📚 Determinants of Products: The Multiplicative Rule det(AB) = det(A)det(B) | 矩阵乘积的行列式：乘法法则及其应用
frequently examined results is the product rule for determinants: the determinant of a matrix product is the product of the individual determinants. This rule allows you to simplify calculations, test invertibility, and understand how linear transformations combine. In this article, we will build up the rule from 2 × 2 and 3 × 3 determinants, verify it in a student-friendly way, and apply it to inverses, transposes and singular matrices.
在 A-Level 数学（尤其是 AQA 考试大纲）中，行列式是矩阵代数的核心内容。其中一个最有力、最常考的结果就是行列式的乘积法则：矩阵乘积的行列式等于各矩阵行列式的乘积。这一法则能帮助你简化计算、判断矩阵是否可逆，并理解线性变换是如何复合的。本文将从 2 × 2 和 3 × 3 行列式入手，用适合学生的方式验证该法则，并将其应用到逆矩阵、转置矩阵和奇异矩阵中。
1. Why Study Determinants of Products? | 为什么研究乘积的行列式
When two square matrices A and B are multiplied, the resulting matrix AB represents the composition of two linear transformations: first apply B, then apply A. Many exam questions ask you to find det(AB) without calculating AB first. Knowing the product rule can save time and reduce arithmetic errors.
当两个方阵 A 和 B 相乘时，所得矩阵 AB 表示两个线性变换的复合：先作用 B，再作用 A。许多考试题会要求你不需要先计算 AB 就直接求 det(AB)。掌握乘积法则可以节省时间，并减少计算错误。
The topic also connects to invertibility, eigenvalues, and the geometry of area and volume scale factors. A solid understanding of det(AB) is therefore essential for both pure and applied questions, and it reappears later in transformation geometry, systems of linear equations, and further maths work on matrices.
这一知识点还与可逆性、特征值以及面积和体积缩放因子的几何意义密切相关。因此，扎实理解 det(AB) 对纯数学和应用题都至关重要，并且它会在后续的变换几何、线性方程组以及进阶数学的矩阵内容中反复出现。
4 or larger, and it extends by repeated application to products of three or more matrices.
这条法则本身表述简短，却具有惊人的普遍性：它对任意两个同阶方阵都成立，无论阶数是 2、3、4 还是更大，并且可以通过反复使用推广到三个或更多矩阵的乘积。
2. Quick Recap: Determinant of a 2 × 2 Matrix | 快速回顾：2 × 2 矩阵的行列式
he first row and c, d in the second row, the determinant is defined by a single formula. The notation det(A) and |A| mean the same thing.
对于第一行为 a、b，第二行为 c、d 的 2 × 2 矩阵 A，其行列式由一个公式定义。记号 det(A) 与 |A| 含义相同。
det(A) = ad − bc
This single number tells you whether A is invertible: A is non-singular if and only if det(A) ≠ 0. Geometrically, |det(A)| is the area scale factor of the transformation represented by A, so a zero determinant means the transformation squashes the plane onto a line or a point.
这个数可以判断 A 是否可逆：当且仅当 det(A) ≠ 0 时 A 为非奇异。从几何上看，|det(A)| 是 A 所代表的变换的面积缩放因子，因此行列式为零意味着该变换把平面压缩成一条直线或一个点。
[2 1; 4 3], then det(A) = (2)(3) − (1)(4) = 6 − 4 = 2. Since det(A) ≠ 0, A is invertible.
例如：若 A = [2 1; 4 3]，则 det(A) = (2)(3) − (1)(4) = 6 − 4 = 2。因为 det(A) ≠ 0，所以 A 可逆。
3. Quick Recap: Determinant of a 3 × 3 Matrix | 快速回顾：3 × 3 矩阵的行列式
e standard formula is the cofactor expansion along the first row, with alternating signs +, −, +.
对于 3 × 3 矩阵 A = [a b c; d e f; g h i]，常用公式是沿第一行展开的代数余子式展开，符号交替为 +、−、+。
det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
nt formed by deleting its row and column, then combine with the sign pattern. The same expansion works along any row or any column, and always gives the same value.
用文字说明：取第一行的每个元素，乘以删去它所在行和列后得到的 2 × 2 行列式，再按符号规律相加。沿任意一行或任意一列展开都成立，并且结果总是相同。
Three shortcuts are worth memorising because they appear constantly in exam questions. If any row or column is entirely zero, the determinant is 0. If the matrix is triangular (upper or lower), the determinant is simply the product of the diagonal entries. If two rows or two columns are identical or proportional, the determinant is again 0.
有三个速算技巧值得牢记，因为它们在考试题中反复出现。若某一行或某一列全为零，行列式为 0。若矩阵为三角矩阵（上三角或下三角），行列式就等于对角线元素的乘积。若两行或两列相同或成比例，行列式同样为 0。
ix A = [2 0 0; 1 3 0; 4 5 4] has det(A) = 2 × 3 × 4 = 24, with no expansion needed at all.
例如，三角矩阵 A = [2 0 0; 1 3 0; 4 5 4] 有 det(A) = 2 × 3 × 4 = 24，完全不需要展开。
4. The Product Rule: det(AB) = det(A) × det(B) | 乘积法则
The central result of this article is stated in one line. If A and B are square matrices of the same order n, then the determinant of the product equals the product of the determinants.
本文的核心结果只需一行即可表述。若 A 与 B 是同阶方阵（阶数为 n），则乘积的行列式等于行列式的乘积。
det(AB) = det(A) × det(B)
an AB = BA, because matrix multiplication is not commutative in general; it only means the two products have the same determinant.
由于右边只是两个普通数的乘积，顺序在这里无关紧要：det(A) × det(B) = det(B) × det(A)。这并不意味着 AB = BA，因为矩阵乘法一般不可交换；它只说明两个乘积的行列式相等。
By repeated application, the rule extends to any finite product: det(A₁A₂A₃…Aₖ) = det(A₁) × det(A₂) × … (Aᵏ) = (det(A))ᵏ, and for a scalar multiple of a matrix of order n, det(kA) = kⁿ × det(A).
通过反复使用，该法则可推广到任意有限乘积：det(A₁A₂A₃…Aₖ) = det(A₁) × det(A₂) × … × det(Aₖ)。特别地，对于矩阵的幂有 det(Aᵏ) = (det(A))ᵏ；而对于 n 阶矩阵的数乘有 det(kA) = kⁿ × det(A)。
5. Verifying the Rule with 2 × 2 Matrices | 用 2 × 2 矩阵验证乘积法则
+ dg, cf + dh], and each entry is a linear combination of the rows and columns of A and B.
一般性的代数证明是一道很好的练习。设 A = [a b; c d]，B = [e f; g h]。则 AB = [ae + bg, af + bh; ce + dg, cf + dh]，其中每个元素都是 A 与 B 的行与列的线性组合。
osite signs, we are left with ad(eh − fg) − bc(eh − fg), which factorises as (ad − bc)(eh − fg) = det(A) × det(B).
对 AB 应用 2 × 2 公式得 det(AB) = (ae + bg)(cf + dh) − (af + bh)(ce + dg)。展开并消去四个以相反符号重复出现的项后，剩下 ad(eh − fg) − bc(eh − fg)，因式分解得 (ad − bc)(eh − fg) = det(A) × det(B)。
17], so det(AB) = (2)(17) − (7)(4) = 34 − 28 = 6, which equals 2 × 3. The rule checks out.
再用数值检验。取 A = [2 1; 4 3]，det(A) = 2；B = [1 2; 0 3]，det(B) = 1 × 3 − 2 × 0 = 3。则 AB = [2 7; 4 17]，于是 det(AB) = (2)(17) − (7)(4) = 34 − 28 = 6，正好等于 2 × 3。法则验证成立。
6. A Worked 3 × 3 Verification | 3 × 3 矩阵验证例题
row: det(A) = 1(1 × 0 − 4 × 6) − 2(0 × 0 − 4 × 5) + 3(0 × 6 − 1 × 5) = −24 + 40 − 15 = 1.
该法则并非 2 × 2 矩阵所特有。考虑 A = [1 2 3; 0 1 4; 5 6 0]。沿第一行展开：det(A) = 1(1 × 0 − 4 × 6) − 2(0 × 0 − 4 × 5) + 3(0 × 6 − 1 × 5) = −24 + 40 − 15 = 1。
he product of the diagonal entries, det(B) = 6. Multiplying, AB = [1 4 9; 0 2 12; 5 12 0].
再取对角矩阵 B = [1 0 0; 0 2 0; 0 0 3]，其行列式等于对角线元素之积，det(B) = 6。相乘得 AB = [1 4 9; 0 2 12; 5 12 0]。
44 + 240 − 90 = 6. Since det(A) × det(B) = 1 × 6 = 6, the product rule is confirmed again.
沿 AB 的第一行展开：det(AB) = 1(2 × 0 − 12 × 12) − 4(0 × 0 − 12 × 5) + 9(0 × 12 − 2 × 5) = −144 + 240 − 90 = 6。由于 det(A) × det(B) = 1 × 6 = 6，乘积法则再次得到验证。
7. Geometric Meaning: Area and Volume Scale Factors | 几何意义：面积与体积缩放因子
A 2 × 2 matrix maps the unit square to a parallelogram, and |det(A)| en B multiplies the area scale factors, so the composite transformation AB scales areas by |det(A)| × |det(B)|, which is precisely |det(AB)|.
一个 2 × 2 矩阵把单位正方形映射为平行四边形，而 |det(A)| 正好是这个平行四边形的面积。先作用 A 再作用 B，面积缩放因子相乘，因此复合变换 AB 把面积缩放 |det(A)| × |det(B)| 倍，这恰好就是 |det(AB)|。
idea becomes volume. A 3 × 3 matrix maps the unit cube to a parallelepiped whose volume is |det(A)|, and composing two such transformations multiplies the volumes, giving |det(AB)| = |det(A)| × |det(B)| once more.
在三维中同样的思想体现为体积。一个 3 × 3 矩阵把单位立方体映射为平行六面体，其体积为 |det(A)|；复合两个这样的变换时体积相乘，于是再次得到 |det(AB)| = |det(A)| × |det(B)|。
This also explains why a zero determinant is so important: the transformation collapses space into a lower-dimensional object with zero area or zero volume, and no inverse transformation can undo that collapse.
这也解释了为什么零行列式如此重要：该变换把空间压缩成面积或体积为零的低维对象，而没有任何逆变换能够撤销这种压缩。
8. det(AB) and Inverse Matrices | det(AB) 与逆矩阵
ersal is the matrix analogue of putting on socks and shoes and then removing them in the opposite order.
若 det(AB) ≠ 0，则 AB 可逆，其逆矩阵把因子的顺序反过来：(AB)⁻¹ = B⁻¹A⁻¹。这种顺序颠倒就像穿袜子和穿鞋，脱下时必须按相反顺序进行。
B)) = 1 / det(AB). For a single matrix, det(A⁻¹) = 1 / det(A), a result worth remembering.
对两边取行列式得 det((AB)⁻¹) = det(B⁻¹) × det(A⁻¹) = 1 / (det(A) × det(B)) = 1 / det(AB)。对于单个矩阵则有 det(A⁻¹) = 1 / det(A)，这个结果值得记住。
A⁻¹ = (1 / det(A)) × [d −b; −c a]
det(A) = 2 then det(A⁻¹) = 1 / 2, and the matrix A⁻¹ has entries divided by 2 throughout.
上式是 2 × 2 逆矩阵公式，仅在 det(A) ≠ 0 时成立。注意其中隐含了乘积法则的检验：若 det(A) = 2，则 det(A⁻¹) = 1 / 2，而矩阵 A⁻¹ 的所有元素都被 2 除。
9. Transposes, Powers and Eigenvalue Links | 行列式、转置、幂与特征值
Determinants behave well with transposes: det(Aᵀ is follows from the definition, because expanding along a row of A is the same as expanding along the corresponding column of Aᵀ.
行列式与转置配合良好：对任意方阵都有 det(Aᵀ) = det(A)。这可以从定义看出，因为沿 A 的某一行展开，与沿 Aᵀ 的对应列展开是同一件事。
Combining this with the product rule gives det(AAᵀ) = det(A) × det(Aᵀ (det(A))ᵏ for every positive integer k, and det(A⁻ᵏ) = 1 / (det(A))ᵏ when A is invertible.
把这一结论与乘积法则结合得 det(AAᵀ) = det(A) × det(Aᵀ) = (det(A))²，对实矩阵而言它总大于或等于零。类似地，对任意
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