Inverse & Invertibility

The last lesson told you whether a matrix could be undone — a non-zero determinant was the green light. This one builds the undoing. Picture A as something that takes every point of space to a new spot, rotating, stretching, shearing. The inverse A⁻¹ is the transformation that puts every point back exactly where it began, like rewinding a video. It exists only when A threw nothing away on the trip out.

Undoing a transformation

The inverse of A is the matrix A⁻¹ for which applying one and then the other leaves you where you started:

A·A⁻¹ = A⁻¹·A = I        (I = the identity, the do-nothing matrix)

This works only if A destroyed no information. If A flattened the plane onto a line — collapsing area to zero — the lost dimension can never be recovered, so no inverse exists. That is precisely the det(A) = 0 case from last lesson: a matrix is invertible exactly when it keeps space “full”. Stretch and rotate the matrix below, then picture the transformation that would push the parallelogram back to a square — that is what A⁻¹ does. Collapse it to a line and nothing can undo it.

The 2×2 formula

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

The determinant ad − bc sits in the denominator — which is exactly why a zero determinant has no inverse: you would be dividing by zero. The thread from last lesson ties off here.

One property, many faces

For an n×n matrix A, the following statements are either all true or all false together — different views of the single property of invertibility:

In words: A is invertible ⇔ det(A) ≠ 0 ⇔ A has full rank ⇔ its columns are independent ⇔ 0 is not an eigenvalue ⇔ Ax = 0 has only the trivial solution x = 0. Spot any one of these in a question and you have all the others for free. One more rule, the mirror of det(AB) = det(A)det(B): the inverse of a product reverses order, (AB)⁻¹ = B⁻¹A⁻¹ — like taking off socks after shoes.

A worked example

Invert A = [[2, 1], [1, 1]] and verify it.

det(A) = (2)(1) − (1)(1) = 2 − 1 = 1        non-zero → the inverse exists

swap the diagonal, negate the off-diagonal, divide by det = 1:
A⁻¹ = (1/1) · [[ 1, −1],   =  [[ 1, −1],
               [−1,  2]]        [−1,  2]]

verify A·A⁻¹:
[[2,1],[1,1]] · [[1,−1],[−1,2]] = [[2−1, −2+2], [1−1, −1+2]] = [[1,0],[0,1]] = I  ✓

The product is the identity, so the inverse is correct.

A question to carry forward

A general matrix shoves vectors around — rotating, shearing, turning them off their original line. But for almost every matrix there are a few special directions it leaves on their own line, merely stretching or shrinking them. Here is the thread onward: which directions does a matrix only stretch, by how much, and why does that pair — direction and stretch — unlock so much of what a matrix is?

In one breath

• The inverse A⁻¹ undoes A: A·A⁻¹ = A⁻¹·A = I; it exists only if A kept space full (threw no dimension away).
• 2×2 formula: A⁻¹ = 1/(ad−bc) · [[d, −b], [−c, a]] — swap the diagonal, negate the off-diagonal, divide by the determinant.
• The determinant is in the denominator, so det(A) = 0 (singular) → no inverse.
• The chain: invertible ⇔ det≠0 ⇔ full rank ⇔ columns independent ⇔ 0 not an eigenvalue ⇔ Ax=0 only trivially. Recognise one, get all.
• Product reverses: (AB)⁻¹ = B⁻¹A⁻¹ (mirror of the determinant product rule).

Practice