矩阵 $\mathbf M$ 为

其中 $k$ 为常数。

(a) 矩阵 $\mathbf M$ 表示依次进行的两个几何变换。

写出每个变换的类型，并明确说明它们的作用顺序。

(b) $x$–$y$ 平面内的三角形 $ABC$ 经 $\mathbf M$ 变换后得到三角形 $DEF$。

用 $k$ 表示将三角形 $DEF$ 变换为三角形 $ABC$ 的单个矩阵。

(c) 求使 $\mathbf M$ 所表示的变换不存在经过原点的不变直线的 $k$ 的取值集合。

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。

先作用的是右侧剪切矩阵，再由左侧矩阵进行旋转。

The sequence is a shear followed by a rotation. The rotation is anticlockwise through $45^\circ$ about the origin.

反转两步次序并分别取逆，最后相乘成题目要求的单一矩阵。

Writing $\mathbf M=\mathbf R\mathbf H$, its inverse is

This single matrix maps $DEF$ back onto $ABC$.

用y=mx建立不变方向的二次方程，其判别式为负才没有实斜率。必须另查竖直线，尤其二次项系数为零的k=1不能混入答案。

Multiplying,

For a point $(x,mx)$ on an invariant line, its image must satisfy $Y=mX$. Cancelling the common scale factor gives

so

When $k\ne1$, no real slopes exist exactly when the discriminant is negative:

The vertical line is invariant only if $k-1=0$, because the image of $(0,y)$ has horizontal coordinate $(k-1)y/\sqrt2$. The value $k=1$ is outside the interval above, and at that value the slope equation also has the real solution $m=-1$.

Thus the required set is

The endpoints are excluded because they give a repeated real invariant direction.

也可利用不变方向就是实特征向量方向，求特征方程是否有实根。矩阵行列式为1，特征值不会为零。

An invariant real line through the origin provides a real eigenvector of $\mathbf M$. Conversely a real eigenvector spans an invariant line. Since $\det\mathbf M=1$, no eigenvalue is zero.

The trace is $(k+2)/\sqrt2$, so the characteristic equation is

There are no real eigenvalues exactly when

Therefore there are no invariant real lines precisely when

At equality a real eigenvalue and eigenvector exist, so neither endpoint is included.