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Inverse of product (AB)⁻¹

Short answer

Since (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I and (B⁻¹A⁻¹)(AB) = B⁻¹(A⁻¹A)B = B⁻¹B = I, B⁻¹A⁻¹ is the inverse of AB. By uniqueness of inverse, (AB)⁻¹ = B⁻¹A⁻¹.

If A and B are two non-singular matrices, then prove that (AB)⁻¹ = B⁻¹A⁻¹.

Non-singular matrices (|A| ≠ 0, |B| ≠ 0) have inverses A⁻¹ and B⁻¹, with AA⁻¹ = A⁻¹A = I. The inverse of a matrix, when it exists, is unique.

Proof

First, AB is non-singular, since |AB| = |A||B| ≠ 0, so (AB)⁻¹ exists.

Multiply AB by B⁻¹A⁻¹ and use associativity:

(AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹
             = A I A⁻¹
             = AA⁻¹
             = I

and in the other order:

(B⁻¹A⁻¹)(AB) = B⁻¹(A⁻¹A)B
             = B⁻¹ I B
             = B⁻¹B
             = I

So B⁻¹A⁻¹ is a matrix whose product with AB, on either side, is I. By the definition and uniqueness of the inverse,

(AB)⁻¹ = B⁻¹A⁻¹. ∎

(This is the "reversal law" for inverses; it extends to any number of factors: (ABC)⁻¹ = C⁻¹B⁻¹A⁻¹.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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