Linear AlgebraCore22 min read

Inverse Matrices

The matrix that undoes a matrix — and exactly when one exists

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Practise the full question bank:

01

The inverse reverses the transformation

If a matrix A is a transformation of space, its inverse A⁻¹ is the transformation that puts everything back. Do A then A⁻¹ and you return to where you started: A⁻¹A = A A⁻¹ = I, the identity.

This is the matrix version of dividing. There is no "÷ A"; instead you multiply by A⁻¹. It is what lets you solve Ax = b in one clean stroke — at least in principle.

02

When an inverse exists

Only square matrices can have inverses, and even then not all do. A matrix is invertible exactly when its determinant is non-zero. When the determinant is zero the matrix is singular: it collapsed space into a lower dimension, merging distinct inputs, and no transformation can un-merge them.

So "invertible", "det ≠ 0", "independent columns", "full rank" and "Ax = b always has a unique solution" are all the same statement — the good case from the previous lessons.

Keep an eye on the det readout. While it is non-zero the grid can be un-transformed back to the original — the matrix is invertible. Push it to 0 and the square collapses to a line: singular, no way back.

03

The 2×2 formula

For a 2×2 matrix there is a formula worth memorising: swap the diagonal, negate the off-diagonal, and divide by the determinant.

A = [ a b ]A⁻¹ = 1/(ad − bc) · [ d −b ][ c d ][ −ca ]undefined when ad − bc = 0 — a singular matrix has no inverse
04

Larger matrices, and a caution

Beyond 2×2 the practical method is Gauss–Jordan: write the matrix beside the identity as [A | I], row-reduce until the left side becomes I, and the right side has turned into A⁻¹.

One professional habit worth forming early: to solve Ax = b you rarely need the inverse at all. Elimination is faster and numerically safer than computing A⁻¹ and multiplying. Reach for the inverse when you truly need the matrix itself, not just one solution.

05

A worked example

Worked example

Find the inverse of A = [[4, 3], [2, 2]].

  1. Determinant: ad − bc = (4)(2) − (3)(2) = 8 − 6 = 2.Non-zero, so an inverse exists.
  2. Swap the diagonal and negate the off-diagonal: [[2, −3], [−2, 4]].The pattern from the 2×2 formula, before dividing.
  3. Divide by the determinant: A⁻¹ = (1/2)[[2, −3], [−2, 4]] = [[1, −1.5], [−1, 2]].Scale every entry by 1/det.

A⁻¹ = [[1, −1.5], [−1, 2]]. A quick check: A·A⁻¹ = [[1, 0], [0, 1]] = I.