Properties of Matrix Inverses | KEVOS® Mathematics

Q: What is the inverse of $A^{k}$?

It is $(A^{-1})^{k}$, by induction from the reversal rule applied to $k$ equal factors. This makes the notation $A^{-k}$ unambiguous and extends the usual index laws to all integer exponents for an invertible matrix.

Overview

Once the inverse of a matrix is defined, the practical question is how it behaves under the operations already available: multiplication, transposition, conjugation and scalar multiplication. The answers form a compact algebra that lets an inverse be rewritten or simplified without ever returning to a row reduction. In a derivation of any length, applying these rules is the difference between two lines of symbol manipulation and a page of arithmetic.

Every proof follows the same pattern, and the pattern is worth learning independently of the results. To show that a matrix $C$ is the inverse of a matrix $X$ , one does not solve for it. One proposes $C$ , computes $C X$ and $X C$ , and checks that both equal the identity. Uniqueness of the inverse then guarantees that the verified candidate is the inverse, so nothing further is required. This candidate-and-verify style avoids division entirely, which matters because matrices cannot be divided and because no entry can be assumed non-zero.

The single result that most often trips up practitioners is the reversal rule for products: $(AB)^{-1} = B^{-1}A^{-1}$ , not $A^{-1}B^{-1}$ . The order reverses because matrix multiplication does not commute, and the everyday analogy is exact — socks go on before shoes, so shoes come off before socks. The rule extends by induction to any number of factors, and it is the reason that inverting a chain of coordinate transformations traverses the chain backwards.

Equally important is what is absent from the list. There is no rule for the inverse of a sum. In general $(A + B)^{-1} \ne A^{-1} + B^{-1}$ , and worse, the sum of two invertible matrices need not be invertible at all. Any argument that distributes inversion over addition is invalid; the correct tools for a perturbed inverse are the Sherman-Morrison and Woodbury identities, which look nothing like a distributive law.

Definition

Matrix Inverse is Unique

MIU

If a square matrix $A$ has an inverse, that inverse is unique. If $B$ and $C$ both satisfy the definition then $B = BI_n = B(AC) = (BA)C = I_nC = C$ .

This result is what licenses the notation $A^{-1}$ and the definite article. It also means that verifying a candidate is a complete proof: anything that behaves like an inverse is the inverse.

Socks and Shoes

If $A$ and $B$ are invertible matrices of size $n$ , then the product $A B$ is invertible and $(AB)^{-1} = B^{-1}A^{-1}$ . The order of the factors reverses.

The proof verifies both $(B^{-1}A^{-1})(AB) = I_n$ and $(AB)(B^{-1}A^{-1}) = I_n$ by regrouping with associativity. The invertibility of the product is part of the conclusion, not an assumption.

Inverse of an Inverse

MIMI

If $A$ is invertible then $A^{-1}$ is invertible and $(A^{-1})^{-1} = A$ . Inversion is an involution on the set of invertible matrices of a given size.

Inverse of a Transpose

MIT

If $A$ is invertible then $A^{t}$ is invertible and $(A^{t})^{-1} = (A^{-1})^{t}$ . Transposition and inversion commute, so the two operations may be applied in either order.

The proof uses the reversal rule for transposes, $(XY)^{t} = Y^{t}X^{t}$ , together with the fact that $I_n$ is symmetric. The same argument applied to the conjugate gives $(A^{*})^{-1} = (A^{-1})^{*}$ for the adjoint.

Inverse of a Scalar Multiple

MISM

If $A$ is invertible and $\alpha$ is a non-zero scalar, then $\alpha A$ is invertible and $(\alpha A)^{-1} = \dfrac{1}{\alpha}A^{-1}$ . The requirement $\alpha \ne 0$ is essential: $0 \cdot A$ is the zero matrix, which is singular for every $n \ge 1$ .

Concepts

Candidate and verify: the standard proof pattern

Every result on this page is proved the same way. Given an expression $X$ whose inverse is wanted, propose a candidate $C$ built from known inverses, then compute $C X$ and $X C$ and show both are $I_n$ . Because the inverse is unique, a successful verification ends the argument. The pattern never divides, never cancels and never assumes an entry is non-zero, which is exactly why it is robust; it also transfers verbatim to linear transformations and to any other setting with an associative product and an identity.

Why the order reverses

Regrouping shows the mechanism. Writing $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}I_nB = B^{-1}B = I_n$ , the inner pair annihilates first and the outer pair follows. Had the candidate been $A^{-1}B^{-1}$ , the interior product would be $B^{-1}A$ , which simplifies to nothing. Interpreting $A$ and $B$ as successive coordinate transformations makes the rule intuitive: to undo "apply $B$ , then apply $A$ ", one must first undo $A$ and then undo $B$ .

Extension to chains and powers

Induction on the number of factors gives $(A_1A_2\cdots A_k)^{-1} = A_k^{-1}\cdots A_2^{-1}A_1^{-1}$ whenever every factor is invertible. Specialising to equal factors gives $(A^{k})^{-1} = (A^{-1})^{k}$ , which justifies defining $A^{-k}$ as either side without ambiguity. With that convention the familiar index laws $A^{p}A^{q} = A^{p+q}$ and $(A^{p})^{q} = A^{pq}$ hold for all integers $p$ and $q$ when $A$ is invertible.

Transposition, conjugation and the adjoint

Because $(XY)^{t} = Y^{t}X^{t}$ reverses order and inversion reverses order, applying both restores the original order: from $AA^{-1} = I_n$ one obtains $(A^{-1})^{t}A^{t} = I_n^{t} = I_n$ , and symmetrically for the other product. Entrywise conjugation preserves order, giving $(\overline{A})^{-1} = \overline{A^{-1}}$ , and combining the two yields $(A^{*})^{-1} = (A^{-1})^{*}$ . A practical corollary: the inverse of a symmetric matrix is symmetric, and the inverse of a Hermitian matrix is Hermitian.

The group of invertible matrices

Taken together, these results say that the invertible matrices of size $n$ are closed under multiplication, contain the identity, are associative and contain an inverse for each member. That is precisely the definition of a group, the general linear group. Nothing here depends on commutativity, and indeed the group is non-abelian for $n \ge 2$ . Recognising the structure explains why so many of these proofs generalise unchanged to invertible linear transformations and to change-of-basis matrices.

The rule that does not exist

Inversion does not distribute over addition. Two invertible matrices can sum to a singular matrix — take $B = -A$ , giving the zero matrix — so $(A + B)^{-1}$ may fail to exist at all, and when it does exist it generally differs from $A^{-1} + B^{-1}$ . The correct machinery for updating an inverse after a low-rank change is the Sherman-Morrison formula and its Woodbury generalisation, which express $(A + \vec{u}\vec{v}^{t})^{-1}$ as $A^{-1}$ plus a rank-one correction. That structure is the reason no simple additive rule can hold.

Equations

Uniqueness of the inverse

EQ-PMI-01

B = B I_{n} = B (A C) = (B A) C = I_{n} C = C

If $B$ and $C$ are both inverses of $A$ they are equal, so the symbol $A^{-1}$ denotes one specific matrix.

Reversal rule for products

EQ-PMI-02

(A B)^{- 1} = B^{- 1} A^{- 1}

Valid whenever $A$ and $B$ are invertible of the same size; the product $A B$ is then invertible as part of the conclusion.

Verification of the reversal rule

EQ-PMI-03

(B^{- 1} A^{- 1}) (A B) = B^{- 1} (A^{- 1} A) B = B^{- 1} I_{n} B = B^{- 1} B = I_{n}

Only associativity and the identity property are used. The companion computation $(AB)(B^{-1}A^{-1}) = I_n$ completes the proof.

Inverse of an inverse and of a chain

EQ-PMI-04

(A^{- 1})^{- 1} = A, (A_{1} A_{2} \dots A_{k})^{- 1} = A_{k}^{-} \dots A_{2}^{-} A_{1}^{-}

Inversion is an involution, and the reversal rule extends to any finite number of invertible factors by induction.

Inverse of a transpose, conjugate and adjoint

EQ-PMI-05

(A^{t})^{- 1} = (A^{- 1})^{t}, (\bar{A})^{- 1} = \bar{A^{- 1}}, (A^{*})^{- 1} = (A^{- 1})^{*}

Inversion commutes with all three operations, so a symmetric matrix has a symmetric inverse and a Hermitian matrix has a Hermitian inverse.

Inverse of a scalar multiple

EQ-PMI-06

(α A)^{- 1} = \frac{1}{α} A^{- 1}, α \neq 0

Scaling a matrix scales its inverse by the reciprocal. The exclusion of $\alpha = 0$ is essential, since the zero matrix is singular.

Failure of an additive rule

EQ-PMI-07

(A + B)^{- 1} \neq A^{- 1} + B^{- 1} in general

Inversion does not distribute over addition, and $A + B$ need not be invertible even when $A$ and $B$ both are.

Variable Definitions

Symbols used on this page
Symbol	Name	Meaning	Domain / type
$A$	Invertible matrix	A square matrix known to possess an inverse	n x n complex matrix
$B$	Second invertible matrix	A square matrix of the same size, also invertible	n x n complex matrix
$A^{- 1}$	Inverse of $A$	The unique matrix with $AA^{-1} = A^{-1}A = I_n$	n x n complex matrix
$A^{t}$	Transpose of $A$	The matrix with rows and columns interchanged	n x n complex matrix
$A^{*}$	Adjoint of $A$	Conjugate transpose $\overline{A}^{\,t}$ ; equals $A^{t}$ when $A$ is real	n x n complex matrix
$Non-zero scalar$	Non-zero scalar	Multiplier applied to a matrix before inversion	nonzero complex number
$I_{n}$	Identity matrix	Multiplicative unit of size $n$ , and the target of every verification	n x n matrix
$k$	Number of factors or power	Length of a product chain, or the exponent in $A^{k}$	positive integer
$n$	Size	Common row and column count of all matrices involved	positive integer

Worked Numerical Example

Problem statement

A planar sensor frame is related to the machine frame by a shear $B$ followed by a scaling-and-skew $A$ . Verify the reversal rule on the composite transformation, confirm the transpose and scalar rules, and demonstrate that no additive rule holds.

State the two transformations and their inverses
Both matrices have determinant $1$ , so the two-by-two formula gives their inverses immediately: swap the diagonal, negate the off-diagonal.
$A = [\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix}], A^{- 1} = [\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix}], B = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}], B^{- 1} = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}]$
Form the composite and invert it directly
The composite is $A B$ , meaning $B$ applied first. Its determinant is $2 \cdot 2 - 3 \cdot 1 = 1$ , so the closed formula applies again.
$A B = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}], (A B)^{- 1} = [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}]$
Compare with the reversal rule
Computing $B^{-1}A^{-1}$ : the first row is $1(1) + (-1)(-1) = 2$ and $1(-1) + (-1)(2) = -3$ ; the second row is $0(1) + 1(-1) = -1$ and $0(-1) + 1(2) = 2$ . This matches the direct inversion exactly.
$B^{- 1} A^{- 1} = [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}] = (A B)^{- 1}$
Show that the wrong order fails
The other arrangement gives $A^{-1}B^{-1}$ with first row $1(1) + (-1)(0) = 1$ and $1(-1) + (-1)(1) = -2$ , second row $-1$ and $1 + 2 = 3$ . This is a different matrix, so the order in the reversal rule is not a stylistic preference.
$A^{- 1} B^{- 1} = [\begin{matrix} 1 & - 2 \\ - 1 & 3 \end{matrix}] \neq (A B)^{- 1}$
Verify the transpose rule
Transposing $B$ gives a lower-triangular matrix whose inverse is again obtained from the closed formula. It agrees with the transpose of $B^{-1}$ , as the rule predicts.
$B^{t} = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}], (B^{t})^{- 1} = [\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}] = (B^{- 1})^{t}$
Verify the scalar rule
Scaling $A$ by $3$ gives determinant $18 - 9 = 9$ , so the inverse is $\tfrac{1}{9}\begin{bmatrix} 3 & -3 \\ -3 & 6 \end{bmatrix}$ . That is exactly $\tfrac{1}{3}A^{-1}$ , confirming the reciprocal scaling.
$(3 A)^{- 1} = [\begin{matrix} \frac{1}{3} & - \frac{1}{3} \\ - \frac{1}{3} & \frac{2}{3} \end{matrix}] = \frac{1}{3} A^{- 1}$
Demonstrate the failure of an additive rule
The sum $A + B$ has determinant $6 - 2 = 4$ , so it happens to be invertible here. Its inverse is nonetheless nothing like the sum of the inverses — the two differ by a factor of four, and no scaling could rescue the identity in general.
$(A + B)^{- 1} = [\begin{matrix} \frac{1}{2} & - \frac{1}{2} \\ - \frac{1}{4} & \frac{3}{4} \end{matrix}], A^{- 1} + B^{- 1} = [\begin{matrix} 2 & - 2 \\ - 1 & 3 \end{matrix}]$
Confirm that sums can be singular
Taking the second matrix to be $-A$ instead makes the sum the zero matrix, which has no inverse for any size. Two invertible matrices can therefore sum to a matrix outside the invertible set altogether, which is the structural reason no additive rule can be formulated.
$A + (- A) = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Result

The reversal rule reproduces the directly computed inverse of the composite transformation, while the reversed-order candidate does not. Practically, this means that undoing a chain of frame transformations requires traversing the chain backwards: sensor-to-machine inverts to machine-to-sensor by inverting each stage and reversing their order. The transpose and scalar rules hold as stated, and the additive rule fails even on this benign pair.

Applications & Industry Use

Robotics and kinematics

Inverting a chain of frame transformations

A manipulator pose is built by composing link transformations from base to tool. Mapping a tool-frame quantity back to the base frame inverts the composite, which by the reversal rule means inverting each link transformation and multiplying them in reverse order — the standard implementation of the inverse kinematic chain.

Computer graphics

Model-view-projection inversion

Screen-space coordinates are unprojected to world space by inverting the composed model, view and projection matrices. Engines apply the reversal rule rather than inverting the composite, because each factor has structure — orthogonal rotation, diagonal scale, simple perspective — that makes its individual inverse nearly free.

Control systems

Similarity transformations of state-space models

Changing state coordinates replaces the system matrix $A$ by $T^{-1}AT$ . The rules on this page show that composing two changes of coordinates corresponds to multiplying the transformation matrices, and that reverting to the original coordinates uses the inverse in the mirrored position — the algebra underlying controllable and modal canonical forms.

Statistics and data analysis

Precision matrices under rescaling

Standardising variables replaces a covariance matrix $\Sigma$ by $D\Sigma D$ with $D$ diagonal. The scalar and product rules give the corresponding precision matrix as $D^{-1}\Sigma^{-1}D^{-1}$ , so the correlation structure can be recovered without re-inverting. Symmetry of the inverse of a symmetric matrix is used implicitly throughout this work.

Cryptography

Composed key matrices

Encrypting with one key matrix and then another is equivalent to encrypting with the product. Decryption applies the inverses in reverse order, and the composition rule shows that the effective key of a two-stage matrix cipher is a single matrix — a structural weakness that motivates non-linear rounds in modern designs.

Structural engineering

Assembly and constraint transformations

Stiffness matrices are transformed between local and global axes by congruence, $K_{g} = R^{t}K_{l}R$ with $R$ orthogonal. The transpose rule guarantees that the transformed flexibility matrix is the congruence transform of the local flexibility, so symmetry is preserved exactly and can be used as a validation check on assembly code.

Design Considerations

Simplify symbolically before computing numerically

An expression such as $(AB)^{-1}C(BA)^{-1}$ can often be reduced to a form requiring one inversion instead of two. Every inversion avoided saves $O(n^3)$ operations and one opportunity for rounding error to enter. Perform the algebra first, then commit to arithmetic.

Preserve structure through the inverse

Symmetry, Hermitian symmetry and realness all survive inversion. Exploit this both for storage — a symmetric inverse needs only its triangle — and as a validation check, since a computed inverse of a symmetric matrix that is visibly asymmetric indicates a coding error rather than rounding.

Never distribute inversion over addition

There is no rule for $(A+B)^{-1}$ , and the sum of invertible matrices may be singular. When an inverse must be updated after a small change, use the Sherman-Morrison identity for a rank-one update or the Woodbury identity for a low-rank one; both cost $O(n^2)$ against the $O(n^3)$ of a fresh inversion.

Invert the factors, not the product

When a composite is built from factors with exploitable structure — orthogonal, triangular, diagonal, permutation — the reversal rule lets each factor be inverted in its own cheap way. Inverting the assembled composite discards that structure and pays the full general cost.

Watch conditioning through a product

The condition number of a product can be as large as the product of the condition numbers. A chain of individually acceptable transformations can therefore compose into a badly conditioned one, so conditioning should be assessed on the composite that is actually inverted, not factor by factor.

Guard the non-zero scalar hypothesis

The rule $(\alpha A)^{-1} = \alpha^{-1}A^{-1}$ silently fails when $\alpha = 0$ . In code that scales matrices by a computed gain, a gain that can pass through zero must be handled explicitly rather than left to produce an infinity or a not-a-number at the moment of inversion.

Standards & Codes

Notation, interchange and numerical standards that govern how this material is written down, stored and computed in production systems.

Applicable standards, conventions and reference implementations
Reference	Title	Relevance to this topic
`ISO 80000-2`	Quantities and units — Part 2: Mathematics	Defines the notation used here for transpose, conjugate, adjoint and inverse, and the convention that superscripts compose from the inside out.
`IEEE 754-2019`	IEEE Standard for Floating-Point Arithmetic	Governs the behaviour of the scalar rule at the boundary: dividing by a computed $\alpha$ that underflows to zero yields an infinity rather than an error, so the non-zero hypothesis must be enforced by the caller.
`LAPACK / BLAS reference`	Linear Algebra PACKage reference implementation	Provides structure-aware inversion routines — `xPOTRI` for symmetric positive definite, `xTRTRI` for triangular — which are the practical expression of the rule that inversion preserves structure.
`ISO/IEC 14882`	Programming languages — C++	Underlies expression-template libraries such as Eigen, in which an expression like `(A*B).inverse()` is rewritten at compile time using precisely these algebraic identities before any arithmetic is performed.
`ISO/IEC 40314`	Mathematical Markup Language (MathML) Version 3.0	Encodes stacked superscripts such as $(A^{-1})^{t}$ unambiguously, so the order of the two operations is machine-readable rather than a matter of typographic interpretation.

Material Selection

For a mathematical topic, "material" is the numeric representation: the scalar field, storage format and precision the computation is built from.

Representation and precision selection
Representation	Select when	Trade-off
Exact rational arithmetic	Verifying an algebraic identity or producing a reference result against which a numerical implementation is tested.	Identities hold exactly, so any discrepancy is a genuine error rather than rounding, but the cost is prohibitive beyond small sizes.
IEEE 754 binary64 (double precision)	Routine engineering computation involving composites, transposes and scalings of moderate size.	The algebraic identities hold only to within rounding, so equality tests must become norm comparisons against a tolerance.
Symmetric packed or triangular storage	The matrix and hence its inverse are symmetric or Hermitian, as for covariance and stiffness matrices.	Halves memory and enables specialised routines, but requires that symmetry be enforced rather than merely expected; an asymmetric input silently violates the storage contract.
Real versus complex scalars	Deciding whether transpose or adjoint is the correct partner operation in a derivation.	Real arithmetic is cheaper and $A^{*} = A^{t}$ , but a derivation written with transposes silently breaks when complex data appears; writing adjoints from the outset costs nothing on real inputs.
Orthogonal or unitary factors	A composite can be arranged so that its factors are orthogonal or unitary, as in QR and singular value decompositions.	Each such factor inverts by transposition at zero arithmetic cost and contributes a condition number of exactly one, at the cost of an initial factorisation.
Low-rank update representation	An inverse must track a matrix that changes by a rank-one or low-rank correction, as in recursive least squares and quasi-Newton methods.	Reduces each update from $O(n^3)$ to $O(n^2)$ , but repeated updates accumulate error and periodic refactorisation is needed to maintain accuracy.

Manufacturing Notes

Implementation notes — how the result is actually produced by hand, by algorithm and by library, including cost and numerical behaviour.

Algebraic simplification is the cheapest optimisation

Rewriting $(AB)^{-1}$ as $B^{-1}A^{-1}$ costs nothing at runtime and can eliminate an entire $O(n^3)$ inversion when one of the factors already has a known inverse. Expression-template libraries perform this rewriting automatically; hand-written code must do it deliberately.

Verifying the identities numerically

To test the reversal rule in floating point, compare $\| (AB)^{-1} - B^{-1}A^{-1} \|$ against a tolerance of order $\varepsilon$ times the product of the operand norms. Exact equality will not hold and its absence is not evidence of an error; a residual many orders of magnitude above the tolerance is.

Structure-preserving routines

For symmetric positive definite matrices, a Cholesky factorisation followed by xPOTRI inverts at roughly half the cost of the general routine and returns a symmetric result by construction. For triangular matrices, xTRTRI inverts in about $\tfrac{1}{3}n^3$ operations. Selecting the routine that matches the structure is worth more than any micro-optimisation.

Rank-one updates in practice

The Sherman-Morrison identity expresses $(A + \vec{u}\vec{v}^{t})^{-1}$ as $A^{-1}$ minus a rank-one term built from $A^{-1}\vec{u}$ and $\vec{v}^{t}A^{-1}$ . Recursive least squares, Kalman filtering and quasi-Newton optimisation all rely on it. Its accuracy degrades as the denominator $1 + \vec{v}^{t}A^{-1}\vec{u}$ approaches zero, which is exactly the neighbourhood in which the updated matrix approaches singularity.

Checking a chain by round trip

For a composite of many transformations, the most informative check is a round trip: apply the composite and then the composed inverse to a set of test vectors and measure the deviation from the input. This exercises the reversal rule end to end and catches an ordering error immediately, which a norm comparison of matrices can obscure.

Failure Modes & Common Mistakes

Failure modes, root causes and prevention
Failure mode / mistake	Impact	Root cause	Prevention & detection
Writing $(AB)^{-1} = A^{-1}B^{-1}$	high	Applying scalar intuition, where multiplication commutes, to matrices where it does not.	Memorise the reversal rule with the socks-and-shoes image and verify order on a small non-commuting pair whenever a derivation depends on it.
Distributing inversion over a sum	high	Treating inversion as a linear operation and writing $(A+B)^{-1} = A^{-1} + B^{-1}$ .	There is no such rule. For an additive perturbation use the Sherman-Morrison or Woodbury identity, and check first whether $A + B$ is invertible at all.
Assuming a product of matrices is invertible	medium	Applying the reversal rule without checking that both factors are invertible; if either is singular the product is singular.	Verify invertibility of each factor before invoking the rule. The invertibility of the product is a conclusion of the theorem, not an independent fact.
Confusing $(A^{-1})^{t}$ with $(A^{t})^{-1}$ in a proof	low	Assuming the two differ because superscripts do not usually commute.	They are equal; the theorem is precisely that these two operations commute. The genuine ordering hazard lies with products, not with transposition.
Ignoring the non-zero scalar hypothesis	medium	Applying $(\alpha A)^{-1} = \alpha^{-1}A^{-1}$ where $\alpha$ is a computed quantity that can vanish.	Guard the scaling factor explicitly. In floating point an underflowed scalar produces an infinity rather than an exception, which propagates silently.
Cancelling a non-invertible factor	high	Deducing $B = C$ from $AB = AC$ when $A$ is singular; the cancellation requires $A^{-1}$ to exist.	Cancellation is left multiplication by $A^{-1}$ in disguise. State the invertibility of $A$ before performing it, or work with the null space instead.
Testing identities with exact equality in code	low	Asserting that two floating-point matrices agree entry for entry after algebraically equivalent computations.	Compare norms of differences against a tolerance scaled by machine epsilon, the matrix size and the operand norms.
Using transpose where adjoint is required	medium	Carrying a real-valued derivation into complex data, where $A^{t} \ne A^{*}$ .	Write adjoints throughout. On real matrices the two coincide, so nothing is lost, and complex inputs then work without modification.

FAQs

Why does the order reverse in

(AB)^{-1} = B^{-1}A^{-1}

Because the inner factors must meet first when the candidate is multiplied against $A B$ : $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B$ collapses cleanly, while $A^{-1}B^{-1}$ leaves the mismatched product $B^{-1}A$ in the middle. Interpreted as transformations, undoing a sequence means undoing the last step first.

Is the product of two invertible matrices always invertible?

Yes, and the reversal-rule theorem proves it as part of the same argument that produces the formula. The converse also holds for square matrices: if a product is invertible then both factors must be, since a product involving a singular factor is singular.

Does the inverse of a symmetric matrix stay symmetric?

Yes. From $(A^{t})^{-1} = (A^{-1})^{t}$ and $A^{t} = A$ it follows that $(A^{-1})^{t} = A^{-1}$ . The same argument with adjoints shows a Hermitian matrix has a Hermitian inverse, which is why covariance and stiffness inverses can be stored in packed triangular form.

What is the inverse of

A^{k}

It is $(A^{-1})^{k}$ , by induction from the reversal rule applied to $k$ equal factors. This makes the notation $A^{-k}$ unambiguous and extends the usual index laws to all integer exponents for an invertible matrix.

(A+B)^{-1}

is not

A^{-1} + B^{-1}

, what is it?

There is no general closed form. When $B$ has low rank the Woodbury identity gives an exact expression as $A^{-1}$ plus a correction whose rank matches that of $B$ ; the rank-one case is the Sherman-Morrison formula. For a general $B$ the only route is to invert the sum directly.

Can I cancel a matrix from both sides of an equation?

Only if it is invertible. From $AB = AC$ one may deduce $B = C$ by left-multiplying by $A^{-1}$ , but if $A$ is singular the conclusion is false in general — a singular $A$ admits distinct $B$ and $C$ whose difference lies in its null space.

Do these rules apply to linear transformations as well as matrices?

Yes, without change. Every proof here uses only associativity of composition, the existence of an identity and the definition of an inverse, all of which hold for invertible linear transformations. That is why the invertible matrices of a given size form a group under multiplication.

References

Beezer, R. A. A First Course in Linear Algebra, Version 0.70. University of Puget Sound, 2006. Section MISLE, Subsection PMI. Licensed under the GNU Free Documentation License v1.2.
ISO 80000-2:2019, Quantities and units — Part 2: Mathematics. International Organization for Standardization.
Horn, R. A. and Johnson, C. R. Matrix Analysis, 2nd edition. Cambridge University Press, 2012.
Golub, G. H. and Van Loan, C. F. Matrix Computations, 4th edition. Johns Hopkins University Press, 2013.
Higham, N. J. Accuracy and Stability of Numerical Algorithms, 2nd edition. Society for Industrial and Applied Mathematics, 2002.
Anderson, E. et al. LAPACK Users' Guide, 3rd edition. Society for Industrial and Applied Mathematics, 1999.

AI Suggested Questions

Prove by induction that the inverse of a product of $k$ invertible matrices reverses the order of all $k$ factors.
Derive the Sherman-Morrison formula and show precisely where an additive rule for inverses would have to fail.
Construct two invertible $3 \times 3$ matrices whose sum is singular, and explain the geometric picture.
Show that the set of invertible $n \times n$ matrices forms a non-abelian group, and identify its identity and inverses.
For a real matrix, when does $A^{-1} = A^{t}$ hold, and what does that class of matrices represent geometrically?
Explain how a compiler-level expression template can use the reversal rule to eliminate a matrix inversion at compile time.

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