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The cofactor matrix consists of the signed minors of a matrix. It is the transpose of the adjugate matrix. The inverse expression applies to invertible matrices; the definition by minors also applies to singular matrices. Cofactors transform oriented hypersurface area vectors in a change of variables formula.

The adjugate matrix is the transpose of the cofactor matrix: $\mathrm{adj}(A{)}_{ij}=(-1{)}^{i+j}\det A_{\hat{j},\hat{i}}$. Cofactor expansion gives the adjugate identityFor an invertible matrix this yields $A^{-1}=\mathrm{adj}(A)/\det A$. Integer entries of $A$ imply integer cofactors, so $\det A=\pm 1$ makes its inverse integral. Conversely, if $A$ and $A^{-1}$ both have integer entries, their determinants are integers and $(\det A)(\det A^{-1})=1$. The only integer units are $\pm 1$. Hence an invertible integer matrix has an integer inverse exactly when it is a unimodular matrix, with determinant $\pm 1$. Invertibility is implicit whenever $A^{-1}$ is used.

The adjugate matrix is the transpose of the cofactor matrix, and it satisfiesPut $M=tI-A$ and substitute the two given expansions into this identity:Comparing coefficients of powers of $t$ givesHere $c_0=1$ because the characteristic polynomial is monic.

Let $M(t_1,\dots ,t_n)=\mathrm{diag}(t_1,\dots ,t_n)-A$. Expanding the determinant along row $i$ shows thatwhere $C_{ii}$ is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal $t_1=\cdots =t_n=t$ yieldsTherefore

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

If $A_{ji}$ denotes the matrix obtained by deleting row $j$ and column $i$, then the adjugate matrix isFor nonsingular $A$,Therefore, when $A$ and $B$ are nonsingular,