Proving the Adjugate Identity

The Identity Is an Entry-by-Entry Statement

The adjugate identity relates a square matrix to its cofactors and determinant. Its importance is not limited to matrices that can be inverted: it holds for every square matrix, including singular ones. The key is to examine the entries of the product individually. A diagonal entry becomes a cofactor expansion of the determinant, and an off-diagonal entry becomes the determinant of a matrix with two equal rows.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), where \(n\geq 1\). For each pair \(i,j\), let \(\mathbf{A}_{ij}\) be the matrix obtained by deleting row \(i\) and column \(j\) from \(\mathbf{A}\). Its determinant is the minor for that position. When \(n=1\), the determinant of the resulting empty matrix is defined to be \(1\), so the sole cofactor is \(1\).

The transpose in this definition determines the indices in the product. By matrix multiplication, the entry in row \(i\), column \(j\) of \(\mathbf{A}\operatorname{adj}(\mathbf{A})\) is

In particular, the cofactor index in this sum begins with \(j\), not \(i\). This is exactly what makes the diagonal and off-diagonal cases correspond to two useful determinant facts: expansion along a row, and the vanishing of a determinant with repeated rows.

Why the Diagonal Entries Equal the Determinant

We use the cofactor expansion of a determinant along any row: expanding \(\det(\mathbf{A})\) along row \(i\) gives \(\sum_{k=1}^{n}a_{ik}C_{ik}\). We also use the determinant property that swapping two rows changes the sign of the determinant. In particular, a matrix with two equal rows has determinant zero: swapping those rows leaves the matrix unchanged but changes the determinant's sign, so its determinant equals its own negative and must be zero over \(\mathbb{R}\).

Proof. Fix indices \(i,j\) between \(1\) and \(n\), and examine the \((i,j)\)-entry of the product. If \(i=j\), the entry formula gives \(\sum_{k=1}^{n}a_{ik}C_{ik}\). By cofactor expansion along row \(i\), this sum equals \(\det(\mathbf{A})\). Thus every diagonal entry of the product is \(\det(\mathbf{A})\).

Now suppose \(i\ne j\). Form a matrix \(\mathbf{B}\) by replacing row \(j\) of \(\mathbf{A}\) with row \(i\) of \(\mathbf{A}\), leaving all other rows unchanged. Rows \(i\) and \(j\) of \(\mathbf{B}\) are equal, so \(\det(\mathbf{B})=0\). Expand \(\det(\mathbf{B})\) along row \(j\). The entries in that row are \(a_{i1},\ldots,a_{in}\). For each column \(k\), the minor obtained by deleting row \(j\) and column \(k\) is the same for \(\mathbf{B}\) as for \(\mathbf{A}\): the replaced row is deleted, and all remaining rows agree. Therefore the corresponding cofactor is \(C_{jk}\), and the expansion gives \(\det(\mathbf{B})=\sum_{k=1}^{n}a_{ik}C_{jk}\). This is the \((i,j)\)-entry of \(\mathbf{A}\operatorname{adj}(\mathbf{A})\), so that entry is zero.

We have shown that every diagonal entry is \(\det(\mathbf{A})\) and every off-diagonal entry is zero. These are precisely the entries of \(\det(\mathbf{A})\mathbf{I}_n\), proving the identity. When \(n=1\), there are no off-diagonal entries; the diagonal argument still applies using the stated empty determinant convention. \(\square\)

Worked Examples

A Companion Identity

The order of the factors can also be reversed. This follows from the identity just proved, together with the behavior of cofactors under transposition.

Proof. Apply the adjugate identity to \(\mathbf{A}^{\top}\): \(\mathbf{A}^{\top}\operatorname{adj}(\mathbf{A}^{\top})=\det(\mathbf{A}^{\top})\mathbf{I}_n\). Transposing a minor exchanges its rows and columns without changing its determinant, so \(C_{ij}(\mathbf{A}^{\top})=C_{ji}(\mathbf{A})\). Consequently \(\operatorname{adj}(\mathbf{A}^{\top})=\operatorname{adj}(\mathbf{A})^{\top}\). Also \(\det(\mathbf{A}^{\top})=\det(\mathbf{A})\), a standard determinant property. Transpose the displayed identity for \(\mathbf{A}^{\top}\); using \((\mathbf{X}\mathbf{Y})^{\top}=\mathbf{Y}^{\top}\mathbf{X}^{\top}\) gives \(\operatorname{adj}(\mathbf{A})\mathbf{A}=\det(\mathbf{A})\mathbf{I}_n\), as required. \(\square\)

Common Indexing Pitfall

Do not forget that the cofactor matrix is transposed to form the adjugate. In the three-by-three example, the correct \((1,2)\)-entry of \(\mathbf{A}\operatorname{adj}(\mathbf{A})\) is \(a_{11}C_{21}+a_{12}C_{22}+a_{13}C_{23}\). It equals \(1(-2)+2(1)+0(4)=0\). A common incorrect calculation uses \(C_{12},C_{22},C_{32}\) instead, giving \(1(2)+2(1)+0(-1)=4\). That calculation uses the entries of the cofactor matrix in the product, not the entries of its transpose. The correction is to first note that \(\operatorname{adj}(\mathbf{A})_{kj}=C_{jk}\), then apply the row-by-column rule for multiplication.

The proof also explains the scope of the identity. No division by \(\det(\mathbf{A})\) occurs, so the result applies whether the determinant is zero or nonzero. The identity is an equality of matrices established by checking every entry; it is not itself an inverse formula. The inverse formula is a further consequence when the determinant is nonzero, and requires that condition.