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The adjugate of a matrix is an operation done on an arbitrary square matrices: For the case of invertible matrices, it happens to rqual ${\displaystyle \operatorname {adj} (A)=\det(A)A^{-1}}$. For the non-invertible cases, see the article. It is not a class of matrix, unless you mean (over algebraically-closed fields) the set of square matrices which are either invertible, zero, or have rank equal to 1 - which are precisely the matrices which occur as adjugates of some matrix. Actually, over general fields, it's also required that the determinant be an ${\displaystyle n-1}$st power, which is guaranteed in the algebraically-closed case, but can fail for certain values of ${\displaystyle n}$ over certain other fields like the reals or finite fields.

To see why those are the only possible values of ${\displaystyle \operatorname {adj} (A)}$, consider the different possible nullities of ${\displaystyle A}$:

- If the nullity of ${\displaystyle A}$ is 0, then ${\displaystyle A}$ is invertible. We get that if ${\displaystyle \operatorname {adj} (B)=A}$ then ${\displaystyle B=\lambda A^{-1}}$. It follows from some tedious calculation that ${\displaystyle \lambda ^{n-1}=\det(A)}$.

- Now consider the case where ${\displaystyle A}$ has nullity 1. In that case, ${\displaystyle A\operatorname {adj} (A)=\det(A)I=0}$, so the columns of ${\displaystyle \operatorname {adj} (A)}$ are all in ${\displaystyle \ker(A)}$. It follows that the rank of ${\displaystyle \operatorname {adj} (A)}$ is at most 1. Since ${\displaystyle \operatorname {adj} (A)}$ has rank at most 1, we get that ${\displaystyle \operatorname {adj} (A)=uv^{T}}$ for two column vectors ${\displaystyle u}$ and ${\displaystyle v}$. We finish this case by noting that it's easy to see that all rank-1 matrices ${\displaystyle uv^{T}}$ arise this way.

- If the nullity of ${\displaystyle A}$ is greater than 1, then a change-of-basis shows that all its minors are 0, so we get ${\displaystyle \operatorname {adj} (A)=0}$. Svennik (talk) 13:06, 14 April 2026 (UTC)[reply]