The determinant of the adjoint of an nxn invertible square matrix A is precisely the determinant of A raised to the power of n minus one. This fundamental property simplifies complex calculations and is a cornerstone in linear algebra.
Understanding the Adjoint and Its Determinant
To fully grasp this concept, let's break down the key terms:
- Adjoint Matrix (Adjugate Matrix): The adjoint of a square matrix A, denoted as $\text{adj}(A)$, is the transpose of its cofactor matrix. The cofactor of an element $a_{ij}$ is $(-1)^{i+j}$ times the determinant of the submatrix formed by removing the i-th row and j-th column.
- Determinant: A scalar value that can be computed from the elements of a square matrix. It provides crucial information about the matrix, such as whether it is invertible (non-zero determinant) or singular (zero determinant).
The Core Formula
For an nxn invertible square matrix A, the determinant of its adjoint is given by the formula:
$$\det(\text{adj}(A)) = (\det(A))^{n-1}$$
Here's what each part signifies:
- A: Represents the original nxn invertible square matrix.
- n: Denotes the order (or dimension) of the square matrix. For example, for a 3x3 matrix, n = 3.
- $\text{adj}(A)$: Is the adjoint matrix of A.
- $\det(\text{adj}(A))$: Is the determinant of the adjoint of A.
- $\det(A)$: Is the determinant of the original matrix A.
This powerful relationship allows us to determine the determinant of the adjoint matrix without needing to calculate the adjoint matrix itself, provided we know the original matrix's determinant and its dimension. This property is specifically applicable when A is an invertible square matrix, meaning $\det(A) \neq 0$.
Why Does This Formula Work?
The formula stems from a key property relating a matrix to its adjoint:
$A \cdot \text{adj}(A) = \det(A) \cdot I$
Where I is the nxn identity matrix. Taking the determinant of both sides:
$\det(A \cdot \text{adj}(A)) = \det(\det(A) \cdot I)$
Using the determinant properties $\det(AB) = \det(A)\det(B)$ and $\det(kI) = k^n$ (where k is a scalar):
$\det(A) \cdot \det(\text{adj}(A)) = (\det(A))^n$
Since A is invertible, $\det(A) \neq 0$, we can divide both sides by $\det(A)$:
$\det(\text{adj}(A)) = \frac{(\det(A))^n}{\det(A)}$
$\det(\text{adj}(A)) = (\det(A))^{n-1}$
This derivation clearly illustrates the validity of the formula for invertible matrices.
Illustrative Examples
Let's explore this formula with practical examples.
Example 1: A 2x2 Invertible Matrix
Consider a 2x2 matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$.
Here, n = 2.
-
Calculate $\det(A)$:
$\det(A) = ad - bc$ -
Calculate $\text{adj}(A)$:
The cofactor matrix $C = \begin{pmatrix} d & -c \ -b & a \end{pmatrix}$.
The adjoint matrix $\text{adj}(A) = C^T = \begin{pmatrix} d & -b \ -c & a \end{pmatrix}$. -
Calculate $\det(\text{adj}(A))$:
$\det(\text{adj}(A)) = (d)(a) - (-b)(-c) = ad - bc$ -
Verify the formula:
According to the formula, $\det(\text{adj}(A)) = (\det(A))^{n-1} = (\det(A))^{2-1} = (\det(A))^1 = \det(A)$.
Indeed, we found $\det(\text{adj}(A)) = ad - bc$, which is equal to $\det(A)$. This confirms the formula for a 2x2 matrix.
Example 2: A 3x3 Invertible Matrix
Let $A = \begin{pmatrix} 1 & 2 & 3 \ 0 & 1 & 4 \ 5 & 6 & 0 \end{pmatrix}$.
Here, n = 3.
-
Calculate $\det(A)$:
Using cofactor expansion along the first column:
$\det(A) = 1 \cdot \det\begin{pmatrix} 1 & 4 \ 6 & 0 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 2 & 3 \ 6 & 0 \end{pmatrix} + 5 \cdot \det\begin{pmatrix} 2 & 3 \ 1 & 4 \end{pmatrix}$
$\det(A) = 1(1 \cdot 0 - 4 \cdot 6) - 0 + 5(2 \cdot 4 - 3 \cdot 1)$
$\det(A) = 1(-24) + 5(8 - 3)$
$\det(A) = -24 + 5(5)$
$\det(A) = -24 + 25 = 1$ -
Apply the formula:
Since $\det(A) = 1$ and n = 3, we use the formula:
$\det(\text{adj}(A)) = (\det(A))^{n-1} = (1)^{3-1} = (1)^2 = 1$.
In this case, the determinant of the adjoint is also 1. This significantly reduces the work compared to computing all cofactors and then the adjoint matrix, only to find its determinant.
Key Properties and Relationships
Understanding the relationship between a matrix, its determinant, and its adjoint is vital in linear algebra. The table below summarizes some key aspects:
| Property | Description |
|---|---|
| Definition | The adjoint of A is the transpose of its cofactor matrix. |
| Invertibility | A matrix A is invertible if and only if $\det(A) \neq 0$. The adjoint plays a role in finding the inverse. |
| Inverse Formula | $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$ (if A is invertible). |
| Determinant of Adjoint | $\det(\text{adj}(A)) = (\det(A))^{n-1}$ for an nxn invertible matrix A. |
| Rank | If $\text{rank}(A) = n-1$, then $\text{rank}(\text{adj}(A))=1$. |
For more detailed information on adjoint matrices and determinants, you can refer to resources like Wikipedia's Adjugate Matrix article or Wolfram MathWorld on Matrix Adjoint.
Conclusion
The determinant of the adjoint of an nxn invertible square matrix A is elegantly given by the formula $(\det(A))^{n-1}$. This property highlights the interconnectedness of fundamental matrix operations and is a powerful tool for analyzing matrix characteristics without cumbersome intermediate steps.
