SME
Linear Algebra · Problem type
LU factorization
Use this for “Factor A = [[2, 1, 1], [4, 3, 3], [8, 7, 9]] as A = LU.”
Diagram
L is 3 × 3 and U is 3 × 3, so A is 3 × 3.
Row 1 of L times column 1 of U gives entry (1, 1): a₁₁ = 1 × 2 + 0 × 0 + 0 × 0 = 2.
Tap an entry of A to see its row and column.
This is an example. Type your own values to start a new problem.
1
0
1
0
0
1
0
0
0
=
Formulas
u₁₁
=
a₁₁
2 = 2
u₁₂
=
a₁₂
1 = 1
u₁₃
=
a₁₃
1 = 1
l₂₁
=
a₂₁
u₁₁
2
=
4
2
l₃₁
=
a₃₁
u₁₁
4
=
8
2
u₂₂
=
a₂₂
−
l₂₁
×
u₁₂
1 = 3 − 2 × 1
u₂₃
=
a₂₃
−
l₂₁
×
u₁₃
1 = 3 − 2 × 1
l₃₂
=
a₃₂
−
l₃₁
×
u₁₂
u₂₂
3
=
7 − 4 × 1
1
u₃₃
=
a₃₃
−
l₃₁
×
u₁₃
−
l₃₂
×
u₂₃
2 = 9 − 4 × 1 − 3 × 1
Assumptions
•
Elimination with no row swaps turns A into the upper triangular U.
•
L has 1s on its diagonal and the multipliers below it: l₃₂ is the number in R₃ → R₃ − l₃₂R₂.
•
A zero pivot (u₁₁ or u₂₂) needs a row swap first, which this page does not do.
Step-by-step
WE KNOW
a₁₁ = 2, a₁₂ = 1, a₁₃ = 1, a₂₁ = 4, a₂₂ = 3, a₂₃ = 3, a₃₁ = 8, a₃₂ = 7, a₃₃ = 9
FIND
U, row 1, column 1 (u₁₁), U, row 1, column 2 (u₁₂), U, row 1, column 3 (u₁₃), multiplier in L, row 2, column 1 (l₂₁), multiplier in L, row 3, column 1 (l₃₁), U, row 2, column 2 (u₂₂), U, row 2, column 3 (u₂₃), multiplier in L, row 3, column 2 (l₃₂), U, row 3, column 3 (u₃₃)
Step 1 · Find U, row 1, column 1 (u₁₁)
Use u₁₁ = a₁₁
Row 1 of U is row 1 of A: elimination never changes the first row.
u₁₁
=
2
Step 2 · Find U, row 1, column 2 (u₁₂)
Use u₁₂ = a₁₂
Row 1 of U is row 1 of A: elimination never changes the first row.
u₁₂
=
1
Step 3 · Find U, row 1, column 3 (u₁₃)
Use u₁₃ = a₁₃
Row 1 of U is row 1 of A: elimination never changes the first row.
u₁₃
=
1
Step 4 · Find multiplier in L, row 2, column 1 (l₂₁)
Use l₂₁ = a₂₁ ÷ u₁₁
The multiplier that clears entry (2, 1): R₂ → R₂ − l₂₁R₁ leaves 0 under the first pivot.
l₂₁
=
4
2
l₂₁
=
2
Step 5 · Find multiplier in L, row 3, column 1 (l₃₁)
Use l₃₁ = a₃₁ ÷ u₁₁
The multiplier that clears entry (3, 1): R₃ → R₃ − l₃₁R₁ leaves 0 under the first pivot.
l₃₁
=
8
2
l₃₁
=
4
Step 6 · Find U, row 2, column 2 (u₂₂)
Use u₂₂ = a₂₂ − l₂₁ × u₁₂
R₂
→
R₂
−
l₂₁R₁
takes
l₂₁
times
row
1
from
entry
(2,
2).
u₂₂
=
3
−
2
×
1
u₂₂
=
1
Step 7 · Find U, row 2, column 3 (u₂₃)
Use u₂₃ = a₂₃ − l₂₁ × u₁₃
R₂
→
R₂
−
l₂₁R₁
takes
l₂₁
times
row
1
from
entry
(2,
3).
u₂₃
=
3
−
2
×
1
u₂₃
=
1
Step 8 · Find multiplier in L, row 3, column 2 (l₃₂)
Use l₃₂ = (a₃₂ − l₃₁ × u₁₂) ÷ u₂₂
After
R₃
→
R₃
−
l₃₁R₁,
entry
(3,
2)
is
a₃₂
−
l₃₁
u₁₂
.
Divide
it
by
the
second
pivot
to
get
the
multiplier
that
clears
it.
l₃₂
=
7 − 4 × 1
1
l₃₂
=
3
Step 9 · Find U, row 3, column 3 (u₃₃)
Use u₃₃ = a₃₃ − l₃₁ × u₁₃ − l₃₂ × u₂₃
Both
row
operations
on
row
3
change
entry
(3,
3):
take
l₃₁
times
row
1
and
l₃₂
times
row
2
of
U.
Then
L
times
U
gives
A
back.
u₃₃
=
9
−
4
×
1
−
3
×
1
u₃₃
=
9
−
4
−
3
=
2
Check: L × U gives A back.
Row 1: [[1 × 2 + 0 × 0 + 0 × 0, 1 × 1 + 0 × 1 + 0 × 0, 1 × 1 + 0 × 1 + 0 × 2]] = [[2, 1, 1]]
Row 2: [[2 × 2 + 1 × 0 + 0 × 0, 2 × 1 + 1 × 1 + 0 × 0, 2 × 1 + 1 × 1 + 0 × 2]] = [[4, 3, 3]]
Row 3: [[4 × 2 + 3 × 0 + 1 × 0, 4 × 1 + 3 × 1 + 1 × 0, 4 × 1 + 3 × 1 + 1 × 2]] = [[8, 7, 9]]
[[1, 0, 0], [2, 1, 0], [4, 3, 1]] × [[2, 1, 1], [0, 1, 1], [0, 0, 2]] = [[2, 1, 1], [4, 3, 3], [8, 7, 9]]
u₃₃
=
2
Check
Put the numbers back in. Both sides should match.
2 = 2 ✓
1 = 1 ✓
1 = 1 ✓
2
=
4
2
✓
4
=
8
2
✓
1 = 3 − 2 × 1 ✓
1 = 3 − 2 × 1 ✓
3
=
7 − 4 × 1
1
✓
2 = 9 − 4 × 1 − 3 × 1 ✓
Topic check
See what you know: 5 questions on this topic
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