📚 Inverse Matrices | 逆矩阵
In AQA A-level Mathematics, inverse matrices are a central tool in linear algebra. They allow you to reverse a matrix transformation and to solve systems of simultaneous equations efficiently.
在 AQA A-level 数学中，逆矩阵是线性代数的核心工具。它可以逆转矩阵变换，并能高效求解线性方程组。
1. What is an Inverse Matrix? | 什么是逆矩阵？
tten A⁻¹ and is defined by the conditions AA⁻¹ = A⁻¹A = I, where I is the identity matrix.
对于方阵 A，其逆矩阵记作 A⁻¹，定义条件是 AA⁻¹ = A⁻¹A = I，其中 I 是单位矩阵。
-level you will mostly work with 2×2 matrices, where the test for invertibility is simple.
只有方阵才可能有逆矩阵，且并非每个方阵都可逆。在 A-level 阶段你主要处理 2×2 矩阵，其可逆性判断非常简单。
2. The 2×2 Inverse Formula | 2×2 逆矩阵公式
If A is the 2×2 matrix given by
若 A 是如下 2×2 矩阵
A = [a b; c d]
then the determinant is
则其行列式为
det(A) = ad − bc
Provided det(A) ≠ 0, the inverse is
当 det(A) ≠ 0 时，其逆矩阵为
A⁻¹ = 1/(ad − bc) [d −b; −c a]
You can remember this as: swap the main diagonal entries a and d, change the signs of the off-diagonal entries b and c, then divide by the determinant.
你可以这样记：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，然后除以行列式。
3. Determinant and Singular Matrices | 行列式与奇异矩阵
hether an inverse exists. If det(A) = 0, the matrix is called singular and has no inverse.
行列式决定逆矩阵是否存在。若 det(A) = 0，该矩阵称为奇异矩阵，没有逆矩阵。
If det(A) ≠ 0, the matrix is non-singular and A⁻¹ exists. Geometrically, a zero determinant means the transformation collapses the plane onto a line or a point, so information is lost and the process cannot be reversed.
若 det(A) ≠ 0，该矩阵为非奇异矩阵，A⁻¹ 存在。从几何上看，行列式为零意味着变换将平面压缩到一条直线或一个点，信息丢失，因此过程不可逆。
4. Verifying an Inverse | 验证逆矩阵
To check that a candidate matrix B is really A⁻¹, calculate both AB and BA. Each product must equal the identity matrix
若要检验某个候选矩阵 B 确实是 A⁻¹，需要计算 AB 和 BA。两个乘积都必须等于单位矩阵
I = [1 0; 0 1]
en enough if A is known to be non-singular, but showing both products is safer in an exam.
在 2×2 情形中，如果已知 A 非奇异，通常只需证明 AB = I 即可，但在考试中同时验证两个乘积更稳妥。
5. Properties of Inverse Matrices | 逆矩阵的性质
Inverse matrices obey several useful rules:
逆矩阵满足以下几条常用规则：
(A⁻¹)⁻¹ = A
(AB)⁻¹ = B⁻¹A⁻¹
(Aᵀ)⁻¹ = (A⁻¹)ᵀ
(kA)⁻¹ = (1/k) A⁻¹, k ≠ 0
det(A⁻¹) = 1/det(A)
Note the reversed order in the inverse of a product: the inverse of AB is B⁻¹A⁻¹, not A⁻¹B⁻¹.
注意乘积逆矩阵的顺序是反向的：AB 的逆为 B⁻¹A⁻¹，而不是 A⁻¹B⁻¹。
6. Solving Matrix Equations | 解矩阵方程
If A is invertible and AX = B, you can multiply both sides on the left by A⁻¹ to obtain
若 A 可逆且 AX = B，你可以在两边同时左乘 A⁻¹，得到
X = A⁻¹B
If the equation is XA = B instead, you must multiply on the right:
如果方程是 XA = B，则必须右乘：
X = BA⁻¹
Because matrix multiplication is not commutative, multiplying on the correct side is essential.
由于矩阵乘法不可交换，在正确的一侧相乘非常重要。
7. Using the Inverse for Simultaneous Equations | 用逆矩阵解线性方程组
A system of equations such as ax + by = e and cx + dy = f can be written in matrix form as
形如 ax + by = e 与 cx + dy = f 的方程组可以写成矩阵形式
A [x; y] = [e; f]
where A = [a b; c d]. If det(A) ≠ 0, multiply both sides on the left by A⁻¹ to get
其中 A = [a b; c d]。若 det(A) ≠ 0，两边同时左乘 A⁻¹，可得
[x; y] = A⁻¹ [e; f]
is gives the unique solution. When det(A) = 0, the system does not have a unique solution.
这样就得到了唯一解。当 det(A) = 0 时，方程组没有唯一解。
8. Inverse of a Product | 乘积的逆
The rule (AB)⁻¹ = B⁻¹A⁻¹ can be proved using the associative law. Consider
规则 (AB)⁻¹ = B⁻¹A⁻¹ 可以用结合律证明。考虑下式
(AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = AA⁻¹ = I
Because the product gives the identity matrix, B⁻¹A⁻¹ is indeed the inverse of AB.
因为乘积得到单位矩阵，所以 B⁻¹A⁻¹ 确实是 AB 的逆矩阵。
9. Geometric Interpretation | 几何解释
sformation of the plane. Its inverse matrix represents the reverse transformation, sending images back to their original positions.
2×2 矩阵表示平面的线性变换。其逆矩阵表示逆变换，将像点送回到原来的位置。
For example, if A represents a rotation or a reflection, A⁻¹ represents the opposite rotation or the same reflection applied again. If the determinant is zero, the transformation cannot be reversed.
例如，若 A 表示旋转或反射，A⁻¹ 就表示相反的旋转或再次施加相同的反射。若行列式为零，变换无法逆转。
10. Common Mistakes | 常见错误
Forgetting to divide by the determinant when writing A⁻¹.
写 A⁻¹ 时忘记除以行列式。
Swapping the off-diagonal entries instead of only changing their signs.
将副对角线元素交换位置，而不是仅改变符号。
Using A⁻¹B⁻¹ instead of B⁻¹A⁻¹ for the inverse of AB.
把 AB 的逆写成 A⁻¹B⁻¹，而正确形式是 B⁻¹A⁻¹。
Multiplying by A⁻¹ on the wrong side when solving XA = B.
在解 XA = B 时在错误的一侧乘以 A⁻¹。
11. Worked Example | 例题
Let A = [3 4; 1 2]. First calculate the determinant:
设 A = [3 4; 1 2]。先计算行列式：
det(A) = 3 × 2 − 4 × 1 = 2
Since det(A) ≠ 0, the inverse exists:
由于 det(A) ≠ 0，逆矩阵存在：
A⁻¹ = 1/2 [2 −4; −1 3] = [1 −2; −0.5 1.5]
To verify, multiply:
验证乘法：
AA⁻¹ = [3 4; 1 2] [1 −2; −0.5 1.5] = [1 0; 0 1]
Now solve the simultaneous equations 3x + 4y = 10 and x + 2y = 4:
现在解方程组 3x + 4y = 10 和 x + 2y = 4：
[x; y] = A⁻¹ [10; 4] = [1 −2; −0.5 1.5] [10; 4] = [2; 1]
Therefore x = 2 and y = 1.
因此 x = 2，y = 1。
12. Exam Tips | 考试技巧
Always state the determinant first and check that it is non-zero before finding A⁻¹.
在求 A⁻¹ 之前，务必先写出行列式并确认它不为零。
Write the full formula in your working; method marks are often available even if an arithmetic slip occurs.
在解题过程中写出完整公式；即使出现计算错误，也可能获得方法分。
Check your inverse by multiplying it by the original matrix when time allows.
时间允许时，用原矩阵乘逆矩阵进行检验。
Give exact answers using fractions where possible, unless the question asks for decimals.
除非题目要求小数，否则尽可能用分数给出精确答案。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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