(a) 求 $a$ 的取值集合，使方程组

有唯一解，并从几何上解释这种情况。

(b) 已知 $a = -\frac{5}{4}$，证明 (a) 中的方程组是不相容的，并从几何上解释这种情况。

矩阵 $\mathbf{A}$ 为

(c) 求矩阵 $\mathbf{P}$ 和对角矩阵 $\mathbf{D}$，使得 $(\mathbf{A} + 3\mathbf{I})^2 = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$。

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。

系数矩阵第三列只有一个非零元素，沿第三列展开行列式即可。行列式非零时，三个平面只有一个公共点。

The determinant of the coefficient matrix is

Thus the solution is unique for $\boxed{a\in\mathbb R\setminus\{-\frac54\}}$. Geometrically, the three planes intersect at exactly one point.

代入后，第二式左边是第一式左边的四倍，而右边不是。再比较第三个平面的法向量，判断它和前两个平面的位置关系。

At $a=-\frac54$, multiplying the first equation by $4$ gives $20x-5y=12$, contradicting the second equation $20x-5y=2$. The first two planes are distinct and parallel. The third plane has a non-zero $z$-coefficient, so it is not parallel to them and intersects both.

先对 $\mathbf A$ 求特征向量。若某个向量对应特征值 $\lambda$，那么 $\mathbf A+3\mathbf I$ 将它乘以 $\lambda+3$，再平方就乘以 $(\lambda+3)^2$。注意 $\mathbf D$ 中元素的顺序须与 $\mathbf P$ 的列一致。

The lower triangular matrix $\mathbf A$ has eigenvalues $5,-5,1$. For a vector $(u,v,w)^{\mathsf T}$, the systems $(\mathbf A-\lambda\mathbf I)(u,v,w)^{\mathsf T}=\mathbf0$ give

For $\lambda=5$, the last two equations are $20u-10v=0$ and $2u-3v-4w=0$, giving $v=2u$, $w=-u$. For $\lambda=-5$, the first and last equations are $10u=0$ and $2u-3v+6w=0$, giving $u=0$, $v=2w$. For $\lambda=1$, the first two equations give $4u=0$ and $20u-6v=0$, so $u=v=0$ and $w$ is free.

The resulting matrix of eigenvectors has determinant $2\ne0$. These eigenvectors form a basis, and $(\mathbf A+3\mathbf I)^2$ acts on them by multiplication by $64,4,16$, respectively. One choice is

This gives $(\mathbf A+3\mathbf I)^2=\mathbf P\mathbf D\mathbf P^{-1}$.