Adjugate Matrix and Matrix Inverse Formula

The adjugate matrix, constructed from cofactors, provides an explicit formula for the inverse of a matrix. This connection reveals the deep relationship between determinants and invertibility.

DefinitionAdjugate Matrix

The adjugate (or classical adjoint) of an $n \times n$ matrix $A$ , denoted $\text{adj}(A)$ , is the transpose of the cofactor matrix: $\text{adj}(A) = [C_{ij}]^T$

That is, the $(i,j)$ -entry of $\text{adj}(A)$ is the cofactor $C_{ji}$ (note the index swap).

Explicitly: $[\text{adj}(A)]_{ij} = C_{ji} = (-1)^{i+j}\det(A_{ji})$

TheoremInverse Formula

For any $n \times n$ matrix $A$ : $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = \det(A) \cdot I$

Consequently, if $\det(A) \neq 0$ , then $A$ is invertible with: $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$

This formula is primarily of theoretical importance; for numerical computation, Gaussian elimination is far more efficient than computing $n^2$ cofactors.

ExampleComputing Inverse via Adjugate

Find the inverse of $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ .

First, $\det(A) = 4 - 6 = -2 \neq 0$ , so $A$ is invertible.

Compute cofactors:

$C_{11} = 4$ , $C_{12} = -3$
$C_{21} = -2$ , $C_{22} = 1$

Thus: $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}^T = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$

Therefore: $A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1.5 \\ 1 & -0.5 \end{bmatrix}$

DefinitionCramer's Rule

Consider the system $A\mathbf{x} = \mathbf{b}$ where $A$ is $n \times n$ with $\det(A) \neq 0$ . Let $A_i(\mathbf{b})$ be the matrix obtained by replacing column $i$ of $A$ with $\mathbf{b}$ .

Then the unique solution is: $x_i = \frac{\det(A_i(\mathbf{b}))}{\det(A)} \quad \text{for } i = 1, 2, \ldots, n$

ExampleCramer's Rule Application

Solve: $\begin{cases} x + 2y = 5 \\ 3x + 4y = 11 \end{cases}$

We have $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ 11 \end{pmatrix}$ , with $\det(A) = -2$ .

$x = \frac{\det\begin{bmatrix} 5 & 2 \\ 11 & 4 \end{bmatrix}}{-2} = \frac{20 - 22}{-2} = \frac{-2}{-2} = 1$

$y = \frac{\det\begin{bmatrix} 1 & 5 \\ 3 & 11 \end{bmatrix}}{-2} = \frac{11 - 15}{-2} = \frac{-4}{-2} = 2$

Remark

While Cramer's rule provides elegant theoretical insight—expressing solutions directly via determinants—it's computationally inefficient for large systems, requiring $(n+1)$ determinant calculations. Gaussian elimination remains the practical algorithm for solving linear systems.