Inverse Matrices and Determinants

Key Terms

Determinant (2×2): det([[a,b],[c,d]]) = ad − bc.
Singular matrix: A matrix with det = 0; has NO inverse; the system of equations it represents has no unique solution.
Inverse A⁻¹: For a 2×2 matrix: A⁻¹ = (1/det A) [[d, −b], [−c, a]] (swap diagonal, negate off-diagonal, divide by det).
Verification: AA⁻¹ = A⁻¹A = I (the identity matrix).
Non-singular: A matrix with det ≠ 0; has a unique inverse.
Product determinant: det(AB) = det(A) × det(B).

Inverse Matrices and Determinants

For A =

a b

c d

Determinant: det(A) = |A| = ad − bc

Inverse (when det(A) ≠ 0):
A⁻¹ = ¹⁄_det(A) ×

d	−b
−c	a

Swap the diagonal elements, negate the off-diagonal elements, divide by det(A).

Key Properties

Property	Statement	Note
Singular matrix	det(A) = 0	No inverse exists
Non-singular matrix	det(A) ≠ 0	Inverse exists
Inverse property	AA⁻¹ = A⁻¹A = I	Always true when inverse exists
det(AB)	det(A) × det(B)	Determinant is multiplicative
det(kA)	k² × det(A) for 2×2	kⁿ × det(A) for n×n

Worked Example 1 — Finding the Determinant

Find det(A) for A =

5	3
2	4

Identify: a = 5, b = 3, c = 2, d = 4

det(A) = ad − bc = (5)(4) − (3)(2) = 20 − 6 = 14

Since det(A) = 14 ≠ 0, the matrix is non-singular and an inverse exists.

Worked Example 2 — Finding the Inverse

Find A⁻¹ for A =

3	2
1	1

Step 1: Find det(A) = (3)(1) − (2)(1) = 3 − 2 = 1

Step 2: Swap diagonal, negate off-diagonal:

Adjusted matrix:

1	−2
−1	3

Step 3: Divide by det(A) = 1:

A⁻¹ =

1	−2
−1	3

Verify: AA⁻¹ = (3)(1)+(2)(−1) | (3)(−2)+(2)(3) | ... = [[1,0],[0,1]] = I ✓

Hot Tip: The memory trick for the inverse formula — swap the main diagonal (a and d swap places), negate the off-diagonal (b and c get minus signs), then divide everything by the determinant. Check: if det = 0, division by zero is undefined, so no inverse exists.

Full Lesson: Inverse Matrices and Determinants

Why Do We Need the Determinant?

For ordinary numbers, every non-zero number has a multiplicative inverse: the inverse of 5 is 1/5 because 5 × (1/5) = 1. For matrices, we want to find A⁻¹ such that AA⁻¹ = I, where I is the identity matrix. The determinant tells us whether such an inverse exists.

The determinant of a 2×2 matrix A is a single number that encodes important information about A. For A =

a	b
c	d

, the determinant is det(A) = ad − bc.

Geometric Meaning

Geometrically, |det(A)| gives the area scaling factor of the linear transformation represented by A. If you apply A to a unit square (area = 1), the resulting shape has area |det(A)|. If det(A) = 3, the transformation triples areas. If det(A) = 0, the transformation collapses everything onto a line (or point) — all area is destroyed, and the transformation cannot be reversed.

Singular Matrices

A matrix is singular when det(A) = 0. This means:

The matrix has no inverse
The rows of the matrix are linearly dependent (one row is a multiple of the other)
Any system of equations Ax = b will either have no solution or infinitely many solutions — never a unique solution
Geometrically, the transformation collapses 2D space onto a 1D line

Computing the Inverse

For a 2×2 matrix A =

a	b
c	d

with det(A) ≠ 0:

Compute det(A) = ad − bc
Form the adjugate matrix by swapping a and d, and negating b and c
Divide every entry by det(A)

Area Application: A transformation matrix T has det(T) = 6. Any region mapped by T will have its area multiplied by |6| = 6. A region of area 5 cm² becomes a region of area 30 cm² after the transformation.

Properties of Determinants

Two important properties you will use:

det(AB) = det(A) × det(B): the determinant of a product equals the product of the determinants.
det(kA) = kⁿ × det(A) for an n×n matrix: scaling a matrix by k scales the determinant by kⁿ. For a 2×2 matrix, det(2A) = 4×det(A).

Lesson Tip: Always compute the determinant first before attempting to find the inverse. If det = 0, stop immediately — the inverse does not exist. When the determinant is a fraction, be careful with the division; it often helps to write A⁻¹ as a scalar multiplied by a matrix of integers.

Mastery Practice

See Answers ➔

Fluency
Find det(A) for A =

3 2

1 4

. Show your working using the formula det(A) = ad − bc.
Fluency
Find det(B) for B =

−5 3

2 −1

. Be careful with the signs when applying the formula.
Fluency
Find A⁻¹ for A =

4 1

3 1

. Show all steps: find det(A), then apply the inverse formula.
Fluency
Find B⁻¹ for B =

2 −3

1 −1

.
Understanding
Show that the matrix

6 2

3 1

is singular. Explain what “singular” means both algebraically and geometrically.
Understanding
If det(A) = 5, find det(3A) for a 2×2 matrix A. Show your reasoning using the property det(kA) = kⁿ × det(A) where n is the size of the matrix.
Understanding
Find the value of k such that

k 2

3 4

is singular.
Understanding
Let A =

2 1

5 3

. Find A⁻¹, then calculate AA⁻¹ and verify that the result is the identity matrix I.
Problem Solving
A transformation matrix T =

1 2

0 1

maps a region R to a region R′. The area of R is 8 cm². Find the area of R′. Explain your answer using the determinant as an area scaling factor.
Problem Solving
A 2×2 matrix has det = 0 but is not the zero matrix.
1. Give a specific example of such a matrix.
2. Verify that your example has det = 0.
3. Explain why having det = 0 causes problems when trying to solve a system of equations AX = B using the matrix inverse method.

−5	3
2	−1

2	−3
1	−1