Inverse Matrices — How to Find and Use the

§01What Is the Inverse of a Matrix?

An inverse "undoes" a matrix, just as dividing by a number undoes multiplication by it.

For a square matrix A, the inverse A⁻¹ is defined by:

Definition of the Inverse

AA⁻¹ = A⁻¹A = I

where I is the identity matrix of the same size.

Not every matrix has an inverse. A matrix that does is called invertible or non-singular. A matrix that doesn't is singular.

Invertibility Criterion

A square matrix A is invertible if and only if det(A) ≠ 0.
Equivalently: A is invertible iff its rows (or columns) are linearly independent, iff its rank equals n, iff 0 is not an eigenvalue.

Why inverses matter The system Ax = b has a unique solution x = A⁻¹b whenever A is invertible. This is conceptually simple — but Gauss-Jordan elimination is often more efficient for large systems.

§02The 2×2 Formula

For 2×2 matrices there is a clean direct formula, memorise it.

2×2 Inverse Formula

If A = [[a, b],[c, d]], then A⁻¹ = 1/(ad−bc) · [[d, −b],[−c, a]]

The scalar 1/(ad − bc) = 1/det(A). If det(A) = 0, A has no inverse.

The pattern is easy to remember: swap the main diagonal entries, negate the off-diagonal entries, and divide by the determinant.

Common error: forgetting to divide by det(A) The matrix [[d,−b],[−c,a]] is the adjugate, not the inverse. Always include the 1/det(A) factor.

§03Gauss-Jordan Elimination — The General Method

For 3×3 and larger matrices, form the augmented matrix [A | I] and row reduce until the left block becomes I.

🔑 Gauss-Jordan Algorithm for A⁻¹

Set up the augmented matrix
Write [A | I], the n×n matrix A alongside the n×n identity.
Row reduce the left block to I
Apply elementary row operations to the entire augmented matrix until the left side becomes the identity matrix.
Read off the inverse
The right block has become A⁻¹.
Check: if the left block won't reach I
If you encounter a zero row on the left, A is singular: no inverse exists.

Why does this work? Each row operation multiplies on the left by an elementary matrix E. Applying the same operations to I accumulates E₁E₂⋯Eₖ = A⁻¹ on the right side.

§04Key Properties of Inverse Matrices

Reversal Law

(AB)⁻¹ = B⁻¹A⁻¹

Transpose

(Aᵀ)⁻¹ = (A⁻¹)ᵀ

Scalar

(kA)⁻¹ = (1/k)A⁻¹ (k≠0)

Inverse of inverse

(A⁻¹)⁻¹ = A

Determinant

det(A⁻¹) = 1/det(A)

Powers

(Aⁿ)⁻¹ = (A⁻¹)ⁿ

The Cofactor / Adjugate Method (3×3)

Adjugate formula

A⁻¹ = (1/det(A)) · adj(A)

adj(A) is the transpose of the cofactor matrix of A. Efficient for 3×3; Gauss-Jordan is faster for larger sizes.

§05Ten Worked Examples

From the 2×2 formula to Gauss-Jordan on 3×3 matrices, every step shown.

Example 012×2 inverse — direct formula

Find A⁻¹ for A = [[3, 1],[5, 2]].

Step 1 — Determinant

det(A) = 3·2 − 1·5 = 6 − 5 = 1

Step 2 — Apply formula

A⁻¹ = (1/1)·[[2,−1],[−5,3]] = [[2,−1],[−5,3]]

Step 3 — Verify: AA⁻¹ = I

[[3,1],[5,2]]·[[2,−1],[−5,3]] = [[6−5, −3+3],[10−10, −5+6]] = [[1,0],[0,1]] ✓

A⁻¹ =

2−1

−53

Example 022×2 — matrix with fractional inverse

Find A⁻¹ for A = [[4, 7],[2, 6]].

Determinant

det(A) = 24 − 14 = 10

Inverse

A⁻¹ = (1/10)·[[6,−7],[−2,4]] = [[0.6, −0.7],[−0.2, 0.4]]

A⁻¹ = [[3/5, −7/10],[−1/5, 2/5]]

Example 03Singular 2×2 — no inverse

Determine if A = [[2, 6],[1, 3]] is invertible.

Determinant check

det(A) = 2·3 − 6·1 = 6 − 6 = 0

Since det(A) = 0, the matrix is singular. No inverse exists. Geometrically, this transformation collapses ℝ² onto a line.

Singular — no inverse exists

Example 043×3 inverse via Gauss-Jordan

Find A⁻¹ for A = [[1,0,0],[1,1,0],[1,1,1]].

Set up augmented matrix [A | I]

[1,0,0 | 1,0,0] [1,1,0 | 0,1,0] [1,1,1 | 0,0,1]

R₂ ← R₂ − R₁ | R₃ ← R₃ − R₁

[1,0,0 | 1,0,0] [0,1,0 | −1,1,0] [0,1,1 | −1,0,1]

R₃ ← R₃ − R₂

[1,0,0 | 1,0,0] [0,1,0 | −1,1,0] [0,0,1 | 0,−1,1]

Left block is now I, so right block is A⁻¹:

A⁻¹ =

100

−110

0−11

Example 053×3 general Gauss-Jordan

Find A⁻¹ for A = [[2,1,0],[1,2,1],[0,1,2]].

det(A) — verify invertibility

det(A) = 2(4−1) − 1(2−0) = 6 − 2 = 4 ≠ 0 ✓

Row reduce [A | I] (steps abbreviated)

After elimination: R₁: swap & eliminate ⟹ left block → I

Result

A⁻¹ = (1/4)·[[3,−2,1],[−2,4,−2],[1,−2,3]]

A⁻¹ = ¼[[3,−2,1],[−2,4,−2],[1,−2,3]]

Example 06Using A⁻¹ to solve a linear system

Solve [[3,1],[5,2]]x = [[5],[8]] using the inverse found in Example 01.

Apply x = A⁻¹b

x = [[2,−1],[−5,3]]·[[5],[8]] = [[10−8],[−25+24]] = [[2],[−1]]

x₁ = 2, x₂ = −1

Example 07Reversal law for products

If A⁻¹ = [[2,1],[1,1]] and B⁻¹ = [[3,0],[0,1]], find (AB)⁻¹.

Apply reversal law: (AB)⁻¹ = B⁻¹A⁻¹

(AB)⁻¹ = [[3,0],[0,1]]·[[2,1],[1,1]] = [[6,3],[1,1]]

Order matters! (AB)⁻¹ = B⁻¹A⁻¹, not A⁻¹B⁻¹. Matrix multiplication is not commutative.

(AB)⁻¹ = [[6,3],[1,1]]

Example 08Inverse of a transpose

Given A = [[1,2],[3,4]], find (Aᵀ)⁻¹.

Method 1 — direct transpose of A⁻¹

det(A) = 4−6 = −2 A⁻¹ = (−1/2)·[[4,−2],[−3,1]] = [[−2,1],[3/2,−1/2]] (Aᵀ)⁻¹ = (A⁻¹)ᵀ = [[−2, 3/2],[1, −1/2]]

(Aᵀ)⁻¹ = [[−2, 3/2],[1, −1/2]]

Example 09Diagonal matrix inverse

Find A⁻¹ for the diagonal matrix A = diag(2, 5, −1).

Rule for diagonal matrices

If A = diag(d₁,d₂,…,dₙ) and all dᵢ ≠ 0, then A⁻¹ = diag(1/d₁, 1/d₂, …, 1/dₙ)

A⁻¹ =

1/200

01/50

00−1

Example 10Inverse from eigendecomposition (preview)

If A = PDP⁻¹ (diagonalisation), express A⁻¹ in terms of P and D.

Derivation

A⁻¹ = (PDP⁻¹)⁻¹ = (P⁻¹)⁻¹D⁻¹P⁻¹ = PD⁻¹P⁻¹

D⁻¹ is easy

If D = diag(λ₁,…,λₙ), then D⁻¹ = diag(1/λ₁,…,1/λₙ) (provided no eigenvalue is zero — i.e. A is invertible)

This is the elegant connection between the inverse and the eigenstructure of a matrix.

A⁻¹ = PD⁻¹P⁻¹

§06Four Common Mistakes

Mistake 1: Applying the 2×2 formula to 3×3 matricesThe swap-and-negate formula only works for 2×2. For larger matrices, use Gauss-Jordan or the cofactor method.

Mistake 2: Writing A⁻¹ = 1/AMatrix inversion is not scalar division. The notation 1/A is meaningless — always write A⁻¹ and verify AA⁻¹ = I.

Mistake 3: Reversing (AB)⁻¹ in the wrong order(AB)⁻¹ = B⁻¹A⁻¹. Many students write A⁻¹B⁻¹ — the correct law reverses the order.

Mistake 4: Assuming an inverse exists without checking det(A)Always verify det(A) ≠ 0 before investing time computing A⁻¹. A single row of zeros or proportional rows means no inverse.

§07Practice Quiz — 10 Questions

Consolidate your understanding before moving on.

Score: 0 / 10

Question 01 — Multiple Choice

A matrix A is invertible if and only if:

det(A) = 1
det(A) ≠ 0
tr(A) ≠ 0
A is symmetric

Question 02 — Fill In

For A = [[2,0],[0,4]], det(A⁻¹) = .

det(A) = 8 and det(A⁻¹) = 1/det(A).

Question 03 — Multiple Choice

(AB)⁻¹ equals:

A⁻¹B⁻¹
B⁻¹A⁻¹
AB
(BA)⁻¹A

Question 04 — Fill In

For A = [[1,0],[0,5]], the (2,2) entry of A⁻¹ is .

Diagonal matrix inverse: reciprocal each diagonal entry.

Question 05 — Multiple Choice

Which matrix has NO inverse?

[[1,0],[0,1]]
[[2,3],[4,6]]
[[5,0],[0,3]]
[[1,2],[3,7]]

Question 06 — Multiple Choice

The Gauss-Jordan method for finding A⁻¹ starts by forming:

[I | A]
[A | I]
[A | A]
[A | 0]

Question 07 — Fill In

For A = [[3,1],[5,2]], det(A) = .

det([[a,b],[c,d]]) = ad − bc.

Question 08 — Multiple Choice

If A has eigenvalue λ = 0, then:

A⁻¹ has eigenvalue 0
A is singular and has no inverse
A = 0
A⁻¹ exists and equals A

Question 09 — Fill In

For A = [[k, 0],[0, k]] (k ≠ 0), the (1,1) entry of A⁻¹ is .

A = kI, so A⁻¹ = (1/k)I. The (1,1) entry is 1/k.

Question 10 — Multiple Choice

If A = PDP⁻¹, what is A⁻¹?

P⁻¹D⁻¹P
PD⁻¹P⁻¹
P⁻¹DP
D⁻¹