LA-5.5

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Adjugate Matrix

The adjugate (classical adjoint) matrix is the transpose of the cofactor matrix. It provides an explicit closed-form formula for matrix inversion: A⁻¹ = adj(A)/det(A). While computationally expensive for large matrices, the adjugate is invaluable for theoretical analysis and symbolic computation.

Cofactor Transpose

Inverse Formula

det(adj A)

Cramer Connection

Learning Objectives

Define the adjugate (classical adjoint) matrix as the transpose of the cofactor matrix
Compute the adjugate of 2×2, 3×3, and larger matrices
Prove and apply the fundamental identity A · adj(A) = det(A) · I
Derive the inverse formula A⁻¹ = adj(A)/det(A) from the adjugate identity
Understand key properties: det(adj(A)), adj(AB), adj(A^T)
Apply adjugate to Cramer's rule and theoretical derivations
Analyze what happens when det(A) = 0 (singular case)
Compare adjugate method vs row reduction for matrix inversion

Prerequisites

Minors and cofactors: M_{ij} and A_{ij} = (-1)^{i+j} M_{ij} (LA-5.1)
Laplace expansion and alien cofactor theorem (LA-5.4)
Matrix inverse concepts and properties (LA-4.3)
Determinant properties: multilinearity, det(AB) = det(A)det(B) (LA-5.2)
Cramer's rule for solving linear systems (LA-5.4)

1. Definition of Adjugate

The adjugate matrix (also called the classical adjoint) is constructed from the cofactors of a matrix. It plays a central role in deriving the inverse formula and connecting to Cramer's rule.

Definition 5.10: Cofactor Matrix

The cofactor matrix of $A = (a_{ij})_{n \times n}$ is the n×n matrix whose $(i,j)$ -entry is the cofactor $A_{ij}$ :

C = (A_{ij})_{n \times n} = \begin{pmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\ A_{21} & A_{22} & \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} & A_{n2} & \cdots & A_{nn} \end{pmatrix}

Definition 5.11: Adjugate Matrix

The adjugate (or classical adjoint) of $A$ is the transpose of the cofactor matrix:

\text{adj}(A) = C^T = (A_{ji})_{n \times n}

Equivalently: $[\text{adj}(A)]_{ij} = A_{ji}$ — note the swapped indices!

Remark 5.19: Why 'Classical Adjoint'?

In older texts, "adjoint" referred to this matrix. Modern usage reserves "adjoint" for the conjugate transpose (A* or A†), so "adjugate" or "classical adjoint" is preferred to avoid confusion.

Example 5.23: 2×2 Adjugate

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ :

Step 1: Compute cofactors:

$A_{11} = (+1) \cdot d = d$
$A_{12} = (-1) \cdot c = -c$
$A_{21} = (-1) \cdot b = -b$
$A_{22} = (+1) \cdot a = a$

Step 2: Form cofactor matrix and transpose:

C = \begin{pmatrix} d & -c \\ -b & a \end{pmatrix} \implies \text{adj}(A) = C^T = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Memory aid: Swap diagonal entries, negate off-diagonal entries.

Example 5.24: Verifying 2×2 Adjugate Identity

Check that $A \cdot \text{adj}(A) = \det(A) \cdot I$ :

\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix} = (ad-bc) I \checkmark

Remark 5.20: Index Convention

The transpose is essential. Without it, the identity would fail:

Cofactor matrix C: $C_{ij} = A_{ij}$
Adjugate: $[\text{adj}(A)]_{ij} = C_{ji} = A_{ji}$

2. The Fundamental Identity

The fundamental identity connects the adjugate with the determinant, providing the foundation for the explicit inverse formula.

Theorem 5.27: Adjugate Identity

For any n×n matrix A:

A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = \det(A) \cdot I

This holds for all matrices, even singular ones (where det(A) = 0).

Proof:

Computing $[A \cdot \text{adj}(A)]_{ij}$ :

[A \cdot \text{adj}(A)]_{ij} = \sum_{k=1}^{n} a_{ik} [\text{adj}(A)]_{kj} = \sum_{k=1}^{n} a_{ik} A_{jk}

Case 1: If $i = j$ :

$\sum_k a_{ik} A_{ik}$ is the Laplace expansion along row i = $\det(A)$

Case 2: If $i \neq j$ :

$\sum_k a_{ik} A_{jk}$ is an alien cofactor sum = 0

Conclusion: $[A \cdot \text{adj}(A)]_{ij} = \delta_{ij} \det(A)$ , which is $\det(A) \cdot I$ .

∎

Remark 5.21: Both Products Equal

The proof for $\text{adj}(A) \cdot A$ is similar, using column expansion instead of row expansion. Both products equal $\det(A) \cdot I$ .

Corollary 5.8: Inverse Formula

If $\det(A) \neq 0$ , dividing both sides by det(A):

A \cdot \frac{\text{adj}(A)}{\det(A)} = I \implies A^{-1} = \frac{1}{\det(A)} \text{adj}(A)

Example 5.25: Inverse via Adjugate

Find $A^{-1}$ for $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ .

Step 1: det(A) = 3(4) - 1(2) = 10

Step 2: adj(A) = (swap diagonal, negate off-diagonal)

\text{adj}(A) = \begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix}

Step 3: A⁻¹ = adj(A)/det(A)

A^{-1} = \frac{1}{10}\begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{pmatrix}

Remark 5.22: Singular Case

When det(A) = 0:

A is singular, so A⁻¹ doesn't exist
But adj(A) still exists and A · adj(A) = 0
The columns of adj(A) are in the null space of A

3. Properties of Adjugate

The adjugate satisfies many elegant algebraic properties that parallel those of the inverse.

Theorem 5.28: Determinant of Adjugate

For an n×n matrix A:

\det(\text{adj}(A)) = (\det(A))^{n-1}

Proof:

Take determinants of both sides of $A \cdot \text{adj}(A) = \det(A) \cdot I$ :

\det(A) \cdot \det(\text{adj}(A)) = \det(A)^n

If det(A) ≠ 0, divide by det(A) to get $\det(\text{adj}(A)) = \det(A)^{n-1}$ .

The formula extends to singular matrices by continuity.

∎

Theorem 5.29: Adjugate of Product

\text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A)

Like the inverse, the adjugate reverses the order of products.

Proof:

For invertible A, B: $(AB)^{-1} = B^{-1}A^{-1}$ , so:

\text{adj}(AB) = \det(AB) \cdot (AB)^{-1} = \det(A)\det(B) \cdot B^{-1}A^{-1}

Meanwhile, $\text{adj}(B)\text{adj}(A) = \det(B)B^{-1} \cdot \det(A)A^{-1}$ , which equals the same.

∎

Theorem 5.30: Adjugate of Transpose

\text{adj}(A^T) = (\text{adj}(A))^T

Theorem 5.31: Scalar Multiple

For scalar c:

\text{adj}(cA) = c^{n-1} \text{adj}(A)

Each cofactor is an (n-1)×(n-1) determinant, so each scales by $c^{n-1}$ .

Theorem 5.32: Adjugate of Adjugate

For invertible A:

\text{adj}(\text{adj}(A)) = (\det(A))^{n-2} A

Example 5.26: Adjugate of Adjugate (3×3)

For a 3×3 matrix with det(A) = 2:

adj(adj(A)) = 2^{3-2} · A = 2 · A

Summary of Properties

adj(I) = I
adj(A⁻¹) = (adj(A))⁻¹ = A/det(A) (for invertible A)
adj(Aᵀ) = (adj(A))ᵀ
adj(AB) = adj(B) · adj(A)
adj(cA) = c^{n-1} adj(A)
det(adj(A)) = det(A)^{n-1}
adj(adj(A)) = det(A)^{n-2} · A

4. Computing the Adjugate

Computing the adjugate requires finding all n² cofactors and then transposing. Here is the step-by-step process.

Algorithm for adj(A)

For each position (i,j), compute the minor $M_{ij}$
Apply the sign: $A_{ij} = (-1)^{i+j} M_{ij}$
Arrange all cofactors in a matrix C
Transpose: adj(A) = Cᵀ

Example 5.27: Complete 3×3 Adjugate

Compute adj(A) for $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ .

Step 1: Compute all 9 cofactors:

A₁₁ = +(1·0 - 4·6) = -24

A₁₂ = -(0·0 - 4·5) = 20

A₁₃ = +(0·6 - 1·5) = -5

A₂₁ = -(2·0 - 3·6) = 18

A₂₂ = +(1·0 - 3·5) = -15

A₂₃ = -(1·6 - 2·5) = 4

A₃₁ = +(2·4 - 3·1) = 5

A₃₂ = -(1·4 - 3·0) = -4

A₃₃ = +(1·1 - 2·0) = 1

Step 2: Form cofactor matrix:

C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}

Step 3: Transpose:

\text{adj}(A) = C^T = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}

Example 5.28: Verification

Verify: det(A) = 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15 = 1

Check A · adj(A) = det(A) · I = 1 · I = I ✓

Remark 5.23: Computational Complexity

Computing adj(A) requires:

n² cofactors, each an (n-1)×(n-1) determinant
Total: O(n² · (n-1)!) ≈ O(n · n!) operations
Compare: row reduction gives A⁻¹ in O(n³)
Use adjugate for theory/symbols, row reduction for numbers

5. Common Mistakes

Forgetting to transpose

adj(A) is the transpose of the cofactor matrix, not the cofactor matrix itself! [adj(A)]ᵢⱼ = Aⱼᵢ

Confusing adjugate with adjoint

In modern usage, "adjoint" often means conjugate transpose (A*). Use "adjugate" or "classical adjoint" for the cofactor transpose.

Wrong sign pattern in cofactors

Remember: Aᵢⱼ = (-1)^{i+j} Mᵢⱼ. The checkerboard pattern starts with + at (1,1).

Thinking adj(AB) = adj(A)adj(B)

Like the inverse, adjugate reverses order: adj(AB) = adj(B) · adj(A)

Using adjugate for large numerical matrices

Adjugate is O(n·n!) vs O(n³) for row reduction. For n > 4, always use row reduction for numerical work.

6. Applications of Adjugate

Despite computational limitations, the adjugate has important theoretical and practical applications.

Theorem 5.33: Connection to Cramer's Rule

Cramer's rule $x_i = \det(A_i)/\det(A)$ can be derived from:

x = A^{-1}b = \frac{1}{\det(A)} \text{adj}(A) \cdot b

The i-th component involves the i-th column of adj(A), which contains cofactors from row i of A.

Remark 5.24: Symbolic Computation

The adjugate formula preserves polynomial structure:

If A has polynomial entries, so does adj(A)
No division is needed to compute adj(A)
Useful in computer algebra systems (Mathematica, Maple, SymPy)

Example 5.29: Parametric Inverse

Find the inverse of $A = \begin{pmatrix} t & 1 \\ 1 & t \end{pmatrix}$ for $t \neq \pm 1$ .

det(A) = t² - 1 = (t-1)(t+1)

adj(A) = (swap diagonal, negate off-diagonal)

A^{-1} = \frac{1}{t^2-1}\begin{pmatrix} t & -1 \\ -1 & t \end{pmatrix}

Remark 5.25: Characteristic Polynomial Connection

For the characteristic matrix $A - \lambda I$ :

(A - \lambda I) \cdot \text{adj}(A - \lambda I) = \det(A - \lambda I) \cdot I = p(\lambda) \cdot I

This identity is used in proving the Cayley-Hamilton theorem.

Theorem 5.34: Singular Matrices

For a singular matrix A (det(A) = 0):

A · adj(A) = 0 (the zero matrix)
Columns of adj(A) are in null(A)
If rank(A) = n-1, then rank(adj(A)) = 1
If rank(A) < n-1, then adj(A) = 0

7. Key Takeaways

Definition

adj(A) = Cᵀ (transpose of cofactor matrix)

[adj(A)]ᵢⱼ = Aⱼᵢ

Fundamental Identity

$A \cdot \text{adj}(A) = \det(A) \cdot I$

Works for all matrices

Inverse Formula

$A^{-1} = \text{adj}(A)/\det(A)$

When det(A) ≠ 0

Key Property

det(adj(A)) = det(A)^{n-1}

Power of n-1, not n

8. Practice Problems

Problem 1

Find adj(A) for $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and verify A · adj(A) = det(A) · I.

Problem 2

If det(A) = 5 for a 4×4 matrix, find det(adj(A)).

Problem 3

Prove: adj(2A) = 2^{n-1} adj(A) for an n×n matrix.

Problem 4

Find adj(adj(A)) for $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ .

Solutions

Solution 1

adj(A) = (swap diagonal, negate off-diagonal) = (4,-3; -1,2)

det(A) = 8-3 = 5

A·adj(A) = (2,3; 1,4)(4,-3; -1,2) = (5,0; 0,5) = 5·I ✓

Solution 2

det(adj(A)) = det(A)^{n-1} = 5^{4-1} = 5³ = 125

Solution 3

Each cofactor of 2A is an (n-1)×(n-1) determinant with entries scaled by 2.

So each cofactor scales by 2^{n-1}, giving adj(2A) = 2^{n-1} adj(A).

Solution 4

det(A) = 3, n = 2

adj(adj(A)) = det(A)^{n-2} · A = 3^0 · A = 1 · A = A

9. Quick Reference

Core Formulas

Adjugate: [adj(A)]ᵢⱼ = Aⱼᵢ (transposed cofactors)
Identity: A · adj(A) = adj(A) · A = det(A) · I
Inverse: A⁻¹ = adj(A) / det(A)
det(adj(A)): det(A)^{n-1}
adj(AB): adj(B) · adj(A)
adj(Aᵀ): (adj(A))ᵀ
adj(cA): c^{n-1} · adj(A)
adj(adj(A)): det(A)^{n-2} · A

2×2 Formula

\text{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Swap diagonal, negate off-diagonal

10. Historical Notes

Arthur Cayley (1821-1895): English mathematician who pioneered matrix theory. He developed the adjugate matrix concept as part of his work on matrix algebra and the theory of invariants. His 1858 paper "A Memoir on the Theory of Matrices" established the foundations.

James Joseph Sylvester (1814-1897): Cayley's colleague who coined many matrix terms. Together they developed much of 19th-century matrix algebra, including determinant theory.

Terminology Evolution: The term "adjoint" was historically used for this matrix. As functional analysis developed in the 20th century, "adjoint" acquired new meanings (conjugate transpose, Hermitian adjoint), leading to adoption of "adjugate" to avoid confusion.

Modern Usage: Today, most advanced texts use "adjugate" for the cofactor transpose, reserving "adjoint" for the conjugate transpose in complex vector spaces. Some older texts still use "classical adjoint" as a compromise.

Etymology

"Adjugate" from Latin "adjungere" (to join to) - refers to how adj(A) is "joined" to A in the identity A·adj(A) = det(A)·I
"Cofactor" from Latin "co-" (together) + "factor" (maker) - the cofactors work together in expansion
"Matrix" from Latin for "womb" or "origin" - introduced by Sylvester in 1850

11. Special Cases

Theorem 5.35: Adjugate of Identity

\text{adj}(I_n) = I_n

All cofactors of I are 0 except diagonals which are 1.

Theorem 5.36: Adjugate of Diagonal Matrix

For diagonal D = diag(d₁, d₂, ..., dₙ):

\text{adj}(D) = \text{diag}\left(\frac{\det(D)}{d_1}, \frac{\det(D)}{d_2}, \ldots, \frac{\det(D)}{d_n}\right)

Example 5.30: Diagonal Case

For D = diag(2, 3, 5):

det(D) = 2 · 3 · 5 = 30

adj(D) = diag(30/2, 30/3, 30/5) = diag(15, 10, 6)

Theorem 5.37: Adjugate of Triangular Matrix

For upper (or lower) triangular A, adj(A) is also upper (or lower) triangular.

Theorem 5.38: Adjugate of Orthogonal Matrix

For orthogonal Q (where Q⁻¹ = Qᵀ):

\text{adj}(Q) = \det(Q) \cdot Q^T = \pm Q^T

Since det(Q) = ±1 for orthogonal matrices.

Example 5.31: Rotation Matrix

For 2D rotation $R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ :

det(R) = cos²θ + sin²θ = 1

adj(R) = det(R) · Rᵀ = 1 · Rᵀ = R⁻¹ = R(-θ)

Remark 5.26: Block Diagonal Matrices

For block diagonal $A = \begin{pmatrix} B & 0 \\ 0 & C \end{pmatrix}$ :

\text{adj}(A) = \begin{pmatrix} \det(C) \cdot \text{adj}(B) & 0 \\ 0 & \det(B) \cdot \text{adj}(C) \end{pmatrix}

12. Proofs and Derivations

Theorem 5.39: Cayley-Hamilton Connection

The adjugate appears in the proof of Cayley-Hamilton theorem. For characteristic polynomial p(λ):

(A - \lambda I) \cdot \text{adj}(A - \lambda I) = p(\lambda) \cdot I

Comparing coefficients of λ leads to p(A) = 0.

Proof:

Outline: The entries of adj(A - λI) are polynomials in λ of degree at most n-1.

Write adj(A - λI) = B₀ + B₁λ + ... + B_{n-1}λ^{n-1} for some matrices Bₖ.

Expand (A - λI)·adj(A - λI) = p(λ)·I and compare powers of λ.

Multiply the k-th equation by A^k and sum to get p(A) = 0.

∎

Remark 5.27: Cramer's Rule Derivation

The solution to Ax = b can be written:

x = A^{-1}b = \frac{1}{\det(A)} \text{adj}(A) \cdot b

The i-th component is:

x_i = \frac{1}{\det(A)} \sum_j [\text{adj}(A)]_{ij} b_j = \frac{1}{\det(A)} \sum_j A_{ji} b_j

This sum equals det(Aᵢ) where Aᵢ has column i replaced by b, giving Cramer's rule.

Example 5.32: Derivation for 2×2

For $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} e \\ f \end{pmatrix}$ :

\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} e \\ f \end{pmatrix} = \frac{1}{ad-bc} \begin{pmatrix} de-bf \\ -ce+af \end{pmatrix}

So x = (de-bf)/(ad-bc) = det(e,b; f,d)/det(A) ✓

Additional Practice

Problem 5

If A is orthogonal with det(A) = 1, prove adj(A) = Aᵀ.

Hint: Use adj(A) = det(A)·A⁻¹ and the fact that A⁻¹ = Aᵀ for orthogonal matrices.

Problem 6

For nilpotent matrix N (where N^k = 0 for some k), what is adj(I - N)?

Hint: Use the geometric series (I - N)⁻¹ = I + N + N² + ...

Problem 7 (Challenge)

Prove: For rank n-1 matrix A, rank(adj(A)) = 1 and adj(A) = uvᵀ for some vectors u, v.

Problem 8

Compute adj(A) for $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{pmatrix}$ (Vandermonde matrix).

Study Tips

Master 2×2: The swap-and-negate pattern is fundamental and appears constantly.
Remember the transpose: adj(A) = Cᵀ, not C. This is the most common error.
Use verification: Always check A·adj(A) = det(A)·I to catch mistakes.
Know when NOT to use: For numerical work with n > 4, use row reduction instead.
Connect concepts: Adjugate links to Cramer, inverse, and Cayley-Hamilton.

Advanced Topics

Theorem 5.40: Rank of Adjugate

For an n×n matrix A:

If rank(A) = n: rank(adj(A)) = n
If rank(A) = n-1: rank(adj(A)) = 1
If rank(A) < n-1: adj(A) = 0

Proof:

When rank(A) = n-1, exactly one (n-1)×(n-1) minor is nonzero, giving rank(adj(A)) = 1.

When rank(A) < n-1, all (n-1)×(n-1) minors are zero, so adj(A) = 0.

∎

Remark 5.28: Geometric Interpretation

For a linear map T: ℝⁿ → ℝⁿ with matrix A:

The rows of adj(A) are normal vectors to the column space of A
For rank n-1, adj(A) = uvᵀ where u ∈ null(Aᵀ) and v ∈ null(A)

Example 5.33: Rank n-1 Case

For $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ (rank 1):

det(A) = 4 - 4 = 0

adj(A) = (4, -2; -2, 1) ≠ 0

Note: rank(adj(A)) = 1 (both rows are multiples of (2, -1))

Verify: A · adj(A) = (1,2; 2,4)(4,-2; -2,1) = (0,0; 0,0) ✓

Theorem 5.41: Adjugate of Similar Matrices

If B = P⁻¹AP (A and B are similar), then:

\text{adj}(B) = P^{-1} \cdot \text{adj}(A) \cdot P

Remark 5.29: Connection to Exterior Algebra

In advanced linear algebra, the adjugate relates to the exterior (wedge) product:

The (n-1)-st exterior power of A gives adj(A)
This connection extends adjugate theory to multilinear algebra

Challenge Problems

Challenge 1

Prove that for any matrix A: A · adj(A) = adj(A) · A (both products equal det(A)·I).

Challenge 2

For 3×3 matrix A with det(A) = 2, find det(adj(adj(adj(A)))).

Hint: Use det(adj(B)) = det(B)^{n-1} iteratively.

Challenge 3

Show that if A² = A (idempotent), then adj(A) is also related to A by a simple formula.

Challenge 4

Prove: tr(adj(A)) equals the sum of (n-1)×(n-1) principal minors of A.

Algorithm

Adjugate Computation Algorithm

Input: n×n matrix A

Output: adj(A)

1. Initialize n×n matrix C

2. FOR i = 1 to n:

FOR j = 1 to n:

M = A with row i, col j deleted

C[i,j] = (-1)^{i+j} · det(M)

3. RETURN Cᵀ

Time complexity: O(n² · T(n-1)) where T(k) is time to compute k×k determinant

What's Next?

With adjugate matrices mastered, you're ready for:

Eigenvalues: Use det(A - λI) = 0 to find eigenvalues
Cayley-Hamilton: Every matrix satisfies its characteristic polynomial
Diagonalization: Express A = PDP⁻¹ using eigenvalues

Skills Mastered

Computing adjugate for 2×2, 3×3, and larger matrices
Applying the fundamental identity A · adj(A) = det(A) · I
Deriving matrix inverse via adjugate
Key properties: det(adj), adj(AB), adj(Aᵀ)

Chapter Summary

This module covered the adjugate matrix, the transpose of the cofactor matrix. The fundamental identity A · adj(A) = det(A) · I leads to the explicit inverse formula A⁻¹ = adj(A)/det(A).

Core Theorems

Quiz Questions

FAQs Answered

Practice Problems

Key Formulas to Remember

• Definition: [adj(A)]ᵢⱼ = Aⱼᵢ (transposed cofactors)
• Identity: A · adj(A) = det(A) · I
• Inverse: A⁻¹ = adj(A) / det(A)
• det(adj(A)): det(A)^{n-1}
• adj(AB): adj(B) · adj(A) (reversed order)

2×2 and 3×3 Quick Reference

2×2 Matrix

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

det(A) = ad - bc

\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

3×3 Process

1. Compute 9 cofactors Aᵢⱼ
2. Form cofactor matrix C
3. Transpose: adj(A) = Cᵀ
4. For inverse: A⁻¹ = adj(A)/det(A)

Verification Checklist

☐ Did you transpose the cofactor matrix?
☐ Did you use the checkerboard sign pattern for cofactors?
☐ Did you delete the correct row AND column for each minor?
☐ Verify: A · adj(A) should equal det(A) · I
☐ Verify: det(adj(A)) should equal det(A)^{n-1}
☐ For inverse: Check A · A⁻¹ = I

Common Adjugate Patterns

Identity Matrix

adj(I) = I

All cofactors are identity cofactors

Diagonal Matrix

adj(diag(d₁,...,dₙ)) = diag(det/d₁,...,det/dₙ)

Each diagonal scaled by det/entry

Orthogonal Matrix

adj(Q) = det(Q) · Qᵀ = ±Qᵀ

Since det(Q) = ±1

Scalar Multiple

adj(cA) = c^{n-1} adj(A)

Power is n-1, not n

Product

adj(AB) = adj(B) · adj(A)

Order reverses like inverse

Transpose

adj(Aᵀ) = (adj(A))ᵀ

Adjugate and transpose commute

Learning Path

Cofactors→Adjugate→Inverse Formula→Cramer's Rule→Eigenvalues

You are here! The adjugate bridges determinant theory to eigenvalue analysis.

Method Comparison

Method	Complexity	Best For	Limitations
Adjugate Formula	O(n·n!)	Symbolic, small matrices	Exponential growth
Row Reduction	O(n³)	Numerical computation	May lose exact values
LU Decomposition	O(n³)	Multiple systems, speed	Requires pivoting
Cayley-Hamilton	O(n³)	Theoretical analysis	Requires char. poly

Complete Worked Examples

Example: Complete 3×3 Inverse via Adjugate

Find A⁻¹ for $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{pmatrix}$

Step 1: Compute determinant

det(A) = 1(1·1 - 1·0) - 2(0·1 - 1·2) + 0 = 1 + 4 = 5

Step 2: Compute cofactors

A₁₁ = +(1-0) = 1

A₁₂ = -(0-2) = 2

A₁₃ = +(0-2) = -2

A₂₁ = -(2-0) = -2

A₂₂ = +(1-0) = 1

A₂₃ = -(0-4) = 4

A₃₁ = +(2-0) = 2

A₃₂ = -(1-0) = -1

A₃₃ = +(1-0) = 1

Step 3: Form adj(A) = Cᵀ

\text{adj}(A) = \begin{pmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{pmatrix}

Step 4: Compute inverse

A^{-1} = \frac{1}{5}\begin{pmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{pmatrix}

Example: Using Adjugate Properties

Given det(A) = 3 for a 4×4 matrix, find det(adj(2A)).

Solution:

Step 1: adj(2A) = 2^{4-1} adj(A) = 8·adj(A)

Step 2: det(adj(2A)) = det(8·adj(A)) = 8^4 · det(adj(A))

Step 3: det(adj(A)) = det(A)^{4-1} = 3³ = 27

Step 4: det(adj(2A)) = 4096 · 27 = 110,592

Key Insights

Theoretical Power

The adjugate gives explicit formulas (not algorithms), making it invaluable for proofs and symbolic manipulation.

Computational Limit

O(n!) complexity makes adjugate impractical for numerical work with n > 4. Use row reduction instead.

Singular Insight

Even when A is singular, adj(A) exists. The columns of adj(A) span the null space of A.

Connection Hub

Adjugate connects inverse, Cramer's rule, and Cayley-Hamilton - a central concept in matrix theory.

Frequently Asked Questions

What's the difference between adjugate and adjoint?

The adjugate (classical adjoint) is the transpose of the cofactor matrix. In modern linear algebra, 'adjoint' usually means the conjugate transpose (A* or A†), so 'adjugate' is used to avoid confusion.

Why is the adjugate useful?

It gives an explicit closed-form formula for the inverse: A^{-1} = adj(A)/det(A). This is valuable for theoretical proofs, symbolic computation, and deriving Cramer's rule.

How do I compute adj(A) for a 3×3 matrix?

Step 1: Compute all 9 cofactors A_{ij} (remember the checkerboard signs). Step 2: Arrange them in a 3×3 matrix. Step 3: Transpose to get adj(A). Remember: [adj(A)]_{ij} = A_{ji}.

What if det(A) = 0?

Then A is singular and A^{-1} doesn't exist. However, adj(A) still exists! The identity becomes A·adj(A) = 0·I = 0. This means columns of adj(A) are in the null space of A.

Why do we transpose the cofactor matrix?

The transpose is needed to make the identity A·adj(A) = det(A)·I work. Without transposing, the (i,j)-entry of the product would involve row i of A with cofactors of row j - not what we want.

Is adj(A) = A^{-1} for orthogonal matrices?

No! For orthogonal Q, we have Q^{-1} = Q^T. The adjugate is adj(Q) = det(Q)·Q^T = ±Q^T (since det(Q) = ±1).

How does adjugate relate to Cramer's rule?

Cramer's rule x_i = det(A_i)/det(A) can be written as x = A^{-1}b = adj(A)b/det(A). The columns of adj(A) contain exactly the information needed for Cramer's formula.

Can I compute adj(A) without computing all cofactors?

For specific applications (like finding one entry of A^{-1}), you might only need certain cofactors. But for the full adj(A), all n² cofactors are required.

What's the computational complexity of adj(A)?

Computing adj(A) directly requires n² cofactors, each an (n-1)×(n-1) determinant. This is O(n² · (n-1)!) ≈ O(n·n!). Row reduction for A^{-1} is only O(n³), vastly faster for large n.

Does adj(AB) = adj(A)adj(B)?

No! Like the inverse, adjugate reverses order: adj(AB) = adj(B)·adj(A). This can be proved from the definition using (AB)^{-1} = B^{-1}A^{-1}.

Adjugate Matrix

1. Definition of Adjugate

2. The Fundamental Identity

3. Properties of Adjugate

Summary of Properties

4. Computing the Adjugate

Algorithm for adj(A)

5. Common Mistakes

Forgetting to transpose

Confusing adjugate with adjoint

Wrong sign pattern in cofactors

Thinking adj(AB) = adj(A)adj(B)

Using adjugate for large numerical matrices

6. Applications of Adjugate

7. Key Takeaways

Definition

Fundamental Identity

Inverse Formula

Key Property

8. Practice Problems

Solutions

9. Quick Reference

Core Formulas

2×2 Formula

10. Historical Notes

Etymology

11. Special Cases

12. Proofs and Derivations

Additional Practice

Study Tips

Advanced Topics

Challenge Problems

Algorithm

Adjugate Computation Algorithm

What's Next?

Skills Mastered

Chapter Summary

Key Formulas to Remember

2×2 and 3×3 Quick Reference

2×2 Matrix

3×3 Process

Verification Checklist

Common Adjugate Patterns

Identity Matrix

Diagonal Matrix

Orthogonal Matrix

Scalar Multiple

Product

Transpose

Learning Path

Method Comparison

Complete Worked Examples

Example: Complete 3×3 Inverse via Adjugate

Example: Using Adjugate Properties

Key Insights

Theoretical Power

Computational Limit

Singular Insight

Connection Hub

Related Topics

Frequently Asked Questions

What's the difference between adjugate and adjoint?

Why is the adjugate useful?

How do I compute adj(A) for a 3×3 matrix?

What if det(A) = 0?

Why do we transpose the cofactor matrix?

Is adj(A) = A^{-1} for orthogonal matrices?

How does adjugate relate to Cramer's rule?

Can I compute adj(A) without computing all cofactors?

What's the computational complexity of adj(A)?

Does adj(AB) = adj(A)adj(B)?