(a) 求 $k$ 的值，使得方程组

没有唯一解。

(b) 已知 $k = -\frac{1}{2}$，证明 (a) 中的方程组不相容。用几何方式解释这种情况。

(c) 改为已知 $k = -1$，证明 (a) 中的方程组同样不相容。用几何方式解释这种情况。

矩阵 $\mathbf{A}$ 定义为

(d) 使用 $\mathbf{A}$ 的特征方程证明 $\mathbf{A}^4 = p\mathbf{A}^2 + q\mathbf{A}$，其中 $p$ 和 $q$ 是需要确定的整数。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

方程组没有唯一解等价于系数矩阵的行列式为零。展开并因式分解。

The system fails to have a unique solution exactly when its coefficient determinant is zero. Expanding along the first row gives

Thus $k=-\frac12$ or $k=-1$.

代入 $k=-\frac12$ 后，用第一式和第二式消去 $y$，再与第一式和第三式之差比较。三个平面两两不平行，但三个平面没有公共点。

At $k=-\frac12$, adding the first and second equations gives $\frac12x+z=3$. Subtracting the first equation from the third gives $x+2z=2$. Doubling the first relation gives $x+2z=6$, contradicting $x+2z=2$. Thus the system is inconsistent.

The three plane normals are pairwise non-parallel, so each pair of planes intersects in a line. The normals span a two-dimensional space because the determinant vanishes. Each line of intersection is therefore perpendicular to that same space of normals. The three lines are distinct and parallel, forming the edges of a triangular prism.

第一、第二式的左边互为相反数，右边却不相容，因此对应两个不同的平行平面。再比较第三个平面的法向量。

For $k=-1$, the first two equations give $x-y-2z=1$ and $-(x-y-2z)=2$, which is impossible. Their planes are distinct and parallel. The third plane has normal $(2,-1,1)$, which is not proportional to $(1,-1,-2)$, so it cuts both parallel planes.

先求特征方程，用 Cayley–Hamilton 定理把 $\mathbf A^3$ 化为低次幂。再乘 $\mathbf A$ 并代回同一关系。

The characteristic determinant is

Hence the characteristic equation is $\lambda^3-3\lambda^2+8\lambda=0$. By the Cayley–Hamilton theorem,

Multiplying by $\mathbf A$ and substituting for $\mathbf A^3$ yields

Therefore $\boxed{p=1,\quad q=-24}$.