ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The adjugate gives the inverse of a $3 \times 3$ rational matrix with determinant $3$

Example

For

$A = (\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}) \in M_{3} (Q),$

one has $\det (A) = 3$ and

$A^{- 1} = \frac{1}{3} (\begin{array}{ccc} 1 & - 2 & 2 \\ 1 & 1 & - 1 \\ - 1 & 2 & 1 \end{array}) .$

Facts & Assumptions

Given: The displayed matrix $A$ .

[F1]

$Q$ is a field (The rationals form a field).

[F2]

The cofactor matrix is $(C_{i j})$ and the adjugate is its transpose (Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring).

[L1]

$A adj (A) = \det (A) I$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

[L2]

If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

Verification

technique · direct

1.1

Computing the nine signed $2 \times 2$ minors gives $cof (A) = (\begin{array}{ccc} 1 & 1 & - 1 \\ - 2 & 1 & 2 \\ 2 & - 1 & 1 \end{array}), adj (A) = (\begin{array}{ccc} 1 & - 2 & 2 \\ 1 & 1 & - 1 \\ - 1 & 2 & 1 \end{array}) .$

F2algebra

2.1

Direct multiplication gives $A adj (A) = 3 I_{3}$ . By [L1], this also confirms $\det (A) = 3$ .

step 1.1L1algebra

3.1

The scalar $3$ is a unit of $Q$ by [F1], so [L2] gives the displayed inverse. Multiplying it by $A$ on either side gives $I_{3}$ .

step 2.1F1L2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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The adjugate gives the inverse of a 3×33×3 rational matrix with determinant 33