Adjugate Matrix & Inverse Matrix: Relationship and Computation | 伴随矩阵与逆矩阵的关系及求法

📚 Adjugate Matrix & Inverse Matrix: Relationship and Computation | 伴随矩阵与逆矩阵的关系及求法

The adjugate matrix (also called the classical adjoint) is a fundamental concept in linear algebra, especially when dealing with inverse matrices. This article explains the definition of the adjugate matrix, its key relationship with the inverse matrix, and provides step-by-step methods for computing both, tailored for IB Mathematics students.

伴随矩阵（也称为经典伴随）是线性代数中的基本概念，尤其在处理逆矩阵时尤为重要。本文将解释伴随矩阵的定义、它与逆矩阵之间的关键关系，并提供分步计算逆矩阵的方法，专为IB数学学生设计。

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1. What is a Cofactor Matrix? | 什么是余子式矩阵？

atrix, we must first understand the cofactor matrix. For an n × n matrix A, the cofactor C_ij is calculated as:

在定义伴随矩阵之前，我们必须先理解余子式矩阵。对于一个 n × n 矩阵 A，余子式 C_ij 的计算公式为：

C_ij = (−1)^i+j × M_ij

where M_ij is the minor of entry a_ij, that is, the determinant of the submatrix obtained by deleting the i-th row and j-th column of A.

其中 M_ij 是元素 a_ij 的余子式，即删除 A 的第 i 行和第 j 列后所得子矩阵的行列式。

Example – 2×2 matrix: For A = [a b; c d], we have:

示例 – 2×2 矩阵：对于 A = [a b; c d]，我们有：

• C₁₁ = d
• C₁₂ = −c
• C₂₁ = −b
• C₂₂ = a

The cofactor matrix is formed by placing each C_ij in position (i, j).

余子式矩阵是将每个 C_ij 放在位置 (i, j) 上组成的矩阵。

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2. Definition of the Adjugate Matrix | 伴随矩阵的定义

The adjugate of a square matrix A, denoted adj(A), is defined as the transpose of the cofactor matrix of A. In other words:

方阵 A 的伴随矩阵，记作 adj(A)，定义为 A 的余子式矩阵的转置。换句话说：

adj(A) = (Cofactor Matrix of A)^T

This means the (i, j) entry of adj(A) is C_ji, not C_ij. The transpose operation is essential — forgetting it is one of the most common student errors.

这意味着 adj(A) 的第 (i, j) 个元素是 C_ji，而不是 C_ij。转置操作至关重要——忘记转置是学生最常见的错误之一。

Worked example: Find adj(A) for A = [2 3; 1 −4].

示例：求 A = [2 3; 1 −4] 的 adj(A)。

Cofactor matrix: [−4 −1; −3 2]

余子式矩阵：[−4 −1; −3 2]

Transpose: adj(A) = [−4 −3; −1 2]

转置后：adj(A) = [−4 −3; −1 2]

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3. The Fundamental Identity: A × adj(A) = det(A) × I | 基本恒等式：A × adj(A) = det(A) × I

The most important relationship between a matrix and its adjugate is:

矩阵与其伴随矩阵之间最重要的关系是：

A × adj(A) = adj(A) × A = det(A) × I

where I is the identity matrix of the same order as A.

其中 I 是与 A 同阶的单位矩阵。

• If det(A) ≠ 0, this identity directly leads to the inverse formula.
• If det(A) = 0, the identity shows that adj(A) exists but A is not invertible.

• 如果 det(A) ≠ 0，这个恒等式直接导出逆矩阵公式。
• 如果 det(A) = 0，恒等式表明 adj(A) 存在但 A 不可逆。

Verification with the 2×2 example: Using A = [2 3; 1 −4], adj(A) = [−4 −3; −1 2], det(A) = (2)(−4) − (3)(1) = −11.

用2×2示例验证：使用 A = [2 3; 1 −4]，adj(A) = [−4 −3; −1 2]，det(A) = (2)(−4) − (3)(1) = −11。

A × adj(A) = [2 3; 1 −4] × [−4 −3; −1 2] = [−11 0; 0 −11] = −11 × I

The result confirms the identity perfectly.

结果完美地验证了该恒等式。

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4. Inverse Matrix Formula Using the Adjugate | 利用伴随矩阵求逆矩阵的公式

When det(A) ≠ 0, the inverse of matrix A is given by:

当 det(A) ≠ 0 时，矩阵 A 的逆矩阵由下式给出：

A⁻¹ = (1 / det(A)) × adj(A)

re matrices of any order, though for 3×3 and above the computation becomes more laborious.

这个公式适用于任意阶数的方阵，尽管对于3×3及以上的矩阵计算会更加繁重。

Example: Find A⁻¹ for A = [2 3; 1 −4].

示例：求 A = [2 3; 1 −4] 的 A⁻¹。

We already have det(A) = −11 and adj(A) = [−4 −3; −1 2]. Thus:

我们已经得到 det(A) = −11 和 adj(A) = [−4 −3; −1 2]。因此：

A⁻¹ = (1/−11) × [−4 −3; −1 2] = [4/11 3/11; 1/11 −2/11]

To check, multiply A × A⁻¹ to confirm the product equals the identity matrix.

为验证结果，将 A 与 A⁻¹ 相乘，确认乘积等于单位矩阵。

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5. Computing the Adjugate for a 3×3 Matrix | 计算3×3矩阵的伴随矩阵

For a 3×3 matrix, the process involves nine cofactors. The general matrix A is:

对于3×3矩阵，计算过程涉及九个余子式。一般矩阵 A 为：

A = [a₁₁ a₁₂ a₁₃; a₂₁ a₂₂ a₂₃; a₃₁ a₃₂ a₃₃]

The cofactor C_ij is computed using 2×2 determinants. For example:

余子式 C_ij 使用2×2行列式计算。例如：

C₁₁ = (a₂₂a₃₃ − a₂₃a₃₂)

C₁₂ = −(a₂₁a₃₃ − a₂₃a₃₁)

C₁₃ = (a₂₁a₃₂ − a₂₂a₃₁)

The cofactor matrix is then transposed to obtain adj(A).

然后将余子式矩阵转置即可得到 adj(A)。

Tip: Use the checkerboard pattern of signs (+ − +; − + −; + − +) carefully. A single sign error can destroy the entire inverse.

提示：要小心使用符号棋盘模式（+ − +；− + −；+ − +）。一个符号错误就会毁掉整个逆矩阵。

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6. 3×3 Worked Example | 3×3完整示例

Let A = [1 2 3; 0 1 4; 5 6 0]. Find A⁻¹.

设 A = [1 2 3; 0 1 4; 5 6 0]。求 A⁻¹。

Step 1: Compute det(A). Using expansion along the first row:

第一步：计算 det(A)。沿第一行展开：

×0 − 4×6) − 2×(0×0 − 4×5) + 3×(0×6 − 1×5) = 1×(−24) − 2×(−20) + 3×(−5) = −24 + 40 − 15 = 1

Since det(A) = 1 ≠ 0, the inverse exists.

因为 det(A) = 1 ≠ 0，所以逆矩阵存在。

Step 2: Compute the cofactor matrix.

第二步：计算余子式矩阵。

C₁₁ = (1×0 − 4×6) = −24	C₁₂ = −(0×0 − 4×5) = 20	C₁₃ = (0×6 − 1×5) = −5
C₂₁ = −(2×0 − 3×6) = 18	C₂₂ = (1×0 − 3×5) = −15	C₂₃ = −(1×6 − 2×5) = 4
C₃₁ = (2×4 − 3×1) = 5	C₃₂ = −(1×4 − 3×0) = −4	C₃₃ = (1×1 − 2×0) = 1

Cofactor matrix = [−24 20 −5; 18 −15 4; 5 −4 1]

余子式矩阵 = [−24 20 −5; 18 −15 4; 5 −4 1]

Step 3: Transpose to get the adjugate.

第三步：转置得到伴随矩阵。

adj(A) = [−24 18 5; 20 −15 −4; −5 4 1]

Step 4: Multiply by 1/det(A). Since det(A) = 1, we have:

第四步：乘以 1/det(A)。因为 det(A) = 1，所以：

A⁻¹ = [−24 18 5; 20 −15 −4; −5 4 1]

This is the final inverse matrix.

这就是最终的逆矩阵。

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7. Properties of the Adjugate Matrix | 伴随矩阵的性质

The adjugate matrix has several important properties that are useful in IB exam problems:

伴随矩阵有几个在IB考试中非常有用的重要性质：

• det(adj(A)) = (det(A))ⁿ⁻¹ for an n × n matrix A.
• adj(A⁻¹) = (adj(A))⁻¹ when A is invertible.
• adj(kA) = kⁿ⁻¹ × adj(A) for a scalar k.
• adj(AB) = adj(B) × adj(A) for invertible matrices A and B (note the reversed order).

• 对于 n × n 矩阵 A，有 det(adj(A)) = (det(A))ⁿ⁻¹。
• 当 A 可逆时，adj(A⁻¹) = (adj(A))⁻¹。
• 对于标量 k，有 adj(kA) = kⁿ⁻¹ × adj(A)。
• 对于可逆矩阵 A 和 B，有 adj(AB) = adj(B) × adj(A)（注意顺序颠倒）。

These properties can significantly reduce computation time when dealing with advanced problems.

这些性质在面对高阶题目时可以显著减少计算时间。

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8. Special Case: 2×2 Formula Shortcut | 特殊情况：2×2矩阵快捷公式

For a 2×2 matrix, there is a well-known shortcut for the inverse:

对于2×2矩阵，存在一个众所周知的逆矩阵快捷公式：

If A = [a b; c d], then A⁻¹ = (1/(ad − bc)) × [d −b; −c a]

This shortcut is equivalent to the adjugate method:

这个快捷公式等价于伴随矩阵方法：

• det(A) = ad − bc
• adj(A) = [d −b; −c a]

Note the pattern: swap the diagonal entries a and d, and change the signs of the off-diagonal entries b and c. Then divide by the determinant.

注意规律：交换对角元素 a 和 d，并改变非对角元素 b 和 c 的符号，然后除以行列式。

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9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students frequently make the following errors when computing adjugates and inverses:

学生在计算伴随矩阵和逆矩阵时经常犯以下错误：

Mistake | 错误	Correction | 纠正方法
Forgetting to transpose the cofactor matrix	Always write adj(A) = (cofactor matrix)ᵀ
Incorrect signs in cofactors	Use the (−1)ⁱ⁺ʲ rule systematically
Dividing by zero when det(A) = 0	Check that det(A) ≠ 0 before computing A⁻¹
Applying the 2×2 shortcut to 3×3 matrices	Use full cofactor expansion for 3×3

忘记转置余子式矩阵

始终写 adj(A) = (余子式矩阵)ᵀ

余子式符号错误

系统地使用 (−1)ⁱ⁺ʲ 规则

当 det(A) = 0 时除以零

在计算 A⁻¹ 前确认 det(A) ≠ 0

把2×2快捷公式用到3×3矩阵上

3×3矩阵需要用完整的余子式展开

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10. When Is a Matrix Invertible? | 矩阵何时可逆？

A square matrix A is invertible if and only if det(A) ≠ 0. The adjugate matrix itself always exists, but the inverse exists only when the determinant is non-zero.

方阵 A 可逆当且仅当 det(A) ≠ 0。伴随矩阵本身总是存在，但只有行列式不为零时逆矩阵才存在。

Equivalent conditions for invertibility:

可逆的等价条件：

• det(A) ≠ 0
• A has full rank
• The rows (or columns) of A are linearly independent
• The system Ax = 0 has only the trivial solution x = 0

• det(A) ≠ 0
• A 满秩
• A 的行（或列）线性无关
• 方程组 Ax = 0 只有零解 x = 0

If any of these conditions fails, A is singular and has no inverse.

如果这些条件中的任何一个不成立，则 A 是奇异的，没有逆矩阵。

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11. Applications in IB Mathematics | 在IB数学中的应用

The adjugate-inverse relationship appears in multiple IB topics:

伴随矩阵与逆矩阵的关系在IB多个主题中出现：

• Solving systems of linear equations using matrix inversion: X = A⁻¹B.
• Transformation matrices in geometry — inverse matrices undo transformations.
• Cryptography with Hill ciphers, where decryption uses the inverse of the key matrix.
• lue problems and diagonalization, where the determinant condition det(A − λI) = 0 is used.

• 使用矩阵求逆求解线性方程组：X = A⁻¹B。
• 几何中的变换矩阵——逆矩阵可以撤销变换。
• Hill密码中的加密解密，解密需要使用密钥矩阵的逆矩阵。
• 特征值问题和对角化，使用行列式条件 det(A − λI) = 0。

Understanding the adjugate method strengthens conceptual understanding beyond just memorizing formulas.

理解伴随矩阵方法可以加深概念理解，而不仅仅是记住公式。

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12. Summary and Revision Checklist | 总结与复习清单

Key points to remember for exams:

考试需要记住的要点：

Concept | 概念	Formula | 公式
Cofactor	Cij = (−1)ⁱ⁺ʲ × Mij
Adjugate	adj(A) = (cofactor matrix)ᵀ
Fundamental identity	A × adj(A) = det(A) × I
Inverse formula	A⁻¹ = (1/det(A)) × adj(A)
2×2 shortcut	A⁻¹ = (1/(ad−bc)) × [d −b; −c a]

anspose process becomes automatic. Always verify your result by checking that A × A⁻¹ = I.

练习2×2和3×3矩阵，直到余子式和转置过程变得熟练自如。务必通过验证 A × A⁻¹ = I 来检查计算结果。

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