Deriving the Adjugate Formula for the Inverse

From the Adjugate Identity to an Inverse

The adjugate is built from the cofactors of a square matrix. Its importance is not only that it packages many cofactor calculations into one matrix: multiplying a matrix by its adjugate produces a scalar multiple of the identity matrix. When that scalar is nonzero, division by it turns the product into the identity. This gives a direct route from the adjugate identity to an explicit formula for the inverse.

Throughout, let \(\mathbf{A}\) be an \(n\times n\) matrix over \(\mathbb{R}\). For each entry \(a_{ij}\), delete row \(i\) and column \(j\) to obtain a smaller matrix, and let its determinant be the minor of \(a_{ij}\). The cofactor \(C_{ij}\) is \((-1)^{i+j}\) times that minor. The adjugate \(\operatorname{adj}(\mathbf{A})\) is the transpose of the matrix of cofactors: its entry in row \(i\), column \(j\) is \(C_{ji}\). For a \(1\times1\) matrix, the determinant of the empty minor is taken to be \(1\).

An inverse is a matrix that multiplies \(\mathbf{A}\) to give the identity. More precisely, \(\mathbf{B}\) is a two-sided inverse of \(\mathbf{A}\) if both \(\mathbf{A}\mathbf{B}=\mathbf{I}_n\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_n\); when it exists, it is denoted \(\mathbf{A}^{-1}\). The adjugate identity already gives both products, but its right-hand side is \((\det\mathbf{A})\mathbf{I}_n\), not necessarily \(\mathbf{I}_n\). The key condition is therefore \(\det\mathbf{A}\ne0\), so that the real scalar \(\det\mathbf{A}\) has a reciprocal.

Proof. Set \(\mathbf{B}=\frac{1}{\det\mathbf{A}}\operatorname{adj}(\mathbf{A})\), which is defined because \(\det\mathbf{A}\ne0\). Pull the scalar through each matrix product and apply the adjugate identity:

In the other order, the other half of the adjugate identity gives

Thus \(\mathbf{B}\) is a two-sided inverse of \(\mathbf{A}\). By the definition of inverse, \(\mathbf{B}=\mathbf{A}^{-1}\), which proves the formula. \(\square\)

Both halves of the identity matter in this derivation: checking only \(\mathbf{A}\mathbf{B}=\mathbf{I}_n\) would show a left inverse, but would not by itself verify the required product in the other order. The adjugate identity supplies both products directly. The formula also depends on the condition \(\det\mathbf{A}\ne0\); it does not define a matrix when the determinant is zero.

Why the Inverse Given by the Formula Is Unique

The formula constructs an inverse, but it is also useful to know that two different matrices cannot both be inverses of the same square matrix. This fact follows from associativity of matrix multiplication and the identity matrix, without needing another determinant calculation.

Proof. Assume \(\mathbf{X}\) and \(\mathbf{Y}\) are both two-sided inverses. Then \(\mathbf{X}\mathbf{A}=\mathbf{I}_n\) and \(\mathbf{A}\mathbf{Y}=\mathbf{I}_n\). Using these equalities and associativity,

Therefore the two inverses are equal. \(\square\)

This uniqueness result explains why the adjugate expression is not merely one possible matrix that happens to work. Once it has been checked on both sides, it must be the inverse. It also gives a useful check on any proposed inverse: if a second candidate is claimed, both candidates must agree.

Worked Examples of the Formula and Its Condition

Using the Inverse to Solve a Square System

The inverse formula also explains how an inverse solves a system \(\mathbf{A}\mathbf{x}=\mathbf{b}\). When \(\det\mathbf{A}\ne0\), left-multiplying by \(\mathbf{A}^{-1}\) gives \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\). The expression is not merely a candidate: it is a solution, and it is the only solution.

Proof. Define \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\). Since \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\), we have \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{A}^{-1}\mathbf{b}=\mathbf{b}\), so a solution exists. If \(\mathbf{y}\) is any other solution, then \(\mathbf{A}\mathbf{y}=\mathbf{b}\). Multiplying on the left by \(\mathbf{A}^{-1}\) yields \(\mathbf{y}=\mathbf{A}^{-1}\mathbf{b}=\mathbf{x}\), so the solution is unique. Substituting the adjugate formula for \(\mathbf{A}^{-1}\) gives the stated expression. \(\square\)

Common Pitfall: Dividing by a Zero Determinant

A common error is to write \(\mathbf{A}^{-1}=\operatorname{adj}(\mathbf{A})/\det\mathbf{A}\) for every square matrix, then try to use the expression even when the determinant is zero. For the singular matrix in the previous example, this would amount to dividing every entry of \(\begin{bmatrix}4&-2\\-2&1\end{bmatrix}\) by \(0\). Those entries are not defined, so the expression is not an inverse. The correction is to check \(\det\mathbf{A}\ne0\) before applying the formula.

It is also important to keep the two matrix products distinct. A scalar multiple of \(\operatorname{adj}(\mathbf{A})\) is shown to be the inverse because the adjugate identity gives both \(\mathbf{A}\operatorname{adj}(\mathbf{A})\) and \(\operatorname{adj}(\mathbf{A})\mathbf{A}\). A calculation in one order should not be silently substituted for a calculation in the other. In the general theorem, both are justified by the identity; in a direct numerical check, write out both products.

The central logic is short but depends on every condition: start with a square matrix, use the adjugate identity, and divide by the determinant only when it is nonzero. That produces a two-sided inverse, and uniqueness ensures that the resulting matrix is precisely \(\mathbf{A}^{-1}\).