(a) 常数 $a$ 和 $b$ 使方程组

没有唯一解。求 $a$，用 $b$ 表示。

(b) 已知 $a=-\frac{1}{4}$ 且 $b=-32$，证明 (a) 中的方程组无解。从几何角度解释这种情况。

矩阵 $\mathbf{A}$ 为

(c) 证明 $\begin{pmatrix}1\\0\\1\end{pmatrix}$ 是 $\mathbf{A}$ 的一个特征向量，并写出对应的特征值。

(d) 求矩阵 $\mathbf{P}$ 和对角矩阵 $\mathbf{D}$，使得 $\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}$。

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。

系数矩阵的行列式必须为零。展开后把含 $a$ 的项合并，同时检查除数为零的情形。

The coefficient determinant is

For example, expanding along the first row gives $6(-32-b)+12(-8a-2)-12(ab-8)$, equal to the expression above. Setting it equal to zero gives $2a(b+8)=-(b+20)$. If $b=-8$, the determinant is $-72$, so the required relation is

也可先用第二式分别消去第一、第三式的 $z$ 项。剩下的两元系数行列式必须为零。

Adding twelve times the second equation to the first, and eight times the second to the third, gives

The second original equation determines $z$ once $x,y$ are known. Thus the condition for the system to fail to have a unique solution is

Expanding and dividing by $6$ gives $2a(b+8)+(b+20)=0$, so $a=-(b+20)/(2(b+8))$, with $b\ne-8$.

比较第二式和第三式。它们的左边互为倍数，右边却不成相同比例。再比较第一式的法向量。

At $a=-\frac14$ and $b=-32$, the third left-hand side is $-8$ times the second. The right-hand sides are $3$ and $2$, and $3\ne -8\cdot2$, Adding eight times the second equation to the third would give $0=3+8(2)=19$. Thus the system is inconsistent. The second and third planes are distinct and parallel. The first plane is not parallel to them because its normal is not proportional to theirs, so it intersects both.

把矩阵与给定向量相乘，比较所得向量与原向量。

We have

Thus the given vector is an eigenvector with eigenvalue $-6$.

由特征方程找出其余特征值。分别解特征向量方程，再按同一顺序组成 $\mathbf P$ 和 $\mathbf D$。

Expanding $\det(\lambda\mathbf I-\mathbf A)$ gives

For $\lambda=2$, the first two rows of the eigenvector system give $v_1=3(v_2+v_3)$ and $v_1=2v_2+v_3$. Hence $v_2=-2v_3$, and $(3,2,-1)^{\mathsf T}$ satisfies all three rows. For $\lambda=6$, they give $v_2=-v_3$ and $v_1=3v_3$, so $(3,-1,1)^{\mathsf T}$ satisfies all three rows. Using the vector from (c) for $\lambda=-6$, take

The eigenvectors form a basis because their eigenvalues are distinct, so $\mathbf A=\mathbf P\mathbf D\mathbf P^{-1}$.