Matrix 2x2 Inverse

2×2 inverse: A⁻¹ = (1/det) · [[d, -b], [-c, a]]

Take A = [[a, b], [c, d]] with det(A) = ad - bc. The inverse works out to A⁻¹ = (1/det) · [[d, -b], [-c, a]]. In plain terms, you swap the diagonal entries, flip the sign on the off-diagonal ones, and divide the whole thing by the determinant. The matrix is invertible if and only if det ≠ 0. When the determinant is zero, A is singular and there's simply no inverse. Here's a worked example with A = [[1, 2], [3, 4]]: det = 1·4 - 2·3 = -2, which gives A⁻¹ = (-1/2) · [[4, -2], [-3, 1]] = [[-2, 1], [1.5, -0.5]]. You can confirm it by checking that A · A⁻¹ = I. The inverse undoes whatever linear transformation A applied, so it lets you solve systems Ax = b ⇒ x = A⁻¹b and run geometric transformations in reverse.

Applications

It shows up when you solve a 2×2 linear system Ax = b head-on, in computer graphics where you need to undo a rotation, scale or shear, and whenever you switch between 2D coordinate systems. Closed-form linear regression on 2 features relies on it too (β = (XᵀX)⁻¹Xᵀy, the formula sklearn uses), along with 2D Kalman filters and plane physics problems involving forces and accelerations.

FAQ

What if det = 0? Then there's no inverse. The two rows are linearly dependent, meaning one is just a multiple of the other, and the matrix squashes the whole plane down onto a single line.

Does (AB)⁻¹ = A⁻¹B⁻¹? No. The right identity is (AB)⁻¹ = B⁻¹A⁻¹, with the order flipped. A couple of others worth remembering: (A⁻¹)⁻¹ = A and det(A⁻¹) = 1/det(A).

Is solving Ax = b with the inverse a good idea? At 2×2 it's perfectly fine. Once the matrices get large, though, computing A⁻¹ and multiplying tends to accumulate error, so Gaussian elimination or LU decomposition holds up better numerically.