Inverse Matrix: Computation Methods & Properties | 逆矩阵的求法及其性质考点

📚 Inverse Matrix: Computation Methods & Properties | 逆矩阵的求法及其性质考点

In linear algebra, the inverse of a matrix is one of the most frequently tested topics in IB Mathematics Analysis and Approaches HL as well as other advanced curricula. A thorough understanding of how to compute inverses and apply their properties is essential for solving systems of linear equations, transforming geometric objects, and simplifying matrix expressions.

在线性代数中，逆矩阵是 IB 数学分析与方法 HL 以及其他高级课程中最常考查的知识点之一。熟练掌握逆矩阵的计算方法及其性质，对于求解线性方程组、几何变换以及化简矩阵表达式都至关重要。

1. Definition and Existence | 定义与存在条件

For a square matrix A of order n, if there exists a square matrix B of the same order such that AB = BA = Iₙ, where Iₙ is the identity matrix of order n, then B is called the inverse of A, denoted as A⁻¹.

对于 n 阶方阵 A，若存在同阶方阵 B，使得 AB = BA = Iₙ，其中 Iₙ 为 n 阶单位矩阵，则称 B 为 A 的逆矩阵，记作 A⁻¹。

The inverse of a matrix exists if and only if the matrix is non-singular. Non-singular means the determinant of A is non-zero: det(A) ≠ 0. If det(A) = 0, the matrix is singular and has no inverse.

矩阵存在逆矩阵当且仅当该矩阵是非奇异的。非奇异意味着矩阵的行列式不为零：det(A) ≠ 0。若 det(A) = 0，则该矩阵为奇异矩阵，不存在逆矩阵。

det(A) = ad − bc ≠ 0 for a 2×2 matrix A = [a b; c d]

It is also important to note that only square matrices can have inverses. Rectangular matrices do not have two-sided inverses in the usual sense, although they may have left or right inverses in special cases.

还需注意，只有方阵才可能有逆矩阵。长方形矩阵在通常意义下没有双侧逆矩阵，尽管在特殊情况下可能存在左逆或右逆。

2. Inverse of a 2×2 Matrix | 二阶矩阵的逆

For a 2×2 matrix A = [a b; c d], the inverse is computed by swapping the diagonal entries a and d, changing the signs of the off-diagonal entries b and c, and then multiplying by 1/det(A).

对于二阶矩阵 A = [a b; c d]，其逆矩阵的计算方法是：交换主对角线元素 a 和 d，改变副对角线元素 b 和 c 的符号，然后乘以 1/det(A)。

A⁻¹ = (1/(ad − bc)) × [d −b; −c a]

As an example, consider A = [2 1; 5 3]. Then det(A) = 2×3 − 1×5 = 6 − 5 = 1, so A⁻¹ = [3 −1; −5 2].

例如，设 A = [2 1; 5 3]，则 det(A) = 2×3 − 1×5 = 6 − 5 = 1，因此 A⁻¹ = [3 −1; −5 2]。

Students must be careful with the order of operations: first compute the determinant, then form the adjugate-like matrix, and finally multiply through by the scalar. A common mistake is to forget the factor 1/det(A) or to incorrectly place the negative signs.

学生在计算时必须注意运算顺序：先计算行列式，再构造伴随型矩阵，最后乘上标量因子。常见错误包括忘记乘以 1/det(A)，或者负号位置摆放错误。

3. Inverse of a 3×3 Matrix: Cofactor Method | 三阶矩阵的逆：代数余子式法

For a 3×3 matrix, the inverse can be found using the cofactor method. The procedure has four steps: compute the matrix of minors, form the matrix of cofactors, transpose it to obtain the adjugate matrix, and divide by the determinant.

对于三阶矩阵，可通过代数余子式法求逆。该过程分为四步：计算余子式矩阵，形成代数余子式矩阵，转置得到伴随矩阵，最后除以行列式。

Let A be a 3×3 matrix. The minor Mᵢⱼ of entry aᵢⱼ is the determinant of the 2×2 matrix obtained by deleting the i-th row and j-th column. The cofactor Cᵢⱼ is then defined as (−1)ⁱ⁺ʲ × Mᵢⱼ.

设 A 为三阶矩阵。元素 aᵢⱼ 的余子式 Mᵢⱼ 是删除第 i 行和第 j 列后所得二阶矩阵的行列式。代数余子式定义为 Cᵢⱼ = (−1)ⁱ⁺ʲ × Mᵢⱼ。

A⁻¹ = (1/det(A)) × adj(A) = (1/det(A)) × (Cᵢⱼ)ᵀ

As a worked example, consider A = [1 2 3; 0 1 4; 5 6 0]. First find det(A) using expansion along the first row: det(A) = 1×(1×0 − 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, the inverse is simply the adjugate matrix.

来看一个完整例子，设 A = [1 2 3; 0 1 4; 5 6 0]。首先沿第一行展开计算行列式：det(A) = 1×(1×0 − 4×6) − 2×(0×0 − 4×5) + 3×(0×6 − 1×5) = 1×(−24) − 2×(−20) + 3×(−5) = −24 + 40 − 15 = 1。由于 det(A) = 1，所以逆矩阵就是伴随矩阵本身。

Computing the cofactors gives C₁₁ = −24, C₁₂ = 20, C₁₃ = −5, C₂₁ = 6, C₂₂ = −15, C₂₃ = 4, C₃₁ = 5, C₃₂ = −4, C₃₃ = 1. Thus A⁻¹ equals the transpose of this cofactor matrix.

计算各代数余子式得到：C₁₁ = −24，C₁₂ = 20，C₁₃ = −5，C₂₁ = 6，C₂₂ = −15，C₂₃ = 4，C₃₁ = 5，C₃₂ = −4，C₃₃ = 1。因此 A⁻¹ 等于这些代数余子式所构成矩阵的转置。

A⁻¹ = [−24 6 5; 20 −15 −4; −5 4 1]

This method works for square matrices of any order, but becomes computationally intensive for n ≥ 4. In such cases, row reduction is preferred.

该方法适用于任意阶方阵，但当 n ≥ 4 时计算量会大幅增加，此时更推荐使用行化简法。

4. Gaussian Elimination / Row Reduction Method | 高斯消元法 / 行化简法

The row reduction method is the most systematic way to compute the inverse of a matrix of any size. The idea is to form an augmented matrix [A | Iₙ] and perform elementary row operations until the left block becomes the identity matrix. The right block will then be A⁻¹.

行化简法是计算任意阶矩阵逆矩阵的最系统方法。其思路是构造增广矩阵 [A | Iₙ]，通过初等行变换将左侧块化为单位矩阵，此时右侧块即为 A⁻¹。

Only three types of elementary row operations are allowed: multiplying a row by a non-zero scalar, swapping two rows, and adding a multiple of one row to another row.

允许使用的初等行变换只有三种：将某行乘以非零标量、交换两行、将某一行的倍数加到另一行上。

Let us illustrate with A = [1 1; 2 3]. Form the augmented matrix [1 1 | 1 0; 2 3 | 0 1]. Subtract 2×row1 from row2 to get [1 1 | 1 0; 0 1 | −2 1]. Then subtract row2 from row1 to get [1 0 | 3 −1; 0 1 | −2 1]. Therefore A⁻¹ = [3 −1; −2 1].

以 A = [1 1; 2 3] 为例说明。构造增广矩阵 [1 1 | 1 0; 2 3 | 0 1]。将第二行减去 2 倍第一行，得到 [1 1 | 1 0; 0 1 | −2 1]。再将第一行减去第二行，得到 [1 0 | 3 −1; 0 1 | −2 1]。因此 A⁻¹ = [3 −1; −2 1]。

If during the process a row of all zeros appears in the left block, the matrix is singular and no inverse exists. This is equivalent to saying that the rows are linearly dependent.

如果在化简过程中左侧块出现全零行，则说明矩阵是奇异的，不存在逆矩阵。这等价于矩阵的行向量线性相关。

5. Key Properties of Inverse Matrices | 逆矩阵的核心性质

The inverse operation satisfies several important algebraic properties that appear frequently in exam questions. These properties allow students to simplify complex matrix expressions efficiently.

逆运算满足若干重要的代数性质，这些性质在考试题目中频繁出现。掌握这些性质可以帮助学生高效地化简复杂的矩阵表达式。

(A⁻¹)⁻¹ = A: the inverse of an inverse returns the original matrix.

(A⁻¹)⁻¹ = A：逆矩阵的逆矩阵等于原矩阵。
(AB)⁻¹ = B⁻¹A⁻¹: the inverse of a product reverses the order of multiplication. This is known as the “shoes-and-socks rule”.

(AB)⁻¹ = B⁻¹A⁻¹：乘积的逆矩阵需要颠倒相乘顺序，这就是所谓的”先穿袜子再穿鞋”法则。
(Aᵀ)⁻¹ = (A⁻¹)ᵀ: the inverse of a transpose equals the transpose of the inverse.

(Aᵀ)⁻¹ = (A⁻¹)ᵀ：转置的逆等于逆的转置。
(kA)⁻¹ = (1/k)A⁻¹ for any non-zero scalar k.

(kA)⁻¹ = (1/k)A⁻¹，其中 k 为任意非零标量。
det(A⁻¹) = 1/det(A): the determinant of the inverse is the reciprocal of the original determinant.

det(A⁻¹) = 1/det(A)：逆矩阵的行列式等于原矩阵行列式的倒数。

Furthermore, a matrix commutes with its own inverse: AA⁻¹ = A⁻¹A = Iₙ. However, in general matrix multiplication is not commutative, so AB ≠ BA and therefore (AB)⁻¹ ≠ A⁻¹B⁻¹ in general.

此外，矩阵与其自身逆矩阵满足交换律：AA⁻¹ = A⁻¹A = Iₙ。但一般而言矩阵乘法不满足交换律，因此通常 AB ≠ BA，(AB)⁻¹ ≠ A⁻¹B⁻¹。

6. Inverse of a Product and Commutativity | 乘积的逆与交换律

The property (AB)⁻¹ = B⁻¹A⁻¹ deserves special attention because it is one of the most commonly tested identities. Students often mistakenly write (AB)⁻¹ = A⁻¹B⁻¹, which is incorrect.

性质 (AB)⁻¹ = B⁻¹A⁻¹ 值得特别注意，因为它是最常考到的恒等式之一。学生经常错误地写成 (AB)⁻¹ = A⁻¹B⁻¹，这是不对的。

To verify the correct formula, multiply AB by B⁻¹A⁻¹: (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIₙA⁻¹ = AA⁻¹ = Iₙ. Similarly, (B⁻¹A⁻¹)(AB) = B⁻¹(A⁻¹A)B = B⁻¹IₙB = Iₙ. This confirms the correct order.

验证正确公式：将 AB 乘以 B⁻¹A⁻¹，得到 (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIₙA⁻¹ = AA⁻¹ = Iₙ。同理，(B⁻¹A⁻¹)(AB) = B⁻¹(A⁻¹A)B = B⁻¹IₙB = Iₙ。这证实了正确的运算顺序。

For a product of three matrices, the pattern continues: (ABC)⁻¹ = C⁻¹B⁻¹A⁻¹. The order of the factors is completely reversed.

对于三个矩阵的乘积，规律继续延伸：(ABC)⁻¹ = C⁻¹B⁻¹A⁻¹。各因子的顺序完全颠倒。

This property is especially useful in solving matrix equations such as AXB = C, where X = A⁻¹CB⁻¹, provided the required inverses exist.

这条性质在求解矩阵方程时特别有用，例如对于 AXB = C，可得 X = A⁻¹CB⁻¹，前提是所需逆矩阵均存在。

7. Inverse of a Transpose and Other Identities | 转置的逆与其他恒等式

The identity (Aᵀ)⁻¹ = (A⁻¹)ᵀ links the concepts of transposition and inversion. This is straightforward to prove by noting that Aᵀ(A⁻¹)ᵀ = (A⁻¹A)ᵀ = Iₙᵀ = Iₙ.

恒等式 (Aᵀ)⁻¹ = (A⁻¹)ᵀ 将转置与求逆联系起来。证明十分直接：Aᵀ(A⁻¹)ᵀ = (A⁻¹A)ᵀ = Iₙᵀ = Iₙ。

For a symmetric matrix A where Aᵀ = A, the inverse is also symmetric, since (A⁻¹)ᵀ = (Aᵀ)⁻¹ = A⁻¹.

对于满足 Aᵀ = A 的对称矩阵 A，其逆矩阵也是对称的，因为 (A⁻¹)ᵀ = (Aᵀ)⁻¹ = A⁻¹。

Another useful identity involves the determinant: det(A⁻¹) = 1/det(A). This follows directly from det(AA⁻¹) = det(Iₙ) = 1 and det(AB) = det(A)det(B).

另一个有用的恒等式涉及行列式：det(A⁻¹) = 1/det(A)。这可由 det(AA⁻¹) = det(Iₙ) = 1 以及 det(AB) = det(A)det(B) 直接推出。

For an orthogonal matrix Q where QᵀQ = QQᵀ = Iₙ, the inverse equals the transpose: Q⁻¹ = Qᵀ. Orthogonal matrices represent rotations and reflections, and this property makes them very convenient to work with in geometry and physics.

对于正交矩阵 Q，满足 QᵀQ = QQᵀ = Iₙ，其逆矩阵等于转置矩阵：Q⁻¹ = Qᵀ。正交矩阵对应旋转和反射变换，这条性质使得它们在几何与物理中非常便于使用。

8. Solving Linear Systems Using Inverses | 用逆矩阵求解线性方程组

A system of linear equations written in matrix form as AX = B can be solved by multiplying both sides on the left by A⁻¹, provided A is invertible. This gives X = A⁻¹B.

将线性方程组写成矩阵形式 AX = B 后，若 A 可逆，则可在等式两边同时左乘 A⁻¹，得到 X = A⁻¹B。

It is crucial to multiply on the left, not on the right, because matrix multiplication is not commutative. If we incorrectly write X = BA⁻¹, the result will generally be wrong.

这里必须左乘而不是右乘，因为矩阵乘法不满足交换律。如果错误地写成 X = BA⁻¹，结果通常是不对的。

Consider the system x + y = 5 and 2x + 3y = 11. In matrix form A = [1 1; 2 3], X = [x; y], and B = [5; 11]. Using A⁻¹ = [3 −1; −2 1] from the earlier example, X = [3 −1; −2 1] × [5; 11] = [3×5 + (−1)×11; (−2)×5 + 1×11] = [15 − 11; −10 + 11] = [4; 1]. Hence x = 4, y = 1.

考虑方程组 x + y = 5，2x + 3y = 11。矩阵形式中 A = [1 1; 2 3]，X = [x; y]，B = [5; 11]。使用前面例子中得到的 A⁻¹ = [3 −1; −2 1]，有 X = [3 −1; −2 1] × [5; 11] = [3×5 + (−1)×11; (−2)×5 + 1×11] = [15 − 11; −10 + 11] = [4; 1]。因此 x = 4，y = 1。

This method is particularly efficient when multiple systems share the same coefficient matrix A but have different right-hand side vectors B. In such cases, A⁻¹ only needs to be computed once.

当多个方程组共享同一个系数矩阵 A，但右端向量 B 不同时，这种方法尤其高效。此时只需计算一次 A⁻¹ 即可。

9. Inverse of Diagonal and Triangular Matrices | 对角矩阵和三角矩阵的逆

Diagonal matrices have especially simple inverses. If D = [d₁ 0; 0 d₂], then D⁻¹ = [1/d₁ 0; 0 1/d₂], provided both d₁ and d₂ are non-zero.

对角矩阵的逆矩阵形式特别简单。若 D = [d₁ 0; 0 d₂]，则 D⁻¹ = [1/d₁ 0; 0 1/d₂]，前提是 d₁ 和 d₂ 均非零。

For a general n×n diagonal matrix D = diag(d₁, d₂, …, dₙ), the inverse is diag(1/d₁, 1/d₂, …, 1/dₙ). This is the quickest inverse to compute in all of linear algebra.

对于一般的 n 阶对角矩阵 D = diag(d₁, d₂, …, dₙ)，其逆矩阵为 diag(1/d₁, 1/d₂, …, 1/dₙ)。这是线性代数中计算最快的逆矩阵。

Upper triangular matrices also have triangular inverses, but the computation is more involved because each diagonal entry must be non-zero and back-substitution is needed. A common exam question asks students to find the inverse of a triangular matrix by inspection when the diagonal entries are simple.

上三角矩阵的逆矩阵也是三角矩阵，但计算过程更为复杂，因为要求每个对角元非零，并且需要进行回代。常见的考试题目会要求学生针对对角元较简单的三角矩阵直接写出其逆矩阵。

It is also worth noting that the inverse of a product of diagonal matrices can be computed either by inverting each factor and reversing the order, or by simply taking reciprocals of the diagonal entries. Both approaches yield the same result.

还需注意，对角矩阵乘积的逆既可以通过对每个因子求逆并颠倒顺序来计算，也可以直接对所有对角元取倒数。两种方法得到的结果相同。

10. Common Mistakes and Exam Pitfalls | 常见错误与考试陷阱

Students frequently lose marks on inverse matrix questions due to a few repeated misconceptions. The first is attempting to compute the inverse without first checking whether det(A) = 0. If the determinant is zero, the matrix is singular and the inverse does not exist.

学生在逆矩阵题目中经常因为几个反复出现的误解而失分。第一个误区是不先检查 det(A) 是否为零就直接计算逆矩阵。若行列式为零，矩阵为奇异矩阵，逆矩阵根本不存在。

The second common mistake is applying row operations incorrectly or inconsistently when using Gaussian elimination. Every operation must be applied to the entire augmented row, including the right-hand block.

第二个常见错误是在使用高斯消元法时行变换运用不当或不一致。每个操作必须作用于整个增广行，包括右侧的分块。

A third error is confusing the inverse with the transpose or the adjugate. For a 2×2 matrix, the inverse is not simply the adjugate unless det(A) = 1. Students should always write the scalar multiplier explicitly.

第三个错误是混淆逆矩阵与转置或伴随矩阵。对于二阶矩阵，逆矩阵并不简单地等于伴随矩阵，除非 det(A) = 1。学生应始终明确写出标量倍数。

Finally, when solving AX = B, students sometimes multiply on the right by A⁻¹ instead of the left. Remember the formula precisely: X = A⁻¹B, not BA⁻¹.

最后一点，在求解 AX = B 时，学生有时会右乘 A⁻¹ 而不是左乘。请务必记准公式：X = A⁻¹B，而不是 BA⁻¹。

11. Worked Exam-Style Question | 典型考题精讲

Let M = [4 2; 3 2]. Find M⁻¹ and verify that M⁻¹M = I₂. Then solve the system 4x + 2y = 10, 3x + 2y = 7.

设 M = [4 2; 3 2]。求 M⁻¹ 并验证 M⁻¹M = I₂。随后求解方程组 4x + 2y = 10，3x + 2y = 7。

First compute det(M) = 4×2 − 2×3 = 8 − 6 = 2. Since det(M) ≠ 0, the inverse exists. Using the 2×2 formula: M⁻¹ = (1/2) × [2 −2; −3 4] = [1 −1; −3/2 2].

首先计算 det(M) = 4×2 − 2×3 = 8 − 6 = 2。由于 det(M) ≠ 0，逆矩阵存在。根据二阶矩阵求逆公式：M⁻¹ = (1/2) × [2 −2; −3 4] = [1 −1; −3/2 2]。

To verify: M⁻¹M = [1 −1; −3/2 2] × [4 2; 3 2] = [1×4 + (−1)×3, 1×2 + (−1)×2; (−3/2)×4 + 2×3, (−3/2)×2 + 2×2] = [4 − 3, 2 − 2; −6 + 6, −3 + 4] = [1 0; 0 1] = I₂. The verification is correct.

验证：M⁻¹M = [1 −1; −3/2 2] × [4 2; 3 2] = [1×4 + (−1)×3, 1×2 + (−1)×2; (−3/2)×4 + 2×3, (−3/2)×2 + 2×2] = [4 − 3, 2 − 2; −6 + 6, −3 + 4] = [1 0; 0 1] = I₂。验证正确。

Now solve the system. Write X = [x; y] and B = [10; 7]. Since MX = B, X = M⁻¹B = [1 −1; −3/2 2] × [10; 7] = [1×10 + (−1)×7; (−3/2)×10 + 2×7] = [10 − 7; −15 + 14] = [3; −1]. Thus x = 3, y = −1.

接下来解方程组。设 X = [x; y]，B = [10; 7]。因为 MX = B，所以 X = M⁻¹B = [1 −1; −3/2 2] × [10; 7] = [1×10 + (−1)×7; (−3/2)×10 + 2×7] = [10 − 7; −15 + 14] = [3; −1]。因此 x = 3，y = −1。

This single problem tests multiple skills: computing a determinant, applying the 2×2 inverse formula, verifying the definition of the inverse, and solving a linear system using matrix inversion.

这道题综合考查了多项技能：计算行列式、套用二阶逆矩阵公式、验证逆矩阵定义以及使用矩阵求逆求解线性方程组。

12. Summary and Revision Strategy | 总结与复习策略

To master inverse matrices for IB examinations, students should practise the 2×2 formula until it becomes automatic, learn the cofactor method for 3×3 matrices, and understand how to use Gaussian elimination for larger or more complex matrices.

为了在 IB 考试中熟练掌握逆矩阵，学生应当反复练习二阶公式直至能自动运用，学习三阶矩阵的代数余子式法，并掌握如何用高斯消元法处理更大或更复杂的矩阵。

Equally important are the algebraic properties: (A⁻¹)⁻¹ = A, (AB)⁻¹ = B⁻¹A⁻¹, (Aᵀ)⁻¹ = (A⁻¹)ᵀ, and det(A⁻¹) = 1/det(A). These identities appear in both Paper 1 and Paper 2 questions.

同样重要的是代数性质：(A⁻¹)⁻¹ = A，(AB)⁻¹ = B⁻¹A⁻¹，(Aᵀ)⁻¹ = (A⁻¹)ᵀ 以及 det(A⁻¹) = 1/det(A)。这些恒等式在 Paper 1 和 Paper 2 中均有出现。

Students should also recognise when a matrix does not have an inverse, and be comfortable solving systems of equations through matrix inversion rather than always resorting to substitution or elimination.

学生还应当能够判断矩阵何时不存在逆矩阵，并习惯通过矩阵求逆来解方程组，而不是总是依赖代入法或消元法。

Finally, always verify your result by checking AA⁻¹ = Iₙ, especially when time permits. This quick check catches most arithmetic errors and is a habit that distinguishes high-scoring students.

最后，条件允许时一定要通过 AA⁻¹ = Iₙ 来验证结果。这个快速检查能发现大多数运算错误，也是高分考生的良好习惯。

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Inverse Matrix: Computation Methods & Properties | 逆矩阵的求法及其性质考点