题目08. 在\(\mathbb{R}^2\)中，先平移\([1,1]^T\)，再旋转\(\frac{\pi}{3}\)，在伸长2倍的映射是：
A: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+\sqrt{3}y+1-\sqrt{3}\\ \sqrt{3}x-y+1+\sqrt{3}\end{pmatrix}\)
B: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x-\sqrt{3}y+1-\sqrt{3}\\ \sqrt{3}x+y+1+\sqrt{3}\end{pmatrix}\)
C: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\sqrt{3}x+y+1-\sqrt{3}\\ x-\sqrt{3}y+1+\sqrt{3}\end{pmatrix}\)
D: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix} \sqrt{3}x-y+1-\sqrt{3}\\ x+\sqrt{3}y+1+\sqrt{3}\end{pmatrix}\)
A: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+\sqrt{3}y+1-\sqrt{3}\\ \sqrt{3}x-y+1+\sqrt{3}\end{pmatrix}\)
B: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x-\sqrt{3}y+1-\sqrt{3}\\ \sqrt{3}x+y+1+\sqrt{3}\end{pmatrix}\)
C: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\sqrt{3}x+y+1-\sqrt{3}\\ x-\sqrt{3}y+1+\sqrt{3}\end{pmatrix}\)
D: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix} \sqrt{3}x-y+1-\sqrt{3}\\ x+\sqrt{3}y+1+\sqrt{3}\end{pmatrix}\)
举一反三
- 题目09. 在\(\mathbb{R}^2\)中先关于\(y=x\)反射，再平移\([1,1]^T\)，再关于\(y=-x\)反射的映射是： A: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix} x+1\\ y+1\end{pmatrix}\) B: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix} x-1\\ y-1\end{pmatrix}\) C: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix} -x+1\\ -y+1\end{pmatrix}\) D: \(f\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix} -x-1\\ -y-1\end{pmatrix}\)
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- 下列矩阵中是单位矩阵的为（ ）. A: $\begin{pmatrix}1&1\\1&1\end{pmatrix}$ B: $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ C: $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ D: $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
- 求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$
- 假设\(x_k\)表示第\(k\)年某校学生喜欢微积分胜于线性代数的百分比，则 \(y_k = 1- x_k\)表示更喜欢线性代数的百分比.在\(k + 1\)年, 原来更喜欢微积分的同学的\(1/5\)想法转变，而原来更喜欢线性代数的同学的\(3/10\)改变了想法.于是有\(\begin{pmatrix}x_{k+1}\\y_{k+1}\end{pmatrix}=A\begin{pmatrix}x_k\\y_k\end{pmatrix}\). 则极限 \(\lim_{k\rightarrow \infty} A^k\begin{pmatrix}0.5\\0.5\end{pmatrix}=\_\_.\) A: \(\begin{pmatrix}0.6\\0.4\end{pmatrix}\) B: \(\begin{pmatrix}0.5\\0.5\end{pmatrix}\) C: \(\begin{pmatrix}0.4\\0.6\end{pmatrix}\) D: \(\begin{pmatrix}0.7\\0.3\end{pmatrix}\)