Is the product of two invertible matrices invertible?
When studying linear algebra, one of the most useful properties to remember is that invertibility is preserved under matrix multiplication. If you have two square matrices (A) and (B) that are each invertible, then their product (AB) is also invertible, and its inverse can be expressed neatly in terms of the inverses of the factors. This fact underpins many proofs, algorithms, and applications ranging from solving systems of linear equations to analyzing transformations in computer graphics. Below we explore why this holds, how to verify it step‑by‑step, and what it means in broader mathematical contexts.
Introduction
A matrix (M) is called invertible (or nonsingular) if there exists another matrix (M^{-1}) such that
[ MM^{-1}=M^{-1}M=I, ]
where (I) is the identity matrix of the same size. Still, the question “is the product of two invertible matrices invertible? ” asks whether the set of invertible matrices is closed under multiplication.
[ (AB)^{-1}=B^{-1}A^{-1}. ]
Notice the reversal of order—a crucial detail that often trips up beginners. The following sections break down the reasoning, provide a concrete verification method, and discuss the intuition behind the result And it works..
Why the Product Remains Invertible
Algebraic Proof
Assume (A) and (B) are (n\times n) invertible matrices. By definition, there exist matrices (A^{-1}) and (B^{-1}) satisfying
[ AA^{-1}=A^{-1}A=I \quad\text{and}\quad BB^{-1}=B^{-1}B=I. ]
Consider the product (AB). Multiply it on the right by (B^{-1}A^{-1}):
[ (AB)(B^{-1}A^{-1}) = A\bigl(BB^{-1}\bigr)A^{-1}=AIA^{-1}=AA^{-1}=I. ]
Similarly, multiply on the left:
[ (B^{-1}A^{-1})(AB)=B^{-1}\bigl(A^{-1}A\bigr)B=B^{-1}IB=B^{-1}B=I. ]
Since a matrix that has both a left and a right inverse is invertible, and that inverse is unique, we conclude that
[ (AB)^{-1}=B^{-1}A^{-1}. ]
Thus the product of two invertible matrices is invertible, and its inverse is obtained by reversing the order of the individual inverses.
Determinant Perspective
Another quick way to see the result uses determinants. Recall that a square matrix is invertible iff its determinant is non‑zero. The determinant of a product equals the product of the determinants:
[ \det(AB)=\det(A)\det(B). ]
If (\det(A)\neq0) and (\det(B)\neq0), then (\det(AB)\neq0) as well, guaranteeing that (AB) is invertible. Worth adding,
[ \det\bigl((AB)^{-1}\bigr)=\frac{1}{\det(AB)}=\frac{1}{\det(A)\det(B)}=\det(A^{-1})\det(B^{-1}), ]
which aligns with the algebraic inverse formula Turns out it matters..
Geometric Interpretation
Think of an invertible matrix as a linear transformation that bijectively maps (\mathbb{R}^n) onto itself—no squashing or collapsing of space. Because of that, applying (B) first reshapes space, then applying (A) reshapes it again. Because each step is reversible, the combined transformation can be undone by first reversing (A) (apply (A^{-1})) and then reversing (B) (apply (B^{-1})). This sequential “undo” explains why the inverse appears in the opposite order.
Step‑by‑Step Verification
If you prefer to check invertibility computationally, follow these steps:
-
Confirm invertibility of each factor
- Compute (\det(A)) and (\det(B)).
- If either determinant is zero, stop— the product cannot be invertible.
- Alternatively, attempt to find (A^{-1}) and (B^{-1}) via Gaussian elimination or a software routine; failure indicates singularity.
-
Form the product
- Compute (C = AB) using standard matrix multiplication (row‑by‑column).
-
Test the product
- Calculate (\det(C)).
- If (\det(C) \neq 0), then (C) is invertible.
- Optionally, verify that (C(B^{-1}A^{-1}) = I) and ((B^{-1}A^{-1})C = I) to confirm the inverse formula.
-
Derive the inverse (if needed)
- Use the rule ((AB)^{-1}=B^{-1}A^{-1}) directly, which is often faster than inverting (C) from scratch.
These steps are especially useful in numerical work where rounding errors may obscure exact zero determinants; in such cases, checking that the product of the computed inverses yields a matrix close to the identity provides a practical sanity check Simple, but easy to overlook..
Frequently Asked Questions
Q1: Does the result hold for non‑square matrices?
No. The concept of an inverse (two‑sided) is defined only for square matrices. For rectangular matrices one can discuss left‑ or right‑inverses, but the simple product rule ((AB)^{-1}=B^{-1}A^{-1}) does not apply in general.
Q2: What if only one matrix is invertible?
If (A) is invertible but (B) is singular, then (\det(B)=0) and consequently (\det(AB)=\det(A)\cdot0=0). Hence the product is singular (non‑invertible). The same holds when (B) is invertible and (A) is singular.
Q3: Can the product be invertible even if the factors are not?
No. If either factor lacks an inverse, its determinant is zero, forcing the determinant of the product to be zero as well. Therefore at least one factor must be singular for the product to be singular; invertibility of the product forces invertibility of each factor.
Q4: How does this property extend to more than two matrices?
By induction, the product of any finite collection of invertible matrices is invertible, and
[ (A_1A_2\cdots A_k)^{-1}=A_k^{-1}\cdots A_2^{-1}A_1^{-1}. ]
The order of inverses is the complete reverse of the original order.
Q5: Are there any exceptions in abstract algebra?
In any ring where the notion of invertibility (having a multiplicative inverse) is defined,
In any ring where the notion of invertibility (having a multiplicative inverse) is defined, the same reversal rule holds for products of invertible elements, provided the ring possesses a multiplicative identity. The proof relies only on associativity and the existence of two‑sided inverses, so it carries over verbatim from matrices to abstract ring elements: if (a) and (b) are invertible, then ((ab)^{-1}=b^{-1}a^{-1}) Most people skip this — try not to..
Exceptions arise when the ambient algebraic structure lacks one of these ingredients. In a rng (a ring without a unit) the very definition of an inverse is ambiguous; one may speak of quasi‑inverses or generalized inverses, but the simple product rule need not hold. In rings with zero divisors, an element can possess a one‑sided inverse without being two‑sided invertible; consequently, a product may admit a left (or right) inverse even though neither factor is two‑sided invertible, yet such one‑sided inverses do not satisfy the full ((ab)^{-1}=b^{-1}a^{-1}) identity. Beyond that, in non‑associative algebras (e.g., octonion algebras) the lack of associativity prevents the straightforward rearrangement of factors, and the inverse of a product is not generally given by the reversed product of inverses Practical, not theoretical..
Thus, while the rule ((AB)^{-1}=B^{-1}A^{-1}) is strong for square matrices over fields—or, more generally, for invertible elements in any unital associative ring—its applicability hinges on the presence of an identity element, associativity, and two‑sided invertibility. When any of these conditions fail, one must examine the specific algebraic context to determine whether a similar reversal property survives.
Conclusion
The invertibility of a matrix product is completely governed by the invertibility of its factors: a product is invertible if and only if each factor is invertible, and its inverse is obtained by reversing the order of the individual inverses. This principle extends naturally to any finite collection of matrices and, more abstractly, to invertible elements in unital associative rings. Computationally, checking the determinants (or attempting to compute the inverses) of the factors offers a quick preliminary test, while forming the product and verifying its determinant—or directly confirming that (B^{-1}A^{-1}) acts as a two‑sided inverse—provides a reliable numerical or symbolic verification. Understanding these conditions helps avoid pitfalls in both theoretical work and practical applications involving matrix computations Easy to understand, harder to ignore..
