SME
Grade 12 Math · Problem type
Inverse of a 2 × 2 matrix
Use this for “Find the inverse of [[4, 7], [2, 6]].”
Diagram
A is 2 × 2 and B is 2 × 2, so AB is 2 × 2.
Row 1 of A times column 1 of B gives entry (1, 1): c₁₁ = 4 × 3/5 + 7 × (−1/5) = 1.
Tap an entry of AB to see its row and column.
This is an example. Type your own values to start a new problem.
−1
=
3
5
−1
5
−7
10
2
5
D
Determinant ad − bc
worked out (not typed)
Formulas
D
=
a
×
d
−
b
×
c
10 = 4 × 6 − 7 × 2
e
=
d
D
3
5
=
6
10
f
=
−
b
D
−
7
10
=
−
7
10
g
=
−
c
D
−
1
5
=
−
2
10
h
=
a
D
2
5
=
4
10
Assumptions
•
For A = [[a, b], [c, d]], swap a and d, change the signs of b and c, then divide by D = ad − bc.
•
A times its inverse is the identity [[1, 0], [0, 1]]: the picture multiplies them.
•
D = 0 means there is no inverse.
Step-by-step
WE KNOW
a = 4, b = 7, c = 2, d = 6
FIND
determinant ad − bc (D), row 1, column 1 of A⁻¹ (e), row 1, column 2 of A⁻¹ (f), row 2, column 1 of A⁻¹ (g), row 2, column 2 of A⁻¹ (h)
Step 1 · Find determinant ad − bc (D)
Use D = a × d − b × c
Multiply down the main diagonal and take away the other diagonal.
D
=
4
×
6
−
7
×
2
D
=
24
−
14
D
=
10
→
A
-1
=
(
1
10
)[[6,
−7],
[−2,
4]]
Step 2 · Find row 1, column 1 of A⁻¹ (e)
Use e = d ÷ D
Swap
a
and
d
:
the
top
left
of
the
inverse
is
d
over
D
.
e
=
6
10
e
=
3
5
Step 3 · Find row 1, column 2 of A⁻¹ (f)
Use f = −b ÷ D
Change
the
sign
of
b
,
then
divide
by
D
.
f
=
−
7
10
f
=
−
7
10
Step 4 · Find row 2, column 1 of A⁻¹ (g)
Use g = −c ÷ D
Change
the
sign
of
c
,
then
divide
by
D
.
g
=
−
2
10
g
=
−
1
5
Step 5 · Find row 2, column 2 of A⁻¹ (h)
Use h = a ÷ D
Swap
a
and
d
:
the
bottom
right
of
the
inverse
is
a
over
D
.
h
=
4
10
h
=
2
5
Check
Put the numbers back in. Both sides should match.
10 = 4 × 6 − 7 × 2 ✓
3
5
=
6
10
✓
−
7
10
=
−
7
10
✓
−
1
5
=
−
2
10
✓
2
5
=
4
10
✓
Topic check
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Systems in three variables, matrices and determinants
Main lesson
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Use this for “Solve 2x + 3y = 13 and x − y = −1 by Cramer’s rule.”