CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11

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The naive quaternionic formula $a d - b c$ is not row-alternating: a matrix with equal rows can have value $2 k \neq 0$

Statement refuted

The commutative-ring formula $a d - b c$ does not define a row-alternating determinant for quaternionic $2 \times 2$ matrices. A matrix with two equal rows can have naive value $2 k \neq 0$ .

Facts & Assumptions

Given: The quaternionic matrix $A = (\begin{array}{cc} i & j \\ i & j \end{array})$ and the naive expression $D (A) = a d - b c$ .

[L1]

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

[L2]

Quaternion multiplication satisfies $i j = k$ , $j i = - k$ , and $k \neq - k$ ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ).

[L3]

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

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 $D (A) = i j - j i = k - (- k) = 2 k$ . If $2 k = 0$ , then $k = - k$ , contrary to [L2]; hence $D (A) \neq 0$ .

step 1.1L2algebra∎

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Dependency tree · two levels

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Sources

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Ch. 4 (standard reference, not scraped)

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