(a) 求使下列方程组有唯一解的 $a$ 的取值范围：

(b) 证明 (a) 中的方程组对所有 $a$ 的取值都相容。

矩阵 $A$ 为

(c) 求矩阵 $P$ 和对角矩阵 $D$，使得

(d) 利用 $A$ 的特征方程证明

其中 $b$ 为待确定的整数。

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。

沿系数矩阵的第三列展开行列式，排除使它为零的参数。

The coefficient determinant is

The solution is unique exactly when $\boxed{a\in\mathbb R\setminus\{-3\}}$.

行列式非零时已有唯一解。只需检查 $a=-3$，此时第一式是第二式的三倍。

For $a\ne-3$, the unique solution from (a) makes the system consistent. At $a=-3$, the first equation is exactly three times the second. The second equation gives $y=2x-1$ and the third becomes $11x+4z=7$. Hence any real $x$, with $y=2x-1$ and $z=(7-11x)/4$, gives a solution. Thus the system is consistent for all real $a$.

也可找一个对所有 $a$ 都成立的解。令 $y=0$，含 $a$ 的项就消失。

Take $y=0$. The second equation then gives $x=\frac12$, which also satisfies the first for any $a$. The third gives $z=\frac38$. Thus $(\frac12,0,\frac38)$ solves the system for every real $a$.

先求下三角矩阵 $A$ 的特征向量。对于特征值 $\lambda$，矩阵 $(14A+24I)^2$ 在对应向量上乘以 $(14\lambda+24)^2$。

The eigenvalues of $A$ are $6,-1,4$. Solving $(A-\lambda I)\mathbf v=\mathbf0$ gives the following choices.

• For $\lambda=6$, $2v_1=7v_2$ and $v_1+5v_2=2v_3$, so take $(14,4,17)^{\mathsf T}$.
• For $\lambda=-1$, $v_1=0$ and $v_3=-v_2$, so take $(0,1,-1)^{\mathsf T}$.
• For $\lambda=4$, $v_1=v_2=0$, so take $(0,0,1)^{\mathsf T}$.

These vectors form a basis. The eigenvalues of $(14A+24I)^2$ on these vectors are $108^2$, $10^2$, and $80^2$, respectively. Hence we can take

用 Cayley–Hamilton 定理把 $14A+24I$ 表成含 $A^2$ 的乘积。再平方，读出括号内的常数。

The characteristic polynomial is

By the Cayley–Hamilton theorem,

Rearranging gives $14A+24I=-A^3+9A^2=-A^2(A-9I)$. The factors commute, so squaring gives

Thus $\boxed{b=-9}$.