Inverse & Invertibility
When a matrix can be undone: the 2x2 inverse formula and the chain of equivalent conditions that all decide whether an inverse exists.
What you'll learn
- The inverse A⁻¹ satisfies AA⁻¹ = I — it undoes the transformation
- The 2x2 inverse formula: 1/(ad − bc) times the swap-and-negate matrix
- The equivalence chain: invertible ⇔ det ≠ 0 ⇔ full rank ⇔ 0 not an eigenvalue
- Products invert in reverse order: (AB)⁻¹ = B⁻¹A⁻¹
Before you start
Last lesson said whether you could undo a matrix. This one is about doing the
undoing. Think of A as a transformation that takes every point in space to a
new spot — rotating, stretching, shearing. The inverse A⁻¹ is the
transformation that puts every point back exactly where it started, like
rewinding a video. It only exists when A didn’t destroy any information along
the way. This is more than theory: the closed-form solution for linear regression
literally inverts a matrix, and “no inverse” is the algebra behind a model that
can’t be fit uniquely.
Intuition — undoing a transformation
The inverse of A is the matrix A⁻¹ for which applying one and then the other
leaves you where you began:
A·A⁻¹ = A⁻¹·A = I (I = the identity matrix)This only works if A did not destroy any information. If A flattened the plane
onto a line (collapsing area to zero), there is no way to recover the lost
dimension — so no inverse exists. That is exactly the det(A) = 0 case from the
previous lesson. A matrix is invertible precisely when it keeps space “full.”
Stretch and rotate the matrix below to deform the unit square, then mentally ask
what transformation would put the parallelogram back to a square — that is what
A⁻¹ does. If you can collapse the matrix to a line (determinant zero), nothing
can undo it; the inverse doesn’t exist.
Mechanism — the 2x2 inverse formula
For a 2x2 matrix there is a formula worth memorising. Swap the diagonal entries, negate the off-diagonal entries, and divide by the determinant:
The determinant ad − bc sits in the denominator — which is why a zero
determinant has no inverse: you would be dividing by zero.
The invertibility equivalence chain
For an n×n matrix A, all of the following statements are either all true or
all false together. They are different views of the same property:
In words: A is invertible if and only if det(A) ≠ 0, if and only if A has
full rank, if and only if its rows/columns are linearly independent, if and
only if 0 is not an eigenvalue, if and only if Ax = 0 has only the trivial
solution x = 0. Recognising any one of these in a question lets you conclude all
the others.
One more product rule, the mirror of det(AB) = det(A)·det(B): the inverse of a
product reverses the order.
(AB)⁻¹ = B⁻¹A⁻¹ (order reverses — like taking off socks then shoes)How GATE asks this
Two flavours. An MCQ gives a list of statements (“A has full rank”, “0 is an eigenvalue of A”, “the columns are dependent”) and asks which guarantee — or rule out — invertibility; you answer straight from the equivalence chain. Or a NAT asks you to compute a specific entry of a 2x2 inverse, where you apply the swap-negate-divide formula and read off one number.
Worked example
Invert A = [[2, 1], [1, 1]] and verify the result.
Step 1 — determinant.
det(A) = (2)(1) − (1)(1) = 2 − 1 = 1It is nonzero, so the inverse exists.
Step 2 — apply the formula. Swap the diagonal (2 and 1), negate the
off-diagonal entries, divide by det = 1:
A⁻¹ = (1/1) · [[ 1, −1], = [[ 1, −1],
[−1, 2]] [−1, 2]]Step 3 — verify A·A⁻¹ = I.
[[2,1],[1,1]] · [[1,−1],[−1,2]]
= [[2·1 + 1·(−1), 2·(−1) + 1·2],
[1·1 + 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.