SME
Linear Algebra
Determinants
Use this for “Find det A by row reduction for A = [[0, 2, 1], [1, 1, 1], [2, 0, 3]].”
Diagram
A swap flips the sign, scaling a row by c scales det by c, and adding a multiple of a row keeps det.
A triangle’s det is its diagonal’s product, so det A = (−1) × 4 = −4.
This is an example. Type your own values to start a new problem.
det
=
D
s
Row swaps
worked out (not typed)
p₁
Pivot 1 on the diagonal
worked out (not typed)
p₂
Pivot 2 on the diagonal
worked out (not typed)
p₃
Pivot 3 on the diagonal
worked out (not typed)
Formulas
s
=
row
swaps
reducing
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
1 = row swaps reducing [[0, 2, 1], [1, 1, 1], [2, 0, 3]]
p₁
=
entry
(1,
1)
of
the
triangle
from
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
1 = entry (1, 1) of the triangle from [[0, 2, 1], [1, 1, 1], [2, 0, 3]]
p₂
=
entry
(2,
2)
of
the
triangle
from
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
2 = entry (2, 2) of the triangle from [[0, 2, 1], [1, 1, 1], [2, 0, 3]]
p₃
=
entry
(3,
3)
of
the
triangle
from
[[
a₁₁
,
a₁₂
,
a₁₃
],
[
a₂₁
,
a₂₂
,
a₂₃
],
[
a₃₁
,
a₃₂
,
a₃₃
]]
2 = entry (3, 3) of the triangle from [[0, 2, 1], [1, 1, 1], [2, 0, 3]]
D
=
(−1)
s
×
p₁
×
p₂
×
p₃
−4
=
(−1)
1
×
1
×
2
×
2
Assumptions
•
A row swap flips the sign of det; adding a multiple of one row to another keeps it.
•
A triangular matrix’s det is the product of its diagonal, the pivots.
•
So with s swaps and no rows scaled, det A = (−1)ˢ × (product of the pivots); det A = 0 means A has no inverse.
Step-by-step
WE KNOW
a₁₁ = 0, a₁₂ = 2, a₁₃ = 1, a₂₁ = 1, a₂₂ = 1, a₂₃ = 1, a₃₁ = 2, a₃₂ = 0, a₃₃ = 3
FIND
row swaps (s), pivot 1 on the diagonal (p₁), pivot 2 on the diagonal (p₂), pivot 3 on the diagonal (p₃), determinant of A (D)
Step 1 · Find row swaps (s)
Use s = row swaps reducing [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Clear under each pivot, one operation a line, with no row scaled. Swap rows only when a pivot is 0, and count the swaps.
Start: [[0, 2, 1], [1, 1, 1], [2, 0, 3]]
R₁ ↔ R₂: [[1, 1, 1], [0, 2, 1], [2, 0, 3]] (det flips sign)
R₃ → R₃ − 2R₁: [[1, 1, 1], [0, 2, 1], [0, −2, 1]] (det kept)
R₃ → R₃ + R₂: [[1, 1, 1], [0, 2, 1], [0, 0, 2]] (det kept)
s
=
1
Step 2 · Find pivot 1 on the diagonal (p₁)
Use p₁ = entry (1, 1) of the triangle from [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Read entry (1, 1) on the diagonal of the last matrix.
p₁
=
1
Step 3 · Find pivot 2 on the diagonal (p₂)
Use p₂ = entry (2, 2) of the triangle from [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Read entry (2, 2) on the diagonal of the last matrix.
p₂
=
2
Step 4 · Find pivot 3 on the diagonal (p₃)
Use p₃ = entry (3, 3) of the triangle from [[a₁₁, a₁₂, a₁₃], [a₂₁, a₂₂, a₂₃], [a₃₁, a₃₂, a₃₃]]
Read entry (3, 3) on the diagonal of the last matrix.
p₃
=
2
Step 5 · Find determinant of A (D)
Use D = (−1)ˢ × p₁ × p₂ × p₃
One swap flips the sign; the other row operations keep det. Multiply the pivots. Cofactors along the first row give the same number.
D
=
(−1)
1
×
1
×
2
×
2
D
=
−1
×
4
=
−4
Check by cofactors along the first row:
det [[0, 2, 1], [1, 1, 1], [2, 0, 3]] = 0 × det [[1, 1], [0, 3]] − 2 × det [[1, 1], [2, 3]] + 1 × det [[1, 1], [2, 0]]
= 0 × (1 × 3 − 1 × 0) − 2 × (1 × 3 − 1 × 2) + 1 × (1 × 0 − 1 × 2)
= 0 × 3 − 2 × 1 + 1 × (−2)
= −4
D
=
−4
Check
Put the numbers back in. Both sides should match.
Row swaps: 1 = 1 ✓
First pivot: 1 = 1 ✓
Second pivot: 2 = 2 ✓
Third pivot: 2 = 2 ✓
−4
=
(−1)
1
×
1
×
2
×
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.