📚 Answers to the Exercises in Support Pack 1 – Chapter 1: Matrix Algebra | 支持包1第1章：矩阵代数习题解答
This revision article provides worked solutions to the exercises from Support Pack 1, Chapter 1 of the AQA A-Level Mathematics course. The chapter focuses on matrix algebra, including matrix operations, determinants, inverse matrices, linear systems, and transformations. Each solution is presented in a step-by-step manner to help you master the key techniques and prepare confidently for examinations.
本篇复习文章提供 AQA A-Level 数学课程中 Support Pack 1 第一章“矩阵代数”的习题解答。本章涉及矩阵运算、行列式、逆矩阵、线性方程组和几何变换等内容。每道题都按步骤给出解答，帮助你掌握核心技巧并自信备考。
1. Matrix Addition and Scalar Multiplication | 矩阵加法与标量乘法
Exercise 1.1. Let A = (1 2; 3 4) and B = (5 6; 7 8). Compute A + B and 3A – 2B.
练习1.1。 设 A = (1 2; 3 4)，B = (5 6; 7 8)。计算 A + B 和 3A – 2B。
Solution. Matrix addition and scalar multiplication are performed entry by entry.
解答。 矩阵加法和标量乘法按元素逐一进行。
A + B = (1+5 2+6; 3+7 4+8) = (6 8; 10 12).
3A – 2B = (3×1-2×5 3×2-2×6; 3×3-2×7 3×4-2×8) = (3-10 6-12; 9-14 12-16) = (-7 -6; -5 -4).
Thus the results are A + B = (6 8; 10 12) and 3A – 2B = (-7 -6; -5 -4).
因此结果为 A + B = (6 8; 10 12)，3A – 2B = (-7 -6; -5 -4)。
2. Matrix Multiplication | 矩阵乘法
Exercise 2.1. Given A = (2 1; 0 3) and B = (1 -2; 4 5), compute AB and BA. Comment on whether AB = BA.
练习2.1。 已知 A = (2 1; 0 3)，B = (1 -2; 4 5)，求 AB 和 BA，并说明 AB 是否等于 BA。
Solution. For two 2×2 matrices, the product is obtained using row-by-column multiplication.
解答。 对于两个2×2矩阵，乘积按行乘以列计算。
AB = (2×1+1×4 2×(-2)+1×5; 0×1+3×4 0×(-2)+3×5) = (2+4 -4+5; 0+12 0+15) = (6 1; 12 15).
BA = (1×2+(-2)×0 1×1+(-2)×3; 4×2+5×0 4×1+5×3) = (2+0 1-6; 8+0 4+15) = (2 -5; 8 19).
Since (6 1; 12 15) ≠ (2 -5; 8 19), we conclude that AB ≠ BA. This demonstrates that matrix multiplication is not commutative in general.
因为 (6 1; 12 15) ≠ (2 -5; 8 19)，所以 AB ≠ BA。这表明矩阵乘法一般不可交换。
3. Determinant of a 2×2 Matrix | 2×2矩阵的行列式
Exercise 3.1. Find the determinant of M = (3 4; 6 8). Is M singular or non-singular?
练习3.1。 求矩阵 M = (3 4; 6 8) 的行列式。M 是奇异矩阵还是非奇异矩阵？
Solution. For a 2×2 matrix M = (a b; c d), det(M) = ad – bc.
解答。 对于2×2矩阵 M = (a b; c d)，det(M) = ad – bc。
det(M) = 3×8 – 4×6 = 24 – 24 = 0.
Because the determinant is 0, the matrix is singular.
因为行列式为0，所以矩阵是奇异的。
4. Determinant of a 3×3 Matrix | 3×3矩阵的行列式
Exercise 4.1. Calculate det(A) for A = (1 2 3; 0 1 4; 2 -1 1).
练习4.1。 计算 A = (1 2 3; 0 1 4; 2 -1 1) 的行列式 det(A)。
Solution. Use expansion along the first row. For each entry, remove its row and column to obtain a 2×2 minor, multiply by the entry and the sign factor (-1)ⁱ⁺ʲ.
解答。 按第一行展开。对每个元素，去掉其所在行与列，得到2×2余子式，乘以该元素及符号因子 (-1)ⁱ⁺ʲ。
det(A) = 1×det(1 4; -1 1) – 2×det(0 4; 2 1) + 3×det(0 1; 2 -1)
= 1×(1×1 – 4×(-1)) – 2×(0×1 – 4×2) + 3×(0×(-1) – 1×2)
= (1+4) – 2×(0-8) + 3×(0-2) = 5 + 16 – 6 = 15.
Thus det(A) = 15.
因此 det(A) = 15。
5. Inverse of a 2×2 Matrix | 2×2矩阵的逆矩阵
Exercise 5.1. Find the inverse of B = (4 3; 2 2), if it exists.
练习5.1。 求 B = (4 3; 2 2) 的逆矩阵（若存在）。
Solution. For B = (a b; c d), the inverse is given by B⁻¹ = 1/(ad-bc) × (d -b; -c a), provided det(B) ≠ 0.
解答。 对于 B = (a b; c d)，逆矩阵为 B⁻¹ = 1/(ad-bc) × (d -b; -c a)，前提是 det(B) ≠ 0。
det(B) = 4×2 – 3×2 = 8 – 6 = 2. Since det(B) ≠ 0, the inverse exists.
det(B) = 4×2 – 3×2 = 8 – 6 = 2。因 det(B) ≠ 0，所以逆矩阵存在。
B⁻¹ = (1/2) × (2 -3; -2 4) = (1 -1.5; -1 2).
You may keep the fractions: B⁻¹ = (1 -3/2; -1 2).
可保留分数：B⁻¹ = (1 -3/2; -1 2)。
6. Solving Linear Equations Using Inverse Matrices | 用逆矩阵解线性方程组
Exercise 6.1. Solve the system of equations by using matrices: 2x + y = 7, 3x – 2y = 0.
练习6.1。 用矩阵方法解方程组：2x + y = 7，3x – 2y = 0。
Solution. Write the system as AX = C, where A = (2 1; 3 -2), X = (x; y), C = (7; 0). Then X = A⁻¹C.
解答。 将方程组写成 AX = C，其中 A = (2 1; 3 -2)，X = (x; y)，C = (7; 0)。则 X = A⁻¹C。
det(A) = 2×(-2) – 1×3 = -4 – 3 = -7.
A⁻¹ = (1/(-7)) × (-2 -1; -3 2) = (2/7 1/7; 3/7 -2/7).
Now compute X = A⁻¹C:
x = (2/7)×7 + (1/7)×0 = 2,
y = (3/7)×7 + (-2/7)×0 = 3.
Thus the solution is x = 2, y = 3.
因此解为 x = 2，y = 3。
7. Singular Matrices | 奇异矩阵
Exercise 7.1. Find the value of k such that the matrix C = (k 2; 4 k) is singular.
练习7.1。 求 k 的值，使矩阵 C = (k 2; 4 k) 为奇异矩阵。
Solution. A matrix is singular if its determinant is zero. For C, det(C) = k×k – 2×4 = k² – 8.
解答。 矩阵奇异当且仅当其行列式为零。对 C，det(C) = k×k – 2×4 = k² – 8。
Set det(C) = 0: k² – 8 = 0 → k = ±√8 = ±2√2.
令 det(C) = 0，得 k² – 8 = 0，故 k = ±√8 = ±2√2。
Therefore the matrix C is singular when k = 2√2 or k = -2√2.
因此当 k = 2√2 或 k = -2√2 时，矩阵 C 是奇异的。
8. Matrix Transformations | 矩阵变换
Exercise 8.1. Find the matrix of the linear transformation that reflects points in the line y = x. Then apply this transformation to the point P(3, -1).
练习8.1。 求关于直线 y = x 反射的线性变换矩阵，并将该变换应用于点 P(3, -1)。
Solution. The reflection in the line y = x swaps the coordinates. Hence the transformation matrix is R = (0 1; 1 0).
解答。 关于直线 y = x 的反射交换坐标，故变换矩阵为 R = (0 1; 1 0)。
To find the image of P, compute R × (3; -1) = (0×3 + 1×(-1); 1×3 + 0×(-1)) = (-1; 3). The reflected point is (-1, 3).
为求 P 的像，计算 R × (3; -1) = (0×3 + 1×(-1); 1×3 + 0×(-1)) = (-1; 3)。反射后的点为 (-1, 3)。
9. Combining Transformations | 复合变换
Exercise 9.1. Let T be the transformation represented by the matrix A = (1 0; 0 -1) (reflection in the x-axis) and let S be the rotation by 90° anticlockwise with matrix B = (0 -1; 1 0). Find the matrix of the composite transformation “apply T first, then S”.
练习9.1。 设 T 为矩阵 A = (1 0; 0 -1) 表示的变换（关于 x 轴的反射），S 为逆时针旋转90°，矩阵为 B = (0 -1; 1 0)。求先应用 T 再应用 S 的复合变换矩阵。
Solution. When applying a transformation T first and then S, the composite matrix is S × A. (Matrix multiplication order: the transform performed first is on the right.)
解答。 先应用 T 再应用 S 时，复合矩阵为 S × A。（矩阵乘法顺序：先执行的变换在右侧。）
S × A = (0 -1; 1 0) × (1 0; 0 -1) = (0×1 + (-1)×0 0×0 + (-1)×(-1); 1×1 + 0×0 1×0 + 0×(-1))
= (0 1; 1 0).
So the composite matrix is (0 1; 1 0), which is the reflection in the line y = x.
因此复合矩阵为 (0 1; 1 0)，即关于直线 y = x 的反射。
10. Inverse Transformations | 逆变换
Exercise 10.1. The matrix M = (2 1; 1 1) represents a linear transformation. Find the matrix representing the inverse transformation, and describe it geometrically.
练习10.1。 矩阵 M = (2 1; 1 1) 表示一个线性变换。求逆变换对应的矩阵，并从几何上描述它。
Solution. The inverse transformation matrix is M⁻¹. First compute det(M) = 2×1 – 1×1 = 1. Since det(M) ≠ 0, the inverse exists.
解答。 逆变换矩阵为 M⁻¹。先计算 det(M) = 2×1 – 1×1 = 1。因 det(M) ≠ 0，所以逆矩阵存在。
M⁻¹ = (1/1) × (1 -1; -1 2) = (1 -1; -1 2).
Geometrically, this inverse transformation undoes the effect of M. It is not a standard reflection or rotation; it is a shear-like linear map with determinant 1, hence it preserves area.
几何上，该逆变换取消 M 的效果。它不是标准反射或旋转，而是行列式为1的剪切类线性映射，因此保面积。
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