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Linear Algebra
Section 2.2 : The idea of elimination
Monday, February 4th
Math 301
Week #3
Tuesday, February 5, 13
How did we solve 2x2 systems?
eliminate
x
3y = 7
(equation 1)
2x
7y = 3
(equation 2)
order is
important!
Apply an elementary row operation to reduce the system
to a one that is easier to solve.
x
3y
y
=
=
7
11
(equation 1)
(equation 2)
(2)(equation 1)
Elementary row
operation
Resulting system is an upper triangular system which we
can solve by back substitution.
Tuesday, February 5, 13
Solve by back substitution
x
3y = 7
y=
11
Start with the last equation, and work backwards
1
1(
Step 1 :
y=
11) = 11
Step 2 :
x = 7 + 3y = 7 + (3)(11) = 40
At each step, we have exactly what we need to know
to find the next variable.
Tuesday, February 5, 13
Now let’s try a 3x3 system
2x + 4y
2z = 2
4x + 9y
2x
3z = 8
3y + 7z = 10
(equation 1)
order is
important!
(equation 2)
(equation 3)
Apply two elementary row operations
2x + 4y
eliminate
2z = 2
(equation 1)
y+z =4
(equation 2)
(2)(equation 1)
y + 5z = 12
(equation 3)
( 1)(equation 1)
Apply one more elementary row operation
2x + 4y
eliminate
Tuesday, February 5, 13
2z = 2
(equation 1)
y+z =4
(equation 2)
4z = 8
(equation 3)
Result is again an
upper triangular
system
(1)(equation 2)
Solve by back substitution
2x + 4y
2z = 2
y+z =4
4z = 8
Start at the bottom and work backwards
Step 1 :
z=
Step 2 :
y=4
Step 3 :
1
2
x=
Each step of back substitution gives
us one more variable, and we can
proceed.
Tuesday, February 5, 13
1
4 (8)
=2
z=4
(2
(4y
2=2
2z)) = . . . =
substitute in known
values for y and z
1
Pivots and multipliers
2x + 4y
2
4x + 9y
2x
2z = 2
(equation 1)
3z = 8
(equation 2)
3y + 7z = 10
(equation 3)
First “pivot”
2x + 4y
Second pivot
1y + z = 4
y + 5z = 12
2x + 4y
Third pivot
Tuesday, February 5, 13
2z = 2
“Pivots” are the leading coefficient
in the elimination row
“Multipliers” are the
coefficient to be
eliminated divided by
the “pivot”
(equation 1)
(equation 2)
(2)(equation 1)
(equation 3)
( 1)(equation 1)
2z = 2
(equation 1)
y+z =4
(equation 2)
4z = 8
4
(equation 3)
“Multipliers”
(1)(equation 2)
Try a 2x2 system
2x + 3y = 1
Pivot
5x
2y = 11
Term to be eliminated
What is the first multiplier?
`21
5
=
2
Coefficient to be
eliminated divided
by the pivot in the
elimination row
What is the new triangular system?
2x + 3y = 1
11
27
y=
2
2
Pivots
(equation 1)
(equation 2)
Use back substitution :
Tuesday, February 5, 13
x=
✓
35
11
5
2
◆
(equation 1)
and y =
27
11
Zero pivots
Elimination can fail if we encounter a zero pivot :
2x + 5y + z = 0
4x + 10y + z = 2
Eliminate
y
Multiplier is `21 = 2.
z=3
2x + 5y + z = 0
zero pivot
Tuesday, February 5, 13
0y
z=2
1y
z=3
(equation 2)
(2)(equation 1)
Next multiplier is 1/0 - not allowed!
Zero pivots - Row exchanges
2x + 5y + z = 0
0y
z=2
1y
z=3
Exchange rows 2 and 3
Check to see if there are any non-zero pivots in one of
the later equations. If so, exchange rows.
2x + 5y + z = 0
We have three
non-zero pivots.
1y
z=3
1z = 2
A full set of non-zero pivots
means the system has exactly
one solution and can be
solved by back substitution.
The failure of elimination was only a temporary failure.
Tuesday, February 5, 13
Zero pivots - no solutions
x
3y = 3
2x
6y = 7
Multiplier is `21 = 2.
Carry out the elementary row operation
x
3y = 3
0y = 1
zero pivot
(equation 2)
(2)(equation 1)
In this case, no row exchanges are possible
and elimination breaks down permanently.
Since the equation 0y = 1 has no solution, the system
has no solution.
Tuesday, February 5, 13
Zero pivots - infinite solutions
x
3y = 3
2x
6y = 6
Multiplier is `21 = 2.
Carry out the elementary row operation
x
3y = 3
0y = 0
zero pivot
(equation 2)
(2)(equation 1)
Again, no row exchanges are possible and
elimination breaks down permanently.
The equation 0y = 0 has an infinite number of solutions
and so the system has an infinite number of solutions.
Tuesday, February 5, 13
Zero pivots
A zero pivot can mean one of three things :
A row exchange is necessary,
• The system has no solution, or
• The system has an infinite number of solutions
•
If a system has no solutions or an
infinite number of solutions, we say the
system is singular.
Be careful! A singular
system does not mean the
system has a “single”
solution!
If a system has exactly one solution, we say it is nonsingular or invertible.
Tuesday, February 5, 13
Analyze the system
For what numbers a does elimination break down (1) temporarily (2) permanently?
ax + 3y =
3
4x + 6y = 6
If a = 0, we can do a row exchange and solve the system and the breakdown is
temporary.
(equation 2)
4x + 6y = 6
3y =
3
(equation 1)
Solve by back substitution to get x = 3 and y =
1.
If a 6= 0, we can apply a row operation using the multiplier `21 = a4 to eliminate
4x and get
ax + 3y = 3
✓
✓ ◆ ◆
4
4
6
(3) y = 6
( 3)
(equation 2) `21 (equation 1)
a
a
If a = 2, the second equation becomes 0y = 12. We have a zero pivot, and the
breakdown is permanent. We cannot solve 0y = 12 and so the system has no
solution.
Tuesday, February 5, 13