矩阵 $\mathbf{M}$ 为

其中 $0<\theta<2\pi$。

(a) 矩阵 $\mathbf{M}$ 表示在 $x$-$y$ 平面上依次进行的两个几何变换。

写出每个变换的类型，并明确说明它们的施行顺序。

(b) 求 $\theta$ 的值，使矩阵 $\mathbf{M}$ 所表示的变换有一条由不变点组成的直线。

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。

矩阵从右向左作用，所以先旋转再错切。

The transformations are a rotation followed by a shear. More precisely, the rotation is anticlockwise through $\theta$ about the origin; the shear fixes the $x$-axis and maps $(0,1)$ to $(2,1)$.

不变点要求两个坐标都保持不变。利用第二坐标方程消去y，得到θ的条件，并回代确认确有整条不变点直线。

Write $c=\cos\theta$ and $s=\sin\theta$. Multiplying the matrices,

A fixed point $(x,y)$ satisfies

Since $0<\theta<2\pi$, $c\ne1$. The second equation gives $y=sx/(1-c)$. A line of fixed points contains a non-zero point; here $x$ cannot be zero, since that would also give $y=0$.

Substitution and division by $x$ give

Expanding and using $c^2+s^2=1$,

Thus $\sqrt2\sin(\theta+\pi/4)=1$. In $0<\theta<2\pi$, the only solution is $\boxed{\theta=\pi/2}$; the alternatives $0$ and $2\pi$ are excluded.

At $\theta=\pi/2$, $\mathbf M=\begin{pmatrix}2&-1\\1&0\end{pmatrix}$, and every point on $y=x$ is fixed, confirming the required line exists.

也可把不变点条件写成齐次方程组，用系数行列式为零找到非零解存在的条件。

A line of fixed points requires non-zero solutions of $(\mathbf M-\mathbf I)\mathbf v=\mathbf0$. Thus the coefficient determinant must vanish:

Hence $\sin\theta+\cos\theta=1$, giving $\theta=\pi/2$ in the stated open interval. At this value, $\mathbf M-\mathbf I=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$, whose solutions form the line $y=x$. Therefore $\boxed{\theta=\pi/2}$.