SME
Linear Algebra · Problem type
Rules for determinants
Use this for “A and B are 3 × 3 with det A = 4 and det B = −2. Find det(2A), det(AB), det A⁻¹ and det Aᵀ.”
Table
Number times A (k)
Determinant of kA (det(kA))
This is an example. Type your own values to start a new problem.
det(
k
A)
=
k
×
det A
=
det(kA)
det(AB)
=
det A
×
(
det B
)
=
det
A⁻¹
=
1
det A
=
1
4
det A⁻¹
,
det
Aᵀ
=
det Aᵀ
Formulas
det(kA)
=
k
n
×
(det
A)
32
=
2
3
×
4
det(AB) = (det A) × (det B)
−8 = 4 × (−2)
(det
A
-1
)
×
(det
A)
=
1
1
4
×
4
=
1
det Aᵀ = det A
4 = 4
Assumptions
•
A and B are n × n. In kA every entry is times k, so each of the n rows is scaled by k.
•
det(AB) = det A × det B, and det Aᵀ = det A: rows and columns play the same part.
•
A⁻¹ exists only when det A ≠ 0, and then det A⁻¹ = 1 ÷ det A, as AA⁻¹ = I has det 1.
Step-by-step
WE KNOW
n = 3, det A = 4, det B = −2, k = 2
FIND
determinant of kA (det(kA)), determinant of AB (det(AB)), determinant of A⁻¹ (det A⁻¹), determinant of Aᵀ (det Aᵀ)
Step 1 · Find determinant of kA (det(kA))
Use det(kA) = kⁿ × (det A)
Each
of
the
n
rows
of
kA
has
a
factor
k
,
and
each
row’s
factor
comes
out
of
det
once:
kⁿ
in
all.
det(kA)
=
2
3
×
4
det(kA) = 8 × 4
det(kA) = 32
Step 2 · Find determinant of AB (det(AB))
Use det(AB) = (det A) × (det B)
The determinant of a product is the product of the determinants (in either order).
det(AB) = 4 × (−2)
det(AB) = −8
Step 3 · Find determinant of A⁻¹ (det A⁻¹)
Use (det A⁻¹) × (det A) = 1
A
×
A
-1
=
I
and
det
I
=
1,
so
det
A
-1
×
det
A
=
1:
divide
1
by
det
A.
det
A
-1
=
1
det A
det
A
-1
=
1
4
det
A
-1
=
1
4
Step 4 · Find determinant of Aᵀ (det Aᵀ)
Use det Aᵀ = det A
Turning rows into columns keeps the determinant: expanding down a column gives the same number as along a row.
det Aᵀ = 4
Check
Put the numbers back in. Both sides should match.
32
=
2
3
×
4
✓
−8 = 4 × (−2) ✓
1
4
×
4
=
1
✓
4 = 4 ✓
Topic check
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