📚 5 Rules for Manipulating Determinants | 行列式操作的五条规则
In AQA A-Level Further Mathematics, determinants provide a powerful shortcut for testing invertibility, solving systems of equations, and understanding linear transformations. This article focuses on the five most important rules for manipulating determinants efficiently and accurately, and shows how to combine them in exam-style calculations.
在 AQA A-Level 进阶数学中，行列式是判断矩阵是否可逆、求解线性方程组以及理解线性变换的重要工具。本文重点讲解行列式操作中最关键的五条规则，并说明如何在考试型计算中把它们组合使用，帮助你高效、准确地算出结果。
1. What Is a Determinant? | 什么是行列式
s only defined for square matrices, and it gives useful information about area scale factor and volume scale factor in linear transformations.
对于两行分别为 (a, b) 与 (c, d) 的 2 × 2 矩阵 A，行列式定义为 det(A) = ad − bc。行列式只对方阵有定义，它可以给出线性变换中面积缩放因子和体积缩放因子的有用信息。
det(A) = ad − bc
The determinant can be positive, negative, or zero, and its sign relates to the orientation of a transformation. A zero determinant means the matrix is singular: it has no inverse, the two rows are linearly dependent, and the corresponding system of equations has either no solution or infinitely many solutions.
行列式可以为正、为负或为零，其符号与线性变换的取向有关。行列式为零意味着矩阵奇异：它没有逆矩阵，各行线性相关，对应的方程组要么无解，要么有无穷多解。
lly use expansion along a row or column; the manipulation rules below often reduce the work substantially.
对于 3 × 3 矩阵，AQA 考生通常使用按行或按列展开；而下面的操作规则往往可以大大减少计算量。
det(A) = a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃ + …
Here M₁₁, M₁₂ and M₁₃ are the minors obtained by deleting one row and one column. The alternating signs +, −, + are essential, and most lost marks in this topic come from sign slips rather than from hard algebra.
其中 M₁₁、M₁₂、M₁₃ 是删去相应行与列后得到的余子式。交替符号 +、−、+ 至关重要，这一专题失分大多来自符号错误，而不是代数本身太难。
2. Why Manipulation Rules Matter | 为什么需要操作规则
ules let you create zeros or simple rows first, which usually turns a long expansion into a single 2 × 2 determinant.
直接展开一个 3 × 3 行列式需要计算三个 2 × 2 行列式，而 4 × 4 需要四个 3 × 3 行列式，计算量增长非常快。五条规则可以让你先制造零元素或简单的行，通常能把冗长的展开压缩成一个 2 × 2 行列式。
There is also a strategic benefit: markers often award method marks for correct row operations, so a clean, visible sequence of operations is safer than a long unlabelled expansion.
这还有策略上的好处：评分时，正确的行变换通常可以获得方法分，因此一条清晰、有标注的操作序列比长篇无标注的展开更稳妥。
3. Rule 1: Transpose Leaves the Determinant Unchanged | 规则 1：转置不改变行列式
Taking the transpose of a matrix does not change its determinant: det(Aᵀ) = det(A).
对矩阵取转置不会改变其行列式：det(Aᵀ) = det(A)。
det(Aᵀ) = det(A)
This means rows and columns are equally important, so any row operation has a matching column operation with the same effect on the determinant.
这意味着行与列的地位是对称的，因此任何行操作都有对应的列操作，且对行列式的效果相同。
In practice you should choose whichever direction makes the entries simpler. If a column already contains two zeros, expand along that column rather than a row, and if a row has a convenient pivot such as 1 or −1, use it to clear the rest of its column.
实际操作中应选择让元素更简单的那一个方向。如果某一列已经有两个零，就沿该列展开而不是沿行展开；如果某一行有 1 或 −1 这样方便的枢轴元素，就用它去清掉同列的其他元素。
4. Rule 2: Swapping Two Rows Changes the Sign | 规则 2：交换两行会改变符号
If two rows (or two columns) of a matrix are interchanged, the determinant changes sign but keeps its magnitude: the new determinant equals −det(A).
如果交换矩阵的两行（或两列），行列式的符号会改变但绝对值不变：新行列式等于 −det(A)。
det(B) = −det(A) (one row swap)
Each single swap multiplies the determinant by −1, so n swaps multiply it by (−1)ⁿ. If you swap twice you return to the original value, which is a quick way to check your working.
每一次单独交换都会把行列式乘以 −1，因此交换 n 次会乘以 (−1)ⁿ。交换两次就回到原来的值，这可以作为检查计算是否出错的快捷方法。
- One swap: det changes sign. | 交换一次：行列式变号。
- Two swaps: det is unchanged. | 交换两次：行列式不变。
- If any two rows are identical, swapping them changes nothing, so det must be 0. | 若有两行完全相同，交换后矩阵不变，因此行列式必为 0。
The last point is worth remembering: it is a two-line proof that a matrix with two equal rows is singular, and it links directly to Rule 5 below.
最后一点值得记住：只需两行推理就能证明"两行相同则矩阵奇异"，它直接通向下面的规则 5。
5. Rule 3: Multiplying a Row by a Scalar | 规则 3：某一行乘以常数
et(B) = k·det(A). Only the chosen row is scaled; the other rows stay exactly as they were.
如果把 A 的某一行所有元素都乘以常数 k 得到 B，那么 det(B) = k·det(A)。只有被选中的那一行被缩放，其余各行保持原样。
det(B) = k · det(A) (one row multiplied by k)
) = kⁿ·det(A), because all n rows are multiplied by k and each one contributes a factor k.
不要把它与"整个矩阵乘以常数"混淆。对于 n × n 矩阵，det(kA) = kⁿ·det(A)，因为 n 行都被乘以 k，每一行都贡献一个因子 k。
- 2 × 2 matrix: det(kA) = k²·det(A). | 2 × 2 矩阵：det(kA) = k²·det(A)。
- 3 × 3 matrix: det(kA) = k³·det(A). | 3 × 3 矩阵：det(kA) = k³·det(A)。
- If a matrix has a common factor k in every entry, factor it out as kⁿ, not as k. | 若矩阵每个元素都有公因子 k，应提取为 kⁿ，而不是 k。
This rule also explains the effect of the operation Rᵢ → kRᵢ: it is not allowed as a "do nothing" simplification, because it genuinely changes the determinant unless k = 1.
这条规则也解释了操作 Rᵢ → kRᵢ 的效果：它不能当作"不改变"的化简来使用，因为只要 k ≠ 1，它确实会改变行列式的值。
6. Rule 4: Adding a Multiple of One Row to Another | 规则 4：把一行的倍数加到另一行
The operation Rᵢ → Rᵢ + kRⱼ useful rule in the whole topic, because it lets you create zeros without paying any price.
操作 Rᵢ → Rᵢ + kRⱼ 完全不改变行列式的值。这是整个专题中最有用的规则，因为它让你制造零元素而不付出任何代价。
Rᵢ → Rᵢ + kRⱼ ⟹ det unchanged
The reason is linearity: the new row is a sum of two rows, so the determinant splits into the original determinant plus k times a determinant with two equal rows, and that second determinant is zero by Rule 2.
原因是线性性：新行是两行之和，因此行列式可拆成原行列式加上 k 倍"两行相同"的行列式，而后者按规则 2 等于零。
g that column. The arithmetic after the clearing step is often a single 2 × 2 determinant.
典型考法是：选一个 1 或 −1 作为枢轴，用它把该列的其他元素全部清成零，再沿这一列展开。清零之后的运算往往只剩一个 2 × 2 行列式。
7. Rule 5: Zero, Equal or Proportional Rows | 规则 5：零行、相同行或成比例的行
More generally, if one row is a linear combination of the others, the determinant is zero.
如果矩阵含有一整行全是零的行，或者有两行完全相同或成比例，那么 det(A) = 0。更一般地，如果某一行可以表示为其他行的线性组合，行列式也为零。
proportional rows ⟹ det(A) = 0 ⟹ A is singular
This rule is a fast route in multiple-choice and "show that" questions. Before starting a long expansion, scan the rows: a zero row, two equal rows or two rows where one is a simple multiple of another settle the answer immediately.
这条规则在选择题和"证明"题中是快速通道。在动手展开之前先扫一遍各行：出现零行、两行相同，或某行是另一行的简单倍数，答案立刻确定。
- Rows (1, 2, 3), (4, 5, 6), (7, 8, 9): the third row is 2 × row 2 − row 1, so det = 0. | 行 (1, 2, 3)、(4, 5, 6)、(7, 8, 9)：第三行等于第二行的 2 倍减去第一行，故 det = 0。
- Equal rows check also gives a quick proof that the rule works. | 相同行的检验也顺带证明了该规则的合理性。
- A determinant equal to 0 is exactly the condition for a set of simultaneous equations to be inconsistent or to have infinitely many solutions. | 行列式为 0 正是方程组无解或有无穷多解的判定条件。
8. Combining the Rules: Triangular Form | 组合规则：化为三角形形式
A powerful strategy is to use Rule 4 repeatedly, plus Rule 2 for the swaps you may need, until the matrix becomes upper triangular. The determinant of a triangular matrix is just the product of the diagonal entries.
一个强有力的策略是反复使用规则 4，必要时配合规则 2 进行交换，直到矩阵化为上三角形。三角形矩阵的行列式就是对角线元素的乘积。
triangular matrix ⟹ det(A) = a₁₁ × a₂₂ × a₃₃ × … × aₙₙ
Remember to track the sign: every row swap you perform on the way contributes a factor of −1, and every row you scale contributes the scaling factor k. Rule 4 operations are free and change nothing.
记得追踪符号：过程中每做一次换行都带来因子 −1，每缩放一行都带来因子 k。规则 4 的操作是免费的，不产生任何变化。
9. Worked Example: Clearing a Column | 例题：清零零列
, 3), (4, 1, 7) and (−2, 3, 1). Instead of expanding immediately, use Rule 4 to create two zeros in the first column.
设 A 为 3 × 3 矩阵，三行分别为 (2, −1, 3)、(4, 1, 7)、(−2, 3, 1)。不要立即展开，先用规则 4 在第一列制造两个零。
- R₂ → R₂ − 2R₁ gives the new row (0, 3, 1). | R₂ → R₂ − 2R₁ 得新行 (0, 3, 1)。
- R₃ → R₃ + R₁ gives the new row (0, 2, 4). | R₃ → R₃ + R₁ 得新行 (0, 2, 4)。
Neither operation changes the determinant, so we may now expand along the first column, where only the leading entry 2 survives.
这两次操作都不改变行列式，因此现在可以沿第一列展开，而该列中只有首元素 2 是非零的。
det(A) = 2 × (3 × 4 − 1 × 2) = 2 × 10 = 20
18 + 42 = 20, which confirms both the value and the efficiency of the manipulation route.
直接展开得到 2 × (1 − 21) + 1 × (4 + 14) + 3 × (12 + 2) = −40 + 18 + 42 = 20，既验证了答案，也说明了操作法的高效。
10. Finding an Unknown for a Singular Matrix | 求使矩阵奇异的未知数
t det(A) = 0 and solve, or use Rule 5 if the conditions for proportional rows are visible.
考试题常给出含未知数 k 的矩阵，要求求出使它奇异的取值。做法是令 det(A) = 0 求解；如果成比例行的条件明显，也可以直接用规则 5。
Consider the matrix with rows (1, 2, 3), (4, 5, 6) and (k, 8, 9). Expanding along the first column uses the alternating signs correctly.
考虑三行分别为 (1, 2, 3)、(4, 5, 6)、(k, 8, 9) 的矩阵。沿第一列展开时要正确使用交替符号。
det(A) = 1 × (45 − 48) − 4 × (18 − 24) + k × (12 − 15) = 21 − 3k
tituting k = 7 back and checking with a row operation is a good habit under exam pressure.
令 21 − 3k = 0 得 k = 7，因此当 k = 7 时矩阵奇异。把 k = 7 代回并用行变换检验，是考试压力下值得养成的习惯。
11. Determinants, Products and Inverses | 行列式、乘积与逆矩阵
Determinants behave well under multiplication, which makes them extremely useful for checking whether an inverse exists and for simplifying powers of matrices.
行列式在乘法下有很好的性质，这使它非常适用于检验逆矩阵是否存在，以及化简矩阵的幂。
det(AB) = det(A) × det(B) det(A⁻¹) = 1 ÷ det(A) det(Aⁿ) = (det(A))ⁿ
- det(AB) = det(A) × det(B), so a product is singular if either factor is singular. | det(AB) = det(A) × det(B)，因此只要有一个因子奇异，乘积就奇异。
- det(A⁻¹) = 1 ÷ det(A), which requires det(A) ≠ 0. | det(A⁻¹) = 1 ÷ det(A)，前提是 det(A) ≠ 0。
- det(Aⁿ) = (det(A))ⁿ, a fast way to evaluate large powers. | det(Aⁿ) = (det(A))ⁿ，是计算大次幂的快捷方法。
- Combined with Rule 1: det(AᵀA) = (det(A))², which is never negative. | 结合规则 1：det(AᵀA) = (det(A))²，永远不会是负数。
12. Area, Volume and Orientation | 面积、体积与取向
For a 2 × 2 matrix, the magnitude |det(A)| is the volume scale factor. The sign records whether orientation is preserved or reversed.
对于 2 × 2 矩阵，|det(A)| 是该变换的面积缩放因子；对于 3 × 3 矩阵，它是体积缩放因子。符号则记录了取向是被保持还是被反转。
