Transcript SOLVING SYSTEMS OF LINEAR EQUATIONS

SOLVING SYSTEMS OF
LINEAR EQUATIONS
• An equation is said to
be linear if every
variable has degree
equal to one (or zero)
• 4x  5 y  2z  8
is a linear equation
1
• 2x  3 y 
is NOT a linear
equation
3
z 8
Review these familiar techniques for solving 2
equations in 2 variables. The same techniques will be
extended to accommodate larger systems.
Times
3
2x  y  5
x  3 y  1
6 x  3 y  15
Add
x  3 y  1
x2
7 x  14
x  3 y  1
Substitute to solve: y=1
L1 represents line one
L2 represents line two
2x  y  5
L1 is replaced by 3L1
6 x  3 y  15
x  3 y  1
x  3 y  1
L1 is replaced by
L1 + L2
7 x  14
x  3 y  1
L1 is replaced by (1/7)L1
x2
x  3 y  1
2x  y  5
x  3 y  1
6 x  3 y  15
x  3 y  1
These systems are said to be EQUIVALENT
because they have the SAME SOLUTION.
x  3 y  1
x2
7 x  14
x  3 y  1
PERFORM ANY OF THESE OPERATIONS ON A
SYSTEM OF LINEAR EQUATIONS TO
PRODUCE AN EQUIVALENT SYSTEM:
• INTERCHANGE two equations (or lines)
• REPLACE Ln with k Ln , k is NOT ZERO
• REPLACE Ln with Ln + cLm
• note: Ln is always part of what replaces it.
EXAMPLES:
L1
L2
L3
is equivalent to
L1
L2
L3
L1
L3
L2
is equivalent to
L1
L2
L3
L1
4L2
L3
is equivalent to
L1
L2
L3 + 2L1
2 x  4 y  8 z  0 Replace L with (1/2) L x  2 y  4 z  0
 x  3y  4z  3
 x  3 y  4z  3
x  3 y  5 z  2 Replace L with L + L
z 1
1
1
3
3
2
x  2 y  4z  0
 x  3 y  4 z  3 Replace L with L + L
2
2
1
x  2 y  4z  0
3
y
z 1
z 1
x  2 y  4 z  0 Replace L with
L + 2L
y
3
z
1
1
1
2
x
+4z
y
z
6
3
1
Replace L1 with
L1 + - 4 L3
x
 4z  6
y
3
z
1
2 x  4 y  8z  0
 x  3 y  4z  3
x  3 y  5 z  2
is EQUIVALENT to
x
y
z
2
3
1
To solve the following system, we look for an equivalent system
whose solution is more obvious. In the process, we manipulate
only the numerical coefficients, and it is not necessary to rewrite
variable symbols and equal signs:
3x
 1x
 3 y  2 z  2
 1y
 1z
0
1x  2 y
 1z
 1
3x
 1x
 3 y  2 z  2
 1y
 1z
0
1x  2 y
 1z
 1
This rectangular arrangement of numbers is called a MATRIX
3 3  2 2
1 1 1
0
1  2  1 1
3 3  2 2
2 1 1 1
0
1  2  1 1
Replace L1 with L1 + 2 L2
1 1 0  2
1  1 1
0
1 2
1 1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
1 1  1 1
0
1 2
1 1
Replace L3 with L3 + L2
1 1
0 2
1  1 1
0 1
0
0
1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
1  1 1
0
1 2
1 1
Replace L2 with L2 + 1L1
1 1 0  2
0 0 1  2
0 1 0 1
1 1
0 21
1  1 1
0
0 1
0
1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
0 0 1  2
0 1 0 1
1 1 0  2
1  1 1
0
1 2
1 1
1 1
0 2
1  1 1
0 1
0
0
1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
0
0 1  2
0 1 0 1
1 1 0  2
1  1 1
0
1 2
1 1
1 1
0 2
1  1 1
0 1
0
0
1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
0 1 0 1
0
0 1  2
1 1 0  2
0
0 1  2
0 1 0 1
1 1 0  2
1  1 1
0
1 2
1 1
Interchange L2 and L3
1 1
0 2
1  1 1
0 1
0
0
1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
1  1 1
0
1 2
1 1
1 1 0  2
0 1 0 1
0
0 1  2
1 1 0  2
0
0 1  2
0 1 0 1
1 1
0 2
1  1 1
0 1
0
0
1
3 3  2 2
1 1 1
0
1  2  1 1
1 1 0  2
1  1 1
0
1 2
1 1
Replace L1 with L1 + -1 L2
1 1 0  2
0 1 0 1
0
0 1  2
1
0
0
1
0
1
0
1
0
0
1
2
Replace L2 with -1 L2
 2 Replace L3 with -1 L3
1 1 0
0
0 1  2
0 1 0 1
1 1
0 2
1  1 1
0 1
0
0
1
3 3  2 2
1 1 1
0
1  2  1 1
The original matrix represents
a system that is equivalent to
this final matrix
whose solution
is obvious
1
0
0
1
0
1
0
1
0
0
1
2
3 3  2 2
1 1 1
0
1  2  1 1
The original matrix represents
a system that is equivalent to
this final matrix
whose solution
is obvious
0
 1
1x
0
0
1y
0
1
0
0
1z
2
0
 1
1x
0
0
1y
0
1
0
0
1z
2
Note the format of the matrix that yields this obvious solution:
The diagonal of ones
The zeros
Whenever possible, aim for this format.
1
0
0
1
0
1
0
1
0
0
1
2