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Determinants over a Ring: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

The Leibniz formula gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$

Example

For every commutative ring, $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc.$

Facts & Assumptions

Given: A $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ over a commutative ring.

[L1]

Determinant is the signed Leibniz sum over $S_2$ (For $n\ge1$ , the determinant over a commutative ring by the Leibniz formula, and $|\det A|$ for a real matrix).

[L2]

Sign is $(-1)$ to the inversion number (Inversions, inversion number, the sign $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$ , and even and odd permutations).

Verification

technique · direct

1.1

The two permutations of $2=\{0,1\}$ are the identity, with sign $1$ , and the swap, with one inversion and sign $-1$ .

L1L2

2.1

Their Leibniz monomials are $a_{00}a_{11}=ad$ and $a_{10}a_{01}=cb=bc$ , so the signed sum is $ad-bc$ .

step 1.1algebra∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

A full $3\times3$ Leibniz expansion lists all six permutations and their signs

Example

For $A=(a_{ij})\in M_3(R)$ , $\det A=a_{00}a_{11}a_{22}+a_{10}a_{21}a_{02}+a_{20}a_{01}a_{12}-a_{10}a_{01}a_{22}-a_{20}a_{11}a_{02}-a_{00}a_{21}a_{12}.$

Facts & Assumptions

Given: A $3\times3$ matrix over a commutative ring.

[L1]

The determinant is the signed Leibniz sum (For $n\ge1$ , the determinant over a commutative ring by the Leibniz formula, and $|\det A|$ for a real matrix).

[L2]

A permutation's sign is determined by the parity of its inversions (Inversions, inversion number, the sign $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$ , and even and odd permutations).

[L3]

A set of size $3$ has $3!=6$ bijections to itself (A finite set $A$ with $\lvert A\rvert = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

Verification

technique · direct

1.1

The even permutations in one-line notation are $[0,1,2]$ , $[1,2,0]$ and $[2,0,1]$ ; the odd ones are $[1,0,2]$ , $[2,1,0]$ and $[0,2,1]$ . These are all six permutations.

L1L2L3

2.1

Substitution yields the displayed six terms. For $A=\begin{pmatrix}1&2&3\\0&1&4\\5&6&0\end{pmatrix}$ they give $0+0+40-0-15-24=1$ , a concrete check of the signs.

step 1.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

An integer matrix of determinant $2$ is invertible over $\mathbb Q$ but not over $\mathbb Z$

Example

The integer matrix $A=\operatorname{diag}(2,1)$ has determinant $2$ . It is invertible as a matrix over $\mathbb Q$ , but it is not invertible as a matrix over $\mathbb Z$ .

Facts & Assumptions

Given: $A=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ .

[L1]

The $2\times2$ determinant is $ad-bc$ (The Leibniz formula gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$ ).

[L2]

An invertible matrix over a commutative ring has unit determinant (An invertible square matrix over a commutative ring has unit determinant).

[L3]

The rationals form a field (The rationals form a field).

[L4]

The integers form a commutative ring (The integers form a commutative ring).

[L5]

The only units of $\mathbb Z$ are $1$ and $-1$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ).

Verification

technique · direct

1.1

Fact [L1] gives $\det A=2$ . Over $\mathbb Q$ , the matrix $\operatorname{diag}(1/2,1)$ multiplies with $A$ in either order to give $I_2$ , so $A$ is invertible there.

L1L2L3L4L5algebra

2.1

Over $\mathbb Z$ , the determinant $2$ is not a unit by [L5]. The contrapositive of [L2] therefore rules out an integer inverse.

step 1.1L2L5∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

Over $\mathbb Z/2$ , an antisymmetric bilinear form need not be alternating

Statement refuted

The false converse is: every antisymmetric bilinear function is alternating. Over $R=\mathbb Z/2$ , define $F:M_2(R)\to R$ on columns $x,y\in R^2$ by $F([x\mid y])=x_0y_0$ . Then $F$ is bilinear and antisymmetric but not alternating.

Facts & Assumptions

Given: The field $R=\mathbb Z/2$ and the displayed function $F$ .

[L1]

Antisymmetric means that swapping columns negates the value, while alternating means vanishing on equal columns (Column-multilinear, alternating, normalized and antisymmetric functions on square matrices over a commutative ring).

[L2]

$\mathbb Z/2$ is a field (For every prime $p$ , the two operations on $\mathbb{Z}/p$ make it a field).

[L3]

In $\mathbb Z/2$ , addition and multiplication are modulo $2$ , so $-1=1$ (For every natural $n$ , $(\mathbb{Z}/n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Counterexample

technique · direct

1.1

The coordinate product is linear in each column. Moreover $F([y\mid x])=y_0x_0=x_0y_0=-F([x\mid y])$ because multiplication is commutative and $-1=1$ in $R$ . Thus $F$ is antisymmetric.

L1L2L3algebra

2.1

For $e_0=(1,0)^{\mathsf T}$ , one has $F([e_0\mid e_0])=1$ , so $F$ does not vanish on equal columns and is not alternating.

step 1.1L1algebra∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

The naive quaternionic formula $ad-bc$ is not row-alternating: a matrix with equal rows can have value $2k\ne0$

Statement refuted

The commutative-ring formula $ad-bc$ does not define a row-alternating determinant for quaternionic $2\times2$ matrices. A matrix with two equal rows can have naive value $2k\ne0$ .

Facts & Assumptions

Given: The quaternionic matrix $A=\begin{pmatrix}i&j\\i&j\end{pmatrix}$ and the naive expression $D(A)=ad-bc$ .

[L1]

The quaternions have basis elements $i,j,k$ and real scalar multiplication (The quaternions $\mathbb{H}$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[L2]

Quaternion multiplication satisfies $ij=k$ , $ji=-k$ , and $k\ne-k$ ( $\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1} = \bar q / N(q)$ for $q \ne 0$ , while $ij = k$ and $ji = -k$ ).

[L3]

Over a commutative ring the genuine $2\times2$ determinant formula is $ad-bc$ (The Leibniz formula gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$ ).

Counterexample

technique · direct

1.1

The two rows of $A$ are both $(i,j)$ , so a row-alternating function would vanish on $A$ .

L1L2L3

2.1

The naive expression instead gives $D(A)=ij-ji=k-(-k)=2k$ . If $2k=0$ , then $k=-k$ , contrary to [L2]; hence $D(A)\ne0$ .

step 1.1L2algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

Row operations track determinant correctly for a singular triangular matrix over $\mathbb Z/6$

Example

Over $R=\mathbb Z/6$ , let $A=\operatorname{diag}(2,1)$ . Then $A$ is not invertible although $\det A=2\ne0$ . A swap changes the determinant to $4=-2$ , scaling the first row by the nonzero nonunit $3$ changes it to $0=3\cdot2$ , and adding four times the second row to the first leaves it equal to $2$ .

Facts & Assumptions

Given: The displayed matrix over $\mathbb Z/6$ .

[L1]

Row swaps, arbitrary scalings and row additions have the three determinant laws without a nonsingularity assumption (For every square matrix, including singular ones, a row swap negates the determinant, scaling a row by any scalar scales it, and row addition leaves it unchanged).

[L2]

A triangular determinant is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[L3]

Arithmetic in $\mathbb Z/6$ is commutative ring arithmetic modulo $6$ (For every natural $n$ , $(\mathbb{Z}/n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L4]

$\mathbb Z/6$ consists of congruence classes modulo $6$ (The congruence class $[a]_n$ and the quotient set $\mathbb{Z}/n$ ).

[L5]

An invertible matrix over a commutative ring has unit determinant (An invertible square matrix over a commutative ring has unit determinant).

Verification

technique · direct

1.1

By [L2], $\det A=2$ . The element $2$ is not a unit modulo $6$ , since every product $2r$ is even and cannot be congruent to $1$ ; therefore [L5] implies that $A$ is not invertible.

L1L2L3L4L5algebra

2.1

Swapping rows gives $\begin{pmatrix}0&1\\2&0\end{pmatrix}$ with determinant $-2=4$ . Scaling row $0$ by $3$ gives $\operatorname{diag}(0,1)$ with determinant $0=3\cdot2$ . Replacing row $0$ by row $0+4R_1$ gives $\begin{pmatrix}2&4\\0&1\end{pmatrix}$ with determinant $2$ .

step 1.1algebra

3.1

These values agree respectively with negation, multiplication by $3$ , and invariance under row addition in [L1]. No cancellation, unit scaling or invertibility was used.

step 2.1L1∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

FALSE: $\det(A+B)=\det(A)+\det(B)$ for all same-sized square matrices

Statement

The following claim is false: for all same-sized square real matrices, $\det(A+B)=\det(A)+\det(B).$

Facts & Assumptions

Given: Over $\mathbb R$ , take $A=B=I_2$ .

[L1]

Matrix addition is entrywise and $I_2$ has diagonal entries $1$ (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L2]

Determinant is multilinear in one selected row or column while all other rows or columns are fixed (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

[L3]

For a $2\times2$ matrix, determinant is $ad-bc$ (The Leibniz formula gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$ ).

[L4]

The real numbers form a field (The reals form a field).

Refutation

technique · direct

1.1

Here $A+B=2I_2$ , so [L3] gives $\det(A+B)=4$ , whereas $\det(A)+\det(B)=1+1=2$ . Since $4\ne2$ in $\mathbb R$ , the claim fails.

L1L3L4algebra

2.1

The property in [L2] does not imply whole-matrix additivity: changing both rows at once produces cross terms. It permits addition only in one selected row or column while all the others remain fixed.

step 1.1L2∎

Sources

Standard references

Recommended treatments; not extraction sources.

Determinants over a Ring: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The Leibniz formula gives det⁡(abcd)=ad−bc\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bcdet(ac​bd​)=ad−bc

Example

Facts & Assumptions

Verification

A full 3×33\times33×3 Leibniz expansion lists all six permutations and their signs

Example

Facts & Assumptions

Verification

An integer matrix of determinant 222 is invertible over Q\mathbb QQ but not over Z\mathbb ZZ

Example

Facts & Assumptions

Verification

Over Z/2\mathbb Z/2Z/2, an antisymmetric bilinear form need not be alternating

Statement refuted

Facts & Assumptions

Counterexample

The naive quaternionic formula ad−bcad-bcad−bc is not row-alternating: a matrix with equal rows can have value 2k≠02k\ne02k=0

Statement refuted

Facts & Assumptions

Counterexample

Row operations track determinant correctly for a singular triangular matrix over Z/6\mathbb Z/6Z/6

Example

Facts & Assumptions

Verification

FALSE: det⁡(A+B)=det⁡(A)+det⁡(B)\det(A+B)=\det(A)+\det(B)det(A+B)=det(A)+det(B) for all same-sized square matrices

Statement

Facts & Assumptions

Refutation

Sources

Standard references

The Leibniz formula gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$

A full $3\times3$ Leibniz expansion lists all six permutations and their signs

An integer matrix of determinant $2$ is invertible over $\mathbb Q$ but not over $\mathbb Z$

Over $\mathbb Z/2$ , an antisymmetric bilinear form need not be alternating

The naive quaternionic formula $ad-bc$ is not row-alternating: a matrix with equal rows can have value $2k\ne0$

Row operations track determinant correctly for a singular triangular matrix over $\mathbb Z/6$

FALSE: $\det(A+B)=\det(A)+\det(B)$ for all same-sized square matrices