Answers to Exercises - Further Pure 4 | 进阶纯数四习题解答

📚 Answers to Exercises – Further Pure 4 | 进阶纯数四习题解答

This article presents worked answers to a selection of typical exercises from the AQA Further Pure 4 syllabus. The questions are chosen to illustrate the most important skills and concepts, and each solution is explained step by step so that you can follow the reasoning and apply it to your own revision.

本文精选了 AQA 进阶纯数四（Further Pure 4）大纲中的典型习题并提供完整解答。所选题目涵盖最重要的技巧与概念，每道题都逐步讲解，帮助你理解推理过程，并将其应用到自己的复习中。

1. Matrix Transformations | 矩阵变换

Exercise: A transformation T is defined by the matrix \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \). Describe geometrically the effect of T.

题目：变换 T 由矩阵 \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) 定义。请从几何上描述 T 的作用。

Answer: The matrix is \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \). Compare with the general rotation matrix \( \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \). Equating entries gives \(\cos\theta=0\) and \(\sin\theta=1\), so \(\theta = 90^\circ\) anticlockwise. Therefore T is a rotation about the origin by \(90^\circ\) anticlockwise.

解答：矩阵为 \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\)。与一般旋转矩阵 \(\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) 比较，得 \(\cos\theta=0\)，\(\sin\theta=1\)，所以 \(\theta = 90^\circ\) 逆时针。因此 T 是绕原点逆时针旋转 \(90^\circ\)。

2. Systems of Linear Equations | 线性方程组

Exercise: Solve the system of equations \(2x + 3y + z = 9\), \(x – y + 2z = 3\), \(3x + 2y – z = 5\) using the inverse matrix method.

题目：用逆矩阵法解方程组 \(2x + 3y + z = 9\)，\(x – y + 2z = 3\)，\(3x + 2y – z = 5\)。

Answer: Write the system as \(A \mathbf{x} = \mathbf{b}\), where \(A = \begin{pmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{pmatrix}\), \(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 9 \\ 3 \\ 5 \end{pmatrix}\).

解答：将方程组写作 \(A \mathbf{x} = \mathbf{b}\)，其中 \(A = \begin{pmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1 \end{pmatrix}\)，\(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)，\(\mathbf{b} = \begin{pmatrix} 9 \\ 3 \\ 5 \end{pmatrix}\)。

Compute the determinant of A: \(\det(A) = 2 \times \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} – 3 \times \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1 \times \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}\)

计算 A 的行列式：\(\det(A) = 2 \times \begin{vmatrix} -1 & 2 \\ 2 & -1 \end{vmatrix} – 3 \times \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1 \times \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}\)

\(\det(A) = 2(1-4) – 3(-1-6) + (2+3) = 2(-3) -3(-7) +5 = -6 +21 +5 =20\)

Now find the adjugate (cofactor transpose) matrix. The matrix of cofactors is:

接着求伴随矩阵（余子式矩阵的转置）。余子式矩阵为：

\(\begin{pmatrix} -3 & 7 & 5 \\ 5 & -5 & 5 \\ 7 & -3 & -5 \end{pmatrix}\)

Transpose to get adjugate: \(\begin{pmatrix} -3 & 5 & 7 \\ 7 & -5 & -3 \\ 5 & 5 & -5 \end{pmatrix}\). Then \(A^{-1} = \frac{1}{20} \begin{pmatrix} -3 & 5 & 7 \\ 7 & -5 & -3 \\ 5 & 5 & -5 \end{pmatrix}\). Multiply by \(\mathbf{b}\):

转置得到伴随矩阵：\(\begin{pmatrix} -3 & 5 & 7 \\ 7 & -5 & -3 \\ 5 & 5 & -5 \end{pmatrix}\)。于是 \(A^{-1} = \frac{1}{20} \begin{pmatrix} -3 & 5 & 7 \\ 7 & -5 & -3 \\ 5 & 5 & -5 \end{pmatrix}\)。乘以 \(\mathbf{b}\)：

\(\mathbf{x} = \frac{1}{20} \begin{pmatrix} -3(9)+5(3)+7(5) \\ 7(9)-5(3)-3(5) \\ 5(9)+5(3)-5(5) \end{pmatrix} = \frac{1}{20} \begin{pmatrix} -27+15+35 \\ 63-15-15 \\ 45+15-25 \end{pmatrix} = \frac{1}{20} \begin{pmatrix} 23 \\ 33 \\ 35 \end{pmatrix}\)

Thus \(x = \frac{23}{20}\), \(y = \frac{33}{20}\), \(z = \frac{35}{20} = \frac{7}{4}\).

所以 \(x = \frac{23}{20}\)，\(y = \frac{33}{20}\)，\(z = \frac{35}{20} = \frac{7}{4}\)。

3. De Moivre’s Theorem and Roots of Unity | 棣莫弗定理与单位根

Exercise: Find the three cube roots of unity, and state their sum and product.

题目：求 1 的三个立方根，并说明它们的和与积。

Answer: The cube roots of unity are solutions of \(z^3 = 1\). Write \(1 = e^{2k\pi i}\) for integer k. Then \(z = e^{2k\pi i/3}\) for \(k = 0,1,2\). So the roots are:

解答：1 的立方根是方程 \(z^3 = 1\) 的解。将 \(1 = e^{2k\pi i}\)（k 为整数），则 \(z = e^{2k\pi i/3}\)，取 \(k = 0,1,2\)。所以三个根为：

\(z_0 = 1\), \(z_1 = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\), \(z_2 = e^{4\pi i/3} = -\frac{1}{2} – \frac{\sqrt{3}}{2}i\).

Sum: \(1 + (-\frac{1}{2} + \frac{\sqrt{3}}{2}i) + (-\frac{1}{2} – \frac{\sqrt{3}}{2}i) = 0\). Product: \(z_0 z_1 z_2 = e^{0} \cdot e^{2\pi i/3} \cdot e^{4\pi i/3} = e^{2\pi i} = 1\). Therefore the sum is 0 and the product is 1.

和：\(1 + (-\frac{1}{2} + \frac{\sqrt{3}}{2}i) + (-\frac{1}{2} – \frac{\sqrt{3}}{2}i) = 0\)。积：\(z_0 z_1 z_2 = e^{0} \cdot e^{2\pi i/3} \cdot e^{4\pi i/3} = e^{2\pi i} = 1\)。因此和为 0，积为 1。

4. Complex Loci | 复数轨迹

Exercise: Sketch and describe the locus \(|z – 2 – i| = 3\).

题目：画出并描述轨迹 \(|z – 2 – i| = 3\)。

Answer: Let \(z = x + yi\). Then \(|z – (2+i)| = 3\) means the distance from \(z\) to the point \((2,1)\) in the complex plane is constant 3. Hence the locus is a circle with centre \((2,1)\) and radius 3. In Cartesian form:

解答：设 \(z = x + yi\)。则 \(|z – (2+i)| = 3\) 表示复数平面上 z 到点 \((2,1)\) 的距离恒为 3。因此轨迹是一个以 \((2,1)\) 为圆心、半径为 3 的圆。直角坐标方程为：

\((x-2)^2 + (y-1)^2 = 9\).

The circle passes through the points \((-1,1)\), \((5,1)\), \((2,4)\), \((2,-2)\). Its centre is \((2,1)\) and radius is 3.

该圆经过点 \((-1,1)\)、\((5,1)\)、\((2,4)\)、\((2,-2)\)。圆心为 \((2,1)\)，半径为 3。

5. First-Order Differential Equations | 一阶微分方程

Exercise: Solve \(x \frac{dy}{dx} + 2y = x^3\) for \(x > 0\), given \(y(1) = 2\).

题目：解 \(x \frac{dy}{dx} + 2y = x^3\)（\(x > 0\)），已知 \(y(1) = 2\)。

Answer: This is a linear first-order ODE. Divide by x: \(\frac{dy}{dx} + \frac{2}{x}y = x^2\). The integrating factor is \(I(x) = e^{\int \frac{2}{x} dx} = e^{2\ln x} = x^2\).

解答：这是一阶线性微分方程。两边除以 x：\(\frac{dy}{dx} + \frac{2}{x}y = x^2\)。积分因子为 \(I(x) = e^{\int \frac{2}{x} dx} = e^{2\ln x} = x^2\)。

Multiply both sides by \(x^2\): \(x^2 \frac{dy}{dx} + 2xy = x^4\). The left side is \(\frac{d}{dx}(x^2 y)\). Integrate:

两边乘以 \(x^2\)：\(x^2 \frac{dy}{dx} + 2xy = x^4\)。左端为 \(\frac{d}{dx}(x^2 y)\)。积分得：

\(x^2 y = \frac{x^5}{5} + C\).

Using \(y(1) = 2\): \(1^2 \cdot 2 = \frac{1}{5} + C\), so \(C = \frac{9}{5}\). Hence \(y = \frac{x^3}{5} + \frac{9}{5x^2}\).

代入 \(y(1) = 2\)：\(1^2 \cdot 2 = \frac{1}{5} + C\)，所以 \(C = \frac{9}{5}\)。因此 \(y = \frac{x^3}{5} + \frac{9}{5x^2}\)。

6. Second-Order Differential Equations | 二阶微分方程

Exercise: Find the general solution of \(y” – 3y’ + 2y = 4e^{x}\).

题目：求 \(y” – 3y’ + 2y = 4e^{x}\) 的通解。

Answer: First solve the homogeneous equation \(y” – 3y’ + 2y = 0\). The auxiliary equation is \(m^2 – 3m + 2 = 0\), giving \(m = 1,2\). So complementary function \(y_c = Ae^{x} + Be^{2x}\).

解答：先解齐次方程 \(y” – 3y’ + 2y = 0\)。特征方程为 \(m^2 – 3m + 2 = 0\)，得 \(m = 1,2\)。因此互补函数 \(y_c = Ae^{x} + Be^{2x}\)。

For the particular integral, since \(e^{x}\) is part of the complementary function, try \(y_p = Cx e^{x}\). Then \(y_p’ = C e^{x}(x+1)\), \(y_p” = C e^{x}(x+2)\). Substitute:

对于特解，因为 \(e^{x}\) 是互补函数的一部分，试设 \(y_p = Cx e^{x}\)。则 \(y_p’ = C e^{x}(x+1)\)，\(y_p” = C e^{x}(x+2)\)。代入原方程：

\(C e^{x}(x+2) – 3C e^{x}(x+1) + 2C x e^{x} = 4 e^{x}\).

Simplify the left: \(C e^{x}[(x+2) – 3(x+1) + 2x] = C e^{x}(x+2 -3x -3 +2x) = C e^{x}(-1) = -C e^{x}\). Hence \(-C = 4\), so \(C = -4\). Particular integral \(y_p = -4x e^{x}\).

化简左边：\(C e^{x}[(x+2) – 3(x+1) + 2x] = C e^{x}(x+2 -3x -3 +2x) = C e^{x}(-1) = -C e^{x}\)。因此 \(-C = 4\)，即 \(C = -4\)。特解为 \(y_p = -4x e^{x}\)。

General solution: \(y = Ae^{x} + Be^{2x} – 4x e^{x}\).

通解为：\(y = Ae^{x} + Be^{2x} – 4x e^{x}\)。

7. Linear Independence and Basis | 线性无关与基

Exercise: Determine whether the vectors \(\mathbf{u} = (1,0,1)\), \(\mathbf{v} = (2,1,0)\), \(\mathbf{w} = (3,1,1)\) are linearly independent. If they are, state the dimension of the subspace they span.

题目：判断向量 \(\mathbf{u} = (1,0,1)\)、\(\mathbf{v} = (2,1,0)\)、\(\mathbf{w} = (3,1,1)\) 是否线性无关。若无关，说明它们张成的子空间的维数。

Answer: Form a matrix with the vectors as rows (or columns) and compute the determinant. Using rows:

解答：以向量为行（或列）构造矩阵并计算行列式。用行向量：

\(\begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{vmatrix} = 1(1\cdot1 – 0\cdot1) – 0(2\cdot1 – 0\cdot3) + 1(2\cdot1 – 1\cdot3) = 1 – 0 + (-1) = 0\).

Since the determinant is zero, the vectors are linearly dependent. To find the dimension of the span, check for a pair of independent vectors. \(\mathbf{u}\) and \(\mathbf{v}\) are clearly independent since one is not a scalar multiple of the other. Also \(\mathbf{w} = \mathbf{u} + \mathbf{v}\) (since \((1,0,1)+(2,1,0)=(3,1,1)\)). So the span has dimension 2.

由于行列式为零，向量线性相关。检查是否有一对线性无关的向量：\(\mathbf{u}\) 与 \(\mathbf{v}\) 显然无关（二者不成倍数）。且 \(\mathbf{w} = \mathbf{u} + \mathbf{v}\)（因为 \((1,0,1)+(2,1,0)=(3,1,1)\)）。所以张成的子空间维数为 2。

8. Eigenvalues and Eigenvectors | 特征值与特征向量

Exercise: Find the eigenvalues and eigenvectors of the matrix \(M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\).

题目：求矩阵 \(M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\) 的特征值和特征向量。

Answer: Solve \(\det(M – \lambda I) = 0\):

解答：解 \(\det(M – \lambda I) = 0\)：

\(\begin{vmatrix} 1-\lambda & 2 \\ 2 & 1-\lambda \end{vmatrix} = (1-\lambda)^2 – 4 = \lambda^2 – 2\lambda -3 = (\lambda -3)(\lambda +1) = 0\).

So eigenvalues are \(\lambda = 3\) and \(\lambda = -1\).

因此特征值为 \(\lambda = 3\) 和 \(\lambda = -1\)。

For \(\lambda = 3\), solve \((M – 3I)\mathbf{x} = 0\): \(\begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0\), so \(-2x + 2y = 0\) i.e. \(y = x\). An eigenvector is \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

对 \(\lambda = 3\)，解 \((M – 3I)\mathbf{x} = 0\)：\(\begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0\)，即 \(-2x + 2y = 0\)，所以 \(y = x\)。特征向量可取 \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\)。

For \(\lambda = -1\), solve \((M + I)\mathbf{x} = 0\): \(\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0\), so \(x + y = 0\) i.e. \(y = -x\). An eigenvector is \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\).

对 \(\lambda = -1\)，解 \((M + I)\mathbf{x} = 0\)：\(\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0\)，即 \(x + y = 0\)，所以 \(y = -x\)。特征向量可取 \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)。

9. Diagonalisation | 对角化

Exercise: Diagonalise the matrix \(M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\), i.e. find an invertible matrix P and a diagonal matrix D such that \(M = P D P^{-1}\).

题目：将矩阵 \(M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\) 对角化，即求可逆矩阵 P 和对角矩阵 D，使得 \(M = P D P^{-1}\)。

Answer: From the previous question, the eigenvalues are 3 and -1, with eigenvectors \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\). Form P by putting the eigenvectors as columns:

解答：由上一题知特征值为 3 和 -1，对应特征向量为 \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) 和 \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)。将特征向量按列排成矩阵 P：

\(P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\), \(D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}\).

Check \(M P = P D\). Compute \(M P = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 3 & 1 \end{pmatrix}\). And \(P D = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 3 & 1 \end{pmatrix}\). Hence \(M = P D P^{-1}\).

验证 \(M P = P D\)。计算 \(M P = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 3 & 1 \end{pmatrix}\)。而 \(P D = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 3 & 1 \end{pmatrix}\)。所以 \(M = P D P^{-1}\)。

10. Lines and Planes in 3D | 三维空间中的直线与平面

Exercise: Find the intersection point of the line \(\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-2}{1}\) and the plane \(x + 2y + z = 6\).

题目：求直线 \(\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-2}{1}\) 与平面 \(x + 2y + z = 6\) 的交点。

Answer: Introduce a parameter t: \(x = 1 + 2t\), \(y = 3 – t\), \(z = 2 + t\). Substitute into the plane equation:

解答：引入参数 t：\(x = 1 + 2t\)，\(y = 3 – t\)，\(z = 2 + t\)。代入平面方程：

\((1 + 2t) + 2(3 – t) + (2 + t) = 1 + 2t + 6 – 2t + 2 + t = 9 + t = 6\).

Thus \(t = -3\). Substitute back: \(x = 1 + 2(-3) = -5\), \(y = 3 – (-3) = 6\), \(z = 2 + (-3) = -1\). The intersection point is \((-5, 6, -1)\).

因此 \(t = -3\)。代回得：\(x = 1 + 2(-3) = -5\)，\(y = 3 – (-3) = 6\)，\(z = 2 + (-3) = -1\)。交点坐标为 \((-5, 6, -1)\)。

11. Vector Product and Area | 向量积与面积

Exercise: Find the area of the triangle with vertices \(A(1,0,1)\), \(B(2,1,0)\), \(C(3,1,1)\).

题目：求顶点为 \(A(1,0,1)\)、\(B(2,1,0)\)、\(C(3,1,1)\) 的三角形的面积。

Answer: The area of triangle ABC is \(\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|\). Compute \(\overrightarrow{AB} = (1,1,-1)\), \(\overrightarrow{AC} = (2,1,0)\).

解答：三角形 ABC 的面积为 \(\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|\)。计算 \(\overrightarrow{AB} = (1,1,-1)\)，\(\overrightarrow{AC} = (2,1,0)\)。

\(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 2 & 1 & 0 \end{vmatrix} = \mathbf{i}(1\cdot0 – (-1)\cdot1) – \mathbf{j}(1\cdot0 – (-1)\cdot2) + \mathbf{k}(1\cdot1 – 1\cdot2) = (1, 2, -1)\).

Its magnitude is \(\sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}\). Therefore the area is \(\frac{\sqrt{6}}{2}\).

其模长为 \(\sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}\)。因此面积为 \(\frac{\sqrt{6}}{2}\)。

12. General Solution of a Linear System | 线性方程组的通解

Exercise: Find the general solution of \(x_1 + 2x_2 – x_3 = 1\) and \(2x_1 + 4x_2 + 3x_3 = 0\).

题目：求方程组 \(x_1 + 2x_2 – x_3 = 1\) 和 \(2x_1 + 4x_2 + 3x_3 = 0\) 的通解。

Answer: Write the augmented matrix and row-reduce:

解答：写出增广矩阵并进行行化简：

\(\begin{pmatrix} 1 & 2 & -1 & 1 \\ 2 & 4 & 3 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 2 & -1 & 1 \\ 0 & 0 & 5 & -2 \end{pmatrix}\)

From the second row, \(5x_3 = -2\), so \(x_3 = -\frac{2}{5}\). Then from the first row, \(x_1 + 2x_2 – (-\frac{2}{5}) = 1\), so \(x_1 + 2x_2 = \frac{3}{5}\). Let \(x_2 = t\) (free parameter). Then \(x_1 = \frac{3}{5} – 2t\). The general solution is:

由第二行得 \(5x_3 = -2\)，所以 \(x_3 = -\frac{2}{5}\)。再由第一行得 \(x_1 + 2x_2 – (-\frac{2}{5}) = 1\)，即 \(x_1 + 2x_2 = \frac{3}{5}\)。令 \(x_2 = t\)（自由参数），则 \(x_1 = \frac{3}{5} – 2t\)。通解为：

\(\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} \\ 0 \\ -\frac{2}{5} \end{pmatrix} + t \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}\), \(t \in \mathbb{R}\).

This represents a line in 3-dimensional space.

这是三维空间中的一条直线。

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Answers to Exercises – Further Pure 4 | 进阶纯数四习题解答