first row of the second column, so it
called the 1, 2-entry.
►We can represent the whole system
with an augmented matrix.
1 2 0 4 
0 1  1 0 


1 3  2 5
Our goal is to create an augmented
matrix like the following:
1 0 0 r 
0 1 0 s 


0 0 1 t 
This represents the system
x1  r
x2  s
x3  t
and shows the solution r , s, t  .
We can get the first matrix into the form
that shows the solution to the system by
using row operations.
Elementary Row Operationss
1. (Replacement) Replace one row by
the sum of itself and another row (or a
multiple of another row).
2. (Interchange) interchange two rows.
3. (Scaling) multiply all entries in a row
by a non-zero constant.
Starting with the augmented matrix of the
system above:
1 2 0 4 
0 1  1 0 


1 3  2 5
Keep 1 in R1 and use it to eliminate the
1 from R3: R3+(-R1) → R3
1 2 0 4 
0 1  1 0


0 1  2 1
We say column 1 is cleared.
Note: The text shows the systems beside
the matrices, but you need only show the
matrices.
Keep the 1 in the 2,2-entry, and use it to
clear the 1 below it.
R3+(-R2) → R3
1 2 0 4 
0 1  1 0


0 0  1 1
Now R3 is almost perfect. Scale it by -1:
1 2 0 4 
0 1  1 0


0 0 1  1
►The matrix is now in triangular form, all
zeroes below the main diagonal.
It is usually more efficient to clear C3
before C2.
1 2 0 4 
0 1 0  1


0 0 1  1
R1+(-2R2) → R1
1 0 0 6 
0 1 0  1


0 0 1  1
This represents the system
x1  6
x2   1
x3  1
The unique solution is a point in R3, the
ordered triple (6, -1, -1).
Check:
►We say two matrices are row
equivalent if there is a series of
elementary row operations that
transforms one matrix into the other.
►Note that each operation is reversible,
so reversing the operations turns the
second matrix into the first.
Existence and Uniqueness
A basic question we ask in mathematics
is “When does an object exist; and if it
exists, is it unique?”
►If a solution exists (one or infinitely
many), the system is called consistent.
If no solution exists, the system is
inconsistent.
You do not need to solve a system to
figure this out. You can look at the
“triangular” form of the matrix.
Example:
1  5 2 7 
0

1
12

1


0
0 0  1 
R3 says that 0 = –1. Since this is false,
there is no solution to the system
represented by the matrix, and the
system is inconsistent.
Example: #20 Determine the value(s) of
h such that the augmented matrix is the
matrix of a consistent system.
h 3 
1
 2 4

6

Start by finding the “triangular” form:
R2+2R1→R2
h
3
1
0 2 h  4

0

If the system is consistent, then
2h  4x2  0
This is true if 2h  4  0 or if x2  0 .
Thus, the system is consistent for all
values of h.
Example:
 2 x4  3
x1
2 x2  x3
0
x3  3 x4  1
 2 x1  3 x2  2 x3  x4  5
has the solution or solution set
1
 7
  , 0, 0, 
3  since
 3
7
3
1
  3
3
00
0
1 
0  3   1
3 
 7 
1 
 2    0  0     5
 3 
 3 4