Grade 12 Math · Problem type

Solving a system with A⁻¹

Use this for “Solve 3x + 2y = 7 and 5x + 4y = 13 with the inverse matrix.”

Diagram

AB2−1−5/23/2×713=12ABc₁₁ = 2 × 7 + (−1) × 13 = 1
A is 2 × 2 and B is 2 × 1, so AB is 2 × 1.
Row 1 of A times column 1 of B gives entry (1, 1): c₁₁ = 2 × 7 + (−1) × 13 = 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.
a
x
+
b
y
=
p
c
x
+
d
y
=
q
x
=
x
,
y
=
y
The picture writes the system as X = A⁻¹B: the inverse of the coefficients times the right sides gives x and y.
D
Determinant ad − bc
worked out (not typed)

Formulas

D
=
a
×
d
−
b
×
c
2 = 3 × 4 − 2 × 5
x
=
d
×
p
−
b
×
q
D
1
=
4 × 7 − 2 × 13
2
y
=
a
×
q
−
c
×
p
D
2
=
3 × 13 − 5 × 7
2

Assumptions

•
The system is AX = B: A holds the coefficients, X = [[x], [y]] and B the right sides.
•
Multiply both sides by A⁻¹ on the left: X = A⁻¹B.
•
A⁻¹ = (1/D)[[d, −b], [−c, a]] with D = ad − bc, so D = 0 means no single solution.

Step-by-step

WE KNOW
a = 3, b = 2, p = 7, c = 5, d = 4, q = 13
FIND
determinant ad − bc (D), solution x (x), solution y (y)
Step 1 · Find determinant ad − bc (D)
Use D = a × d − b × c
A
-1
needs
the
determinant:
down
the
main
diagonal,
take
away
the
other.
D
=
3
×
4
−
2
×
5
D
=
12
−
10
D
=
2
→
A
-1
=
(
1
2
)[[4,
−2],
[−5,
3]]
Step 2 · Find solution x (x)
Use x = (d × p − b × q) ÷ D
Row
1
of
A
-1
times
B:
d
times
p
,
take
away
b
times
q
,
all
over
D
.
x
=
4 × 7 − 2 × 13
2
x
=
28 − 26
2
x
=
2
2
x
=
1
Step 3 · Find solution y (y)
Use y = (a × q − c × p) ÷ D
Row
2
of
A
-1
times
B:
a
times
q
,
take
away
c
times
p
,
all
over
D
.
y
=
3 × 13 − 5 × 7
2
y
=
39 − 35
2
y
=
4
2
y
=
2
Check
Put the numbers back in. Both sides should match.
2 = 3 × 4 − 2 × 5 ✓
1
=
4 × 7 − 2 × 13
2
✓
2
=
3 × 13 − 5 × 7
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

Refresh

Related lessons

© 2026 SME. All rights reserved.