📚 Support Pack 1 Answers: Matrix Algebra (Chapter 1) | 支持包1练习答案：矩阵代数（第1章）
This article provides fully worked answers to the Chapter 1 Matrix Algebra exercises in AQA A-Level Further Mathematics Support Pack 1. All matrices are written in the form [row 1; row 2; row 3] with a semicolon separating rows, and column vectors are written as [x; y; z]. Carefully check the order of multiplication and the use of determinants when finding inverses.
本文提供 AQA A-Level 进阶数学支持包 1 中第 1 章矩阵代数练习的完整答案解析。所有矩阵均以 [第一行; 第二行; 第三行] 的形式书写，分号分隔行，列向量写作 [x; y; z]。在求逆矩阵和进行乘法时，请特别注意乘法顺序与行列式的使用。
1. Addition, Subtraction and Scalar Multiplication | 矩阵加减与数乘
Exercise: Given A = [3 0; −1 2] and B = [1 4; 2 −3], find A + B and 2A − 3B.
练习：已知 A = [3 0; −1 2]，B = [1 4; 2 −3]，求 A + B 和 2A − 3B。
olution: Add corresponding entries to obtain A + B = [3+1 0+4; −1+2 2+(−3)] = [4 4; 1 −1].
解答：将对应元素相加，得到 A + B = [3+1 0+4; −1+2 2+(−3)] = [4 4; 1 −1]。
. Then subtract corresponding entries: 2A − 3B = [6−3 0−12; −2−6 4−(−9)] = [3 −12; −8 13].
对于 2A − 3B，先分别进行数乘：2A = [6 0; −2 4]，3B = [3 12; 6 −9]。再相减对应元素：2A − 3B = [6−3 0−12; −2−6 4−(−9)] = [3 −12; −8 13]。
A + B = [4 4; 1 −1], 2A − 3B = [3 −12; −8 13]
关键答案：A + B = [4 4; 1 −1]，2A − 3B = [3 −12; −8 13]。
2. Matrix Multiplication | 矩阵乘法
A = [1 2; 0 3] and B = [4 1; 2 5], find AB and BA. Comment on the order of multiplication.
练习：已知 A = [1 2; 0 3]，B = [4 1; 2 5]，求 AB 和 BA，并说明乘法顺序的影响。
ultiply rows of A by columns of B: AB = [1×4+2×2 1×1+2×5; 0×4+3×2 0×1+3×5] = [8 11; 6 15].
解答：对于 AB，用 A 的行乘以 B 的列：AB = [1×4+2×2 1×1+2×5; 0×4+3×2 0×1+3×5] = [8 11; 6 15]。
ultiply rows of B by columns of A: BA = [4×1+1×0 4×2+1×3; 2×1+5×0 2×2+5×3] = [4 11; 2 19].
对于 BA，用 B 的行乘以 A 的列：BA = [4×1+1×0 4×2+1×3; 2×1+5×0 2×2+5×3] = [4 11; 2 19]。
Since AB ≠ BA, matrix multiplication is not commutative in general.
由于 AB ≠ BA，矩阵乘法一般不满足交换律。
AB = [8 11; 6 15], BA = [4 11; 2 19]
关键答案：AB = [8 11; 6 15]，BA = [4 11; 2 19]。
3. Transpose and Symmetric Matrices | 转置与对称矩阵
Exercise: Given A = [1 2 3; 4 5 6], write down Aᵀ. Also determine whether B = [1 3; 3 2] is symmetric.
练习：已知 A = [1 2 3; 4 5 6]，写出 Aᵀ。并判断 B = [1 3; 3 2] 是否为对称矩阵。
Solution: The transpose swaps rows and columns, so Aᵀ = [1 4; 2 5; 3 6].
解答：转置交换行与列，因此 Aᵀ = [1 4; 2 5; 3 6]。
To test symmetry, compare B with its transpose. Here Bᵀ = [1 3; 3 2], so Bᵀ = B and B is symmetric.
判断对称性时，将 B 与其转置比较。这里 Bᵀ = [1 3; 3 2]，因此 Bᵀ = B，所以 B 是对称矩阵。
Aᵀ = [1 4; 2 5; 3 6], B is symmetric
关键答案：Aᵀ = [1 4; 2 5; 3 6]，B 为对称矩阵。
4. Determinant of a 2×2 Matrix | 2×2 矩阵的行列式
Exercise: Find det A for A = [3 1; 4 2] and det B for B = [−2 5; 1 6].
练习：求矩阵 A = [3 1; 4 2] 和 B = [−2 5; 1 6] 的行列式。
tion: For a 2×2 matrix [a b; c d], det = ad − bc. Therefore det A = 3×2 − 1×4 = 6 − 4 = 2.
解答：对于 2×2 矩阵 [a b; c d]，行列式为 det = ad − bc。因此 det A = 3×2 − 1×4 = 6 − 4 = 2。
For B, det B = (−2)×6 − 5×1 = −12 − 5 = −17.
对于 B，det B = (−2)×6 − 5×1 = −12 − 5 = −17。
det A = 2, det B = −17
关键答案：det A = 2，det B = −17。
5. Determinant of a 3×3 Matrix | 3×3 矩阵的行列式
Exercise: Find the determinant of A = [1 2 3; 0 1 4; 5 6 0].
练习：求矩阵 A = [1 2 3; 0 1 4; 5 6 0] 的行列式。
and along the first row. The determinant is 1×(1×0 − 4×6) − 2×(0×0 − 4×5) + 3×(0×6 − 1×5).
解答：沿第一行展开。行列式为 1×(1×0 − 4×6) − 2×(0×0 − 4×5) + 3×(0×6 − 1×5)。
This simplifies to 1×(0 − 24) − 2×(0 − 20) + 3×(0 − 5) = −24 + 40 − 15 = 1.
化简得 1×(0 − 24) − 2×(0 − 20) + 3×(0 − 5) = −24 + 40 − 15 = 1。
det A = 1
关键答案：det A = 1。
6. Inverse of a 2×2 Matrix | 2×2 矩阵的逆矩阵
Exercise: Find the inverse of A = [4 1; 3 2].
练习：求矩阵 A = [4 1; 3 2] 的逆矩阵。
late det A = 4×2 − 1×3 = 8 − 3 = 5. Since the determinant is non-zero, the inverse exists.
解答：先计算 det A = 4×2 − 1×3 = 8 − 3 = 5。由于行列式不为零，逆矩阵存在。
inverse is 1/(ad−bc) × [d −b; −c a]. Thus A⁻¹ = 1/5 × [2 −1; −3 4] = [2/5 −1/5; −3/5 4/5].
对于 2×2 矩阵 [a b; c d]，逆矩阵为 1/(ad−bc) × [d −b; −c a]。因此 A⁻¹ = 1/5 × [2 −1; −3 4] = [2/5 −1/5; −3/5 4/5]。
A⁻¹ = [2/5 −1/5; −3/5 4/5]Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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