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How to get to Reduced Row Echelon Form
Use rows with leading 1’s Get the leading 1’s starting at the pivot Use the pivot to get zeros below
and/or above Use rows with zeros on top of each
other as you move right
Interchange two rows
Multiply a row throughby a non-zero constant
Add a multiple of onerow to another.
Allowable Row Operations Notation
Ri ↔Rj
kRi → Ri
kRi + Rj → Rj
PIVOT
1 1 2 7
3 2 1 10
1 3 1 2
-3R1 + R2 → R2
R1 + R3 → R3
Use this one, called the pivot, to get zeros below.
1 1 2 7
3 2 1 10
1 3 1 2
-3R1 + R2 → R2
1 1 2 7
0 5 7 31
1 3 1 2
1 1 2 7
0 5 7 31
0 2 3 5
R1 + R3 → R3
1 1 2 7
0 5 7 31
1 3 1 2
New pivot location
1 1 2 7
0 5 7 31
0 2 3 5
-2R3 + R2 → R2
We want a 1 here but we also want to avoid fractions
New pivot
1 1 2 7
0 1 13 41
0 2 3 5
Use this new pivot to get zeros above and below
R2 + R1 → R1
-2R2 + R3 → R3
New pivot location
1 0 11 34
0 1 13 41
0 0 29 87 Now get a 1 here.
R3 → R31
29
New pivot
1 0 11 34
0 1 13 41
0 0 1 3
Use this new pivot to make zeros above.
11R3 + R1→ R1
13R3 + R2→ R2
1 0 0 1
0 1 0 2
0 0 1 3
1
2
3
1
2
3
x
x
x
1 3 4
1 2 4
7 5
2 3 18 12
x x x
x x x
Solve the system:
Augmented matrix:
1 0 1 7 5
2 3 0 18 12
1 2 3 4x x x x
There are two equations and
four unknowns so we need a 2 x 5
matrix.
1 0 1 7 5
2 3 0 18 12-2R1 + R2 → R2
1 0 1 7 5
0 3 2 4 2R2 → R2
13
42 23 3 3
1 0 1 7 5
0 1←RREF
42 23 3 3
1 0 1 7 5
0 1
Expressing the solution:
If a system is consistent, any
column that does not contain a leading one will have a parameter.
1 2 3 4x x x x
Parameters required
for x3 and x4
42 23 3 3
1 0 1 7 5
0 1
Expressing the solution:
1 2 3 4x x x x
s t
3
4
x s
x t
LET
42 23 3 3
1 0 1 7 5
0 1
Converting back to equations in terms of s and t: s t
1
42 22 3 3 3
7 5x s t
x s t
142 2
2 3 3 3
3
4
5 7x s t
x s t
x s
x t
Solving for x1 and x2 leads to the final solution: