矩阵 $\mathbf{A}$ 为

(a) 证明 $\mathbf{A}$ 的特征方程为 $\lambda^3-4\lambda^2-20\lambda+48=0$，并由此求 $\mathbf{A}$ 的特征值。

(b) 求矩阵 $\mathbf{P}$ 和对角矩阵 $\mathbf{D}$，使得 $\mathbf{A}^5=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}$。

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。

写出 $\lambda\mathbf I-\mathbf A$ 的行列式并沿第一行展开。得到三次式后，试整数根，再因式分解。

Expanding along the first row gives

Setting the determinant to zero gives the stated characteristic equation and eigenvalues $\boxed{2,-4,6}$.

对每个特征值，从 $\mathbf A-\lambda\mathbf I$ 中选两行独立的行向量，求它们的叉积作为特征向量。五次幂矩阵的对应特征值为 $\lambda^5$。

For example, the cross product for $\lambda=2$ is

For $\lambda=2$, the first two rows of $\mathbf A-2\mathbf I$ have cross product $(24,-24,24)$, so take $(1,-1,1)^{\mathsf T}$. For $\lambda=-4$, the second and third rows have cross product $(-120,0,-240)$, so take $(1,0,2)^{\mathsf T}$. For $\lambda=6$, the first two rows have cross product $(-20,-40,-40)$, so take $(1,2,2)^{\mathsf T}$. The matrices have rank two, so these vectors satisfy all the rows.

The three eigenvectors form a basis. Since the corresponding eigenvalues of $\mathbf A^5$ are $32,-1024,7776$, take

Then $\mathbf A^5=\mathbf P\mathbf D\mathbf P^{-1}$.