SME
Linear Algebra · Problem type
The inverse by row reduction
Use this for “Reduce [A | I] to find A⁻¹ for A = [[2, 1, 0], [1, 1, 0], [0, 0, 3]].”
Diagram
[A | I] → [I | A⁻¹]: the 5 row operations that turn A into I turn I into A⁻¹.
The left block is I, so the right block is A⁻¹; check: A × A⁻¹ = I.
This is an example. Type your own values to start a new problem.
−1
=
1
3
D
Determinant of A
worked out (not typed)
Formulas
D
=
det
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
3 = det [[2, 1, 0], [1, 1, 0], [0, 0, 3]]
b₁₁
=
a₂₂
×
a₃₃
−
a₂₃
×
a₃₂
D
1
=
1 × 3 − 0 × 0
3
b₁₂
=
a₃₂
×
a₁₃
−
a₃₃
×
a₁₂
D
−1
=
0 × 0 − 3 × 1
3
b₁₃
=
a₁₂
×
a₂₃
−
a₁₃
×
a₂₂
D
0
=
1 × 0 − 0 × 1
3
b₂₁
=
a₃₁
×
a₂₃
−
a₃₃
×
a₂₁
D
−1
=
0 × 0 − 3 × 1
3
b₂₂
=
a₁₁
×
a₃₃
−
a₁₃
×
a₃₁
D
2
=
2 × 3 − 0 × 0
3
b₂₃
=
a₂₁
×
a₁₃
−
a₂₃
×
a₁₁
D
0
=
1 × 0 − 0 × 2
3
b₃₁
=
a₂₁
×
a₃₂
−
a₂₂
×
a₃₁
D
0
=
1 × 0 − 1 × 0
3
b₃₂
=
a₃₁
×
a₁₂
−
a₃₂
×
a₁₁
D
0
=
0 × 1 − 0 × 2
3
b₃₃
=
a₁₁
×
a₂₂
−
a₁₂
×
a₂₁
D
1
3
=
2 × 1 − 1 × 1
3
Assumptions
•
Row operations that turn A into I turn I into A⁻¹: reduce [A | I] to [I | A⁻¹].
•
A has an inverse exactly when det A ≠ 0; otherwise a row of zeros appears in the left block.
•
Each entry of A⁻¹ is also a cofactor over det A, a check on the reduction.
Step-by-step
WE KNOW
a₁₁ = 2, a₁₂ = 1, a₁₃ = 0, a₂₁ = 1, a₂₂ = 1, a₂₃ = 0, a₃₁ = 0, a₃₂ = 0, a₃₃ = 3
FIND
determinant of A (D), inverse A⁻¹, row 1, column 1 (b₁₁), inverse A⁻¹, row 1, column 2 (b₁₂), inverse A⁻¹, row 1, column 3 (b₁₃), inverse A⁻¹, row 2, column 1 (b₂₁), inverse A⁻¹, row 2, column 2 (b₂₂), inverse A⁻¹, row 2, column 3 (b₂₃), inverse A⁻¹, row 3, column 1 (b₃₁), inverse A⁻¹, row 3, column 2 (b₃₂), inverse A⁻¹, row 3, column 3 (b₃₃)
Step 1 · Find determinant of A (D)
Use D = det [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Along the first row: each entry times its 2 × 2 minor, signs +, −, +. As det A ≠ 0, A has an inverse.
D
=
det
[[2,
1,
0],
[1,
1,
0],
[0,
0,
3]]
det [[2, 1, 0], [1, 1, 0], [0, 0, 3]] = 2 × det [[1, 0], [0, 3]] − 1 × det [[1, 0], [0, 3]] + 0 × det [[1, 1], [0, 0]]
= 2 × (1 × 3 − 0 × 0) − 1 × (1 × 3 − 0 × 0) + 0 × (1 × 0 − 1 × 0)
= 2 × 3 − 1 × 3 + 0 × 0
= 3
D
=
3
Step 2 · Find inverse A⁻¹, row 1, column 1 (b₁₁)
Use b₁₁ = (a₂₂ × a₃₃ − a₂₃ × a₃₂) ÷ D
Reduce
[A
|
I]
until
the
left
block
is
I:
the
right
block
is
then
A
-1
.
Read
its
entry
(1,
1).
[A | I] = [[2, 1, 0 | 1, 0, 0], [1, 1, 0 | 0, 1, 0], [0, 0, 3 | 0, 0, 1]]
R₁
→
(
1
2
)R₁:
[[1,
1
2
,
0
|
1
2
,
0,
0],
[1,
1,
0
|
0,
1,
0],
[0,
0,
3
|
0,
0,
1]]
R₂
→
R₂
−
R₁:
[[1,
1
2
,
0
|
1
2
,
0,
0],
[0,
1
2
,
0
|
−
1
2
,
1,
0],
[0,
0,
3
|
0,
0,
1]]
R₂
→
2R₂:
[[1,
1
2
,
0
|
1
2
,
0,
0],
[0,
1,
0
|
−1,
2,
0],
[0,
0,
3
|
0,
0,
1]]
R₁
→
R₁
−
(
1
2
)R₂:
[[1,
0,
0
|
1,
−1,
0],
[0,
1,
0
|
−1,
2,
0],
[0,
0,
3
|
0,
0,
1]]
R₃
→
(
1
3
)R₃:
[[1,
0,
0
|
1,
−1,
0],
[0,
1,
0
|
−1,
2,
0],
[0,
0,
1
|
0,
0,
1
3
]]
A
-1
=
[[1,
−1,
0],
[−1,
2,
0],
[0,
0,
1
3
]]
Check:
[[2,
1,
0],
[1,
1,
0],
[0,
0,
3]]
×
[[1,
−1,
0],
[−1,
2,
0],
[0,
0,
1
3
]]
=
[[1,
0,
0],
[0,
1,
0],
[0,
0,
1]]
b₁₁
=
1
Step 3 · Find inverse A⁻¹, row 1, column 2 (b₁₂)
Use b₁₂ = (a₃₂ × a₁₃ − a₃₃ × a₁₂) ÷ D
Read entry (1, 2): row 1, column 2 of the right block.
b₁₂
=
−1
Step 4 · Find inverse A⁻¹, row 1, column 3 (b₁₃)
Use b₁₃ = (a₁₂ × a₂₃ − a₁₃ × a₂₂) ÷ D
Read entry (1, 3): row 1, column 3 of the right block.
b₁₃
=
0
Step 5 · Find inverse A⁻¹, row 2, column 1 (b₂₁)
Use b₂₁ = (a₃₁ × a₂₃ − a₃₃ × a₂₁) ÷ D
Read entry (2, 1): row 2, column 1 of the right block.
b₂₁
=
−1
Step 6 · Find inverse A⁻¹, row 2, column 2 (b₂₂)
Use b₂₂ = (a₁₁ × a₃₃ − a₁₃ × a₃₁) ÷ D
Read entry (2, 2): row 2, column 2 of the right block.
b₂₂
=
2
Step 7 · Find inverse A⁻¹, row 2, column 3 (b₂₃)
Use b₂₃ = (a₂₁ × a₁₃ − a₂₃ × a₁₁) ÷ D
Read entry (2, 3): row 2, column 3 of the right block.
b₂₃
=
0
Step 8 · Find inverse A⁻¹, row 3, column 1 (b₃₁)
Use b₃₁ = (a₂₁ × a₃₂ − a₂₂ × a₃₁) ÷ D
Read entry (3, 1): row 3, column 1 of the right block.
b₃₁
=
0
Step 9 · Find inverse A⁻¹, row 3, column 2 (b₃₂)
Use b₃₂ = (a₃₁ × a₁₂ − a₃₂ × a₁₁) ÷ D
Read entry (3, 2): row 3, column 2 of the right block.
b₃₂
=
0
Step 10 · Find inverse A⁻¹, row 3, column 3 (b₃₃)
Use b₃₃ = (a₁₁ × a₂₂ − a₁₂ × a₂₁) ÷ D
Read entry (3, 3): row 3, column 3 of the right block.
b₃₃
=
1
3
Check
Put the numbers back in. Both sides should match.
3 = det [[2, 1, 0], [1, 1, 0], [0, 0, 3]] ✓
1
=
1 × 3 − 0 × 0
3
✓
−1
=
0 × 0 − 3 × 1
3
✓
0
=
1 × 0 − 0 × 1
3
✓
−1
=
0 × 0 − 3 × 1
3
✓
2
=
2 × 3 − 0 × 0
3
✓
0
=
1 × 0 − 0 × 2
3
✓
0
=
1 × 0 − 1 × 0
3
✓
0
=
0 × 1 − 0 × 2
3
✓
1
3
=
2 × 1 − 1 × 1
3
✓
Topic check
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