Grade 12 Math · Problem type

Determinant of a 3 × 3 matrix

Use this for “Find the determinant of [[2, 0, 1], [1, 3, 2], [1, 1, 4]].”

Diagram

201132114+ 2 × (3 × 4 − 2 × 1)= + 2 × 10201132114− 0 × (1 × 4 − 2 × 1)= − 0 × 2201132114+ 1 × (1 × 1 − 3 × 1)= + 1 × (−2)D = 20 + 0 − 2 = 18
D = 2 × 10 − 0 × 2 + 1 × (−2) = 18
Each first-row entry times the 2 × 2 left when its row and column are struck, signs +, −, +
This is an example. Type your own values to start a new problem.
=
D

Formulas

D
=
a
×
(
e
×
k
−
f
×
h
)
−
b
×
(
d
×
k
−
f
×
g
)
+
c
×
(
d
×
h
−
e
×
g
)
18 = 2 × (3 × 4 − 2 × 1) − 0 × (1 × 4 − 2 × 1) + 1 × (1 × 1 − 3 × 1)

Assumptions

•
Expand along the first row: each entry times the 2 × 2 determinant left when its row and column are crossed out.
•
The signs go +, −, + along the row.
•
D = 0 means the matrix has no inverse and its system has no single solution.

Step-by-step

WE KNOW
a = 2, b = 0, c = 1, d = 1, e = 3, f = 2, g = 1, h = 1, k = 4
FIND
determinant (D)
Step 1 · Find determinant (D)
Use D = a × (e × k − f × h) − b × (d × k − f × g) + c × (d × h − e × g)
Each first-row entry times its 2 × 2 minor, with signs +, −, +.
D
=
2
×
(3
×
4
−
2
×
1)
−
0
×
(1
×
4
−
2
×
1)
+
1
×
(1
×
1
−
3
×
1)
D
=
2
×
(12
−
2)
−
0
×
2
−
2
D
=
2
×
10
−
0
×
2
−
2
D
=
20
−
2
D
=
18
Check
Put the numbers back in. Both sides should match.
18 = 2 × (3 × 4 − 2 × 1) − 0 × (1 × 4 − 2 × 1) + 1 × (1 × 1 − 3 × 1) ✓
Topic check
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