國立台北科技大學108學年度碩士班招生考試
系所組別: 2150電機工程系碩士班戊組
第一節 線性代數
解答: $$\cases{x_1+3x_2+10x_3=18\\ -2x_1+7x_2+32x_3=29\\ -x_1+3x_2+14x_3= 12\\ x_1+x_2+2x_3=8} \Rightarrow \left[\begin{matrix}1 & 3 & 10\\-2 & 7 & 32\\-1 & 3 & 14\\1 & 1 & 2\end{matrix} \right] \begin{bmatrix} x_1 \ \\x_2\\x_3 \end{bmatrix} = \begin{bmatrix} 18 \ \\29\\ 12\\8 \end{bmatrix} \\ \Rightarrow \text{augmented matrix }\left[\begin{matrix}1 & 3 & 10 & 18\\-2 & 7 & 32 & 29\\-1 & 3 & 14 & 12\\1 & 1 & 2 & 8\end{matrix}\right] \xrightarrow{2R_1+R_2 \to R_2, R_1+R_3 \to R_3, -R_1+R_4\to R_4} \\ \left[ \begin{matrix}1 & 3 & 10 & 18\\0 & 13 & 52 & 65\\0 & 6 & 24 & 30\\0 & -2 & -8 & -10\end{matrix} \right] \xrightarrow{R_3/6\to R_3, -R_4/2 \to R_4} \left[ \begin{matrix}1 & 3 & 10 & 18\\0 & 13 & 52 & 65\\0 & 1 & 4 & 5\\0 & 1 & 4 & 5\end{matrix} \right] \\\xrightarrow{-3R_3+R_1 \to R_1, -13R_3+R_2 \to R_2, -R_3+R_4 \to R_4} \left[\begin{matrix}1 & 0 & -2 & 3\\0 & 0 & 0 & 0\\0 & 1 & 4 & 5\\0 & 0 & 0 & 0\end{matrix} \right] \xrightarrow{R_2 \leftrightarrow R_3} \left[\begin{matrix}1 & 0 & -2 & 3\\0 & 1 & 4 & 5\\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{matrix} \right] \\ \Rightarrow \cases{x_1-2x_3=3\\ x_2+4x_3=5} \Rightarrow (x_1,x_2,x_3,x_4) \in \bbox[red, 2pt]{ \{(2k+3,-4k+5,k) \mid k \in \mathbb R\}}$$
解答: $$\textbf{(a)}\;\begin{vmatrix}1 & 1& 3\\ 1& 2& 4\\ 1 & 3& a \end{vmatrix} =0 \Rightarrow a=5 \Rightarrow \left[ \begin{array}{rrr|r}1 & 1& 3& 2\\ 1& 2& 4 & 3\\ 1 & 3& 5 &b\end{array} \right] \xrightarrow{R_2-R_1\to R_2, R_3-R_1\to R_3} \left[ \begin{array}{rrr|r}1 & 1& 3& 2\\ 0& 1& 1 & 1\\ 0 & 2& 2 &b-2\end{array} \right] \\ \text{infinitely many sollutions }\Rightarrow {1\over 2}={1\over b-2} \Rightarrow b=4 \Rightarrow \bbox[red, 2pt]{\cases{a=5\\ b=4}} \\ \textbf{(b)}\; \text{ trivially,} \bbox[red, 2pt]{\cases{a=5\\ b\ne 4}}$$
解答: $$\text{Let }T:E\to F \text{ and }T=\begin{bmatrix}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{bmatrix} \Rightarrow T(v_i)=u_i,i=1,2,3 \\ \Rightarrow \cases{\begin{bmatrix}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{bmatrix} \begin{bmatrix}1 \\1\\ 1 \end{bmatrix} = \begin{bmatrix}1 \\1\\ 0 \end{bmatrix} \\[1ex] \begin{bmatrix}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{bmatrix} \begin{bmatrix}2 \\3\\ 2 \end{bmatrix} = \begin{bmatrix}1 \\2\\ 0 \end{bmatrix} \\[1ex] \begin{bmatrix}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{bmatrix} \begin{bmatrix}1 \\5\\ 4 \end{bmatrix} = \begin{bmatrix}1 \\2\\ 1 \end{bmatrix} } \Rightarrow \cases{\begin{bmatrix} 1 &1 & 1 \\2 & 3 & 2\\ 1 & 5 & 4 \end{bmatrix} \begin{bmatrix}a_1 \\a_2\\ a_3 \end{bmatrix} = \begin{bmatrix}1 \\1\\ 1 \end{bmatrix} \\[1ex] \begin{bmatrix} 1 &1 & 1 \\2 & 3 & 2\\ 1 & 5 & 4 \end{bmatrix} \begin{bmatrix}b_1 \\b_2\\ b_3 \end{bmatrix} = \begin{bmatrix}1 \\2\\ 2 \end{bmatrix} \\[1ex] \begin{bmatrix} 1 &1 & 1 \\2 & 3 & 2\\ 1 & 5 & 4 \end{bmatrix} \begin{bmatrix}c_1 \\c_2\\ c_3 \end{bmatrix} = \begin{bmatrix}0 \\0\\ 1 \end{bmatrix} }\\ \begin{bmatrix} 1 &1 & 1 \\2 & 3 & 2\\ 1 & 5 & 4 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \\-2 & 1 & 0 \\\frac{7}{3} & \frac{-4}{3} & \frac{1}{3} \end{bmatrix} \Rightarrow \cases{\begin{bmatrix}a_1 \\a_2\\ a_3 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \\-2 & 1 & 0 \\\frac{7}{3} & \frac{-4}{3} & \frac{1}{3} \end{bmatrix} \begin{bmatrix}1 \\1\\ 1 \end{bmatrix}= \begin{bmatrix}\frac{2}{3} \\-1 \\ \frac{4}{3} \end{bmatrix} \\[1ex] \begin{bmatrix}b_1 \\b_2\\ b_3 \end{bmatrix} =\begin{bmatrix} \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \\-2 & 1 & 0 \\\frac{7}{3} & \frac{-4}{3} & \frac{1}{3} \end{bmatrix} \begin{bmatrix}1 \\2\\ 2 \end{bmatrix}= \begin{bmatrix} \frac{2}{3} \\0 \\\frac{1}{3}\end{bmatrix}\\[1ex] \begin{bmatrix}c_1 \\c_2\\ c_3 \end{bmatrix} =\begin{bmatrix} \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \\-2 & 1 & 0 \\\frac{7}{3} & \frac{-4}{3} & \frac{1}{3} \end{bmatrix} \begin{bmatrix}0 \\0\\ 1 \end{bmatrix} =\begin{bmatrix} \frac{-1}{3} \\0 \\\frac{1}{3} \end{bmatrix} } \\ \Rightarrow T= \bbox[red, 2pt]{\begin{bmatrix} \frac{2}{3} & -1 & \frac{4}{3} \\\frac{2}{3} & 0 & {1\over 3} \\ -{1\over 3} & 0 & {1\over 3}\end{bmatrix} } \\x=3v_1+2v_2-v_3=(3,3,3)+(4,6,4)-(1,5,4)=(6,5,3) =au_1+ bu_2 +cu_3 \\ \Rightarrow \cases{a+b+c=6\\ a+2b+2c= 5\\ c=3} \Rightarrow (a,b,c)= \bbox[red, 2pt]{(7,-4,3)}$$
解答: $$\lim_{n\to\infty} {a_n\over a_{n-1}} =\alpha \Rightarrow \lim_{n\to\infty} {a_n\over a_{n-1}} =\lim_{n\to\infty} {a_{n-1}+a_{n-2}\over a_{n-1}} = 1+\lim_{n\to\infty} {a_{n-2}\over a_{n-1}} =1+{1\over \alpha} \\ \Rightarrow \alpha=1+{1\over \alpha} \Rightarrow \alpha^2-\alpha -1=0 \Rightarrow \alpha={ 1+\sqrt 5\over 2}, \bbox[red, 2pt]{Q.E.D.}$$
解答: $$A=[v_1 \mid v_2\mid v_3] =\begin{bmatrix}1 & 2& 5 \\-1 & 1& -4\\ -1 & 4& -3\\ 1& -4& 7\\ 1& 2& 1 \end{bmatrix} \Rightarrow \cases{v_1=(1,-1,-1,1,1)^T\\ v_2=(1,1,4,-4,2)^T \\ v_3=(5, -4,-3 ,7,1)^T} \\\text{ by Gram-Schmidt processing} \Rightarrow \cases{e_1={1\over \sqrt 5}(1,-1,-1,1,1)^T \\e_2={1\over 2}(1,0,1,-1,1)^T\\ e_3={1\over 2}(1,0,1,1,-1)^T} \\ \Rightarrow Q=[e_1 \mid e_2\mid e_3] =\begin{bmatrix}{\sqrt 5\over 5} & {1\over 2}& {1\over 2} \\ -{\sqrt 5\over 5} & 0 & 0\\ -{\sqrt 5\over 5} & {1\over 2}& {1\over 2} \\ {\sqrt 5\over 5} & -{1\over 2}& {1\over 2} \\ {\sqrt 5\over 5}& {1\over 2}& -{1\over 2}\end{bmatrix} \\ \Rightarrow R=Q^TA= \begin{bmatrix}\sqrt 5 & -\sqrt 5& 4\sqrt 5\\ 0 & 6 & -2\\ 0 & 0 & 4 \end{bmatrix} \Rightarrow A=QR=\bbox[red, 2pt]{ \begin{bmatrix}{\sqrt 5\over 5} & {1\over 2}& {1\over 2} \\ -{\sqrt 5\over 5} & 0 & 0\\ -{\sqrt 5\over 5} & {1\over 2}& {1\over 2} \\ {\sqrt 5\over 5} & -{1\over 2}& {1\over 2} \\ {\sqrt 5\over 5}& {1\over 2}& -{1\over 2} \end{bmatrix} \begin{bmatrix} \sqrt 5 & -\sqrt 5& 4\sqrt 5\\ 0 & 6 & -2\\ 0 & 0 & 4 \end{bmatrix} }$$
解答: $$<1,1>=\int_{-1}^1 1\,dx =2 \Rightarrow \Vert 1\Vert =\sqrt 2 \Rightarrow e_1={1\over \Vert 1\Vert } ={1\over \sqrt 2}\\ \cases{<{1\over \sqrt 2},x> =\int_{-1}^1 {1\over \sqrt 2}x\,dx = 0 \\ <x,x>= \int_{-1}^1 x^2\,dx={2\over 3}}\Rightarrow e_2={x-<{1\over \sqrt 2},x>{1\over \sqrt 2} \over \Vert x-<{1\over \sqrt 2},x>{1\over \sqrt 2} \Vert} ={x\over \Vert x\Vert} ={\sqrt 3\over \sqrt 2}x\\ \cases{<{1\over \sqrt 2}, x^2>={\sqrt 2\over 3} \\ <\sqrt{3\over 2}x,x^2>=0} \Rightarrow e_3={ x^2-<{1\over \sqrt 2},x^2>{1\over \sqrt 2} -<\sqrt{3\over 2}x,x^2>\sqrt{3\over 2}x\over \Vert x^2-<{1\over \sqrt 2},x^2>{1\over \sqrt 2} -<\sqrt{3\over 2}x,x^2>\sqrt{3\over 2}x\Vert} \\={x^2-{1\over 3}\over \Vert x^2-{1\over 3}\Vert }={3\sqrt 5\over 2\sqrt 2}(x^2-{1\over 3})\\ \Rightarrow e_4={x^3-<{1\over \sqrt 2},x^3>{1\over \sqrt 2} -<\sqrt{3\over 2}x,x^3>\sqrt{3\over 2}x -<{3\sqrt 5\over 2\sqrt 2}(x^2-{1\over 3}), x^3>{3\sqrt 5\over 2\sqrt 2}(x^2-{1\over 3}) \over \Vert x^3-<{1\over \sqrt 2},x^3>{1\over \sqrt 2} -<\sqrt{3\over 2}x, x^3>\sqrt{3\over 2}x -<{3\sqrt 5\over 2\sqrt 2}(x^2-{1\over 3}), x^3>{3\sqrt 5\over 2\sqrt 2}(x^2-{1\over 3}) \Vert } \\={x^3 -<\sqrt{3\over 2}x, x^3>\sqrt{3\over 2}x \over \Vert x^3 -<\sqrt{3\over 2}x,x^3> \sqrt{3\over 2}x \Vert } ={x^3-{1\over 5}x \over \Vert x^3-{1\over 5}x\Vert} ={5\sqrt 5\over 2\sqrt 2}(x^3-{1\over 5}x)\\ \Rightarrow \text{an orthonormal basis: }\bbox[red, 2pt]{{1\over \sqrt 2}, {\sqrt 3\over \sqrt 2}x, {3\sqrt 5\over 2\sqrt 2}(x^2-{1\over 3}), {5\sqrt 5\over 2\sqrt 2}(x^3-{1\over 5}x)}$$
解答: $$A=\left[\begin{matrix}1 & 2 & 0 & 1 & -1\\2 & 1 & 3 & 1 & 0\\-1 & 0 & -2 & 0 & 1\\0 & 0 & 0 & 2 & 8\end{matrix}\right] \Rightarrow rref(A)= \left[ \begin{matrix} 1 & 0 & 2 & 0 & -1\\0 & 1 & -1 & 0 & -2\\0 & 0 & 0 & 1 & 4\\0 & 0 & 0 & 0 & 0\end{matrix}\right] \\ \left[ \begin{matrix} 1 & 0 & 2 & 0 & -1\\0 & 1 & -1 & 0 & -2\\0 & 0 & 0 & 1 & 4\\0 & 0 & 0 & 0 & 0\end{matrix} \right] \begin{bmatrix}x_1 \\x_2 \\ x_3\\ x_4\\ x_5 \end{bmatrix} =0 \Rightarrow \cases{x_1+2x_3-x_5=0\\ x_2-x_3-2x_5=0\\ x_4+4x_5=0} \Rightarrow x=x_3 \begin{pmatrix}-2\\ 1\\ 1\\0\\ 0 \end{pmatrix}+ x_5 \begin{pmatrix}1\\ 2\\0\\-4 \\ 1 \end{pmatrix} \\ \Rightarrow \text{a basis of ker}(T) =\bbox[red, 2pt]{\left\{ \begin{pmatrix}-2\\ 1\\ 1\\0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 2\\0\\-4 \\ 1 \end{pmatrix}\right\}}$$
解答: $$\left[ \begin{matrix}2 & -4 & -2 & 3 \\6 & -9 & -5 & 8 \\2 & -7 & -3 & 9 \\4 & -2 & -2 & -1 \\-6 & 3 & 3 & 4 \end{matrix} \right] =\left[ \begin{matrix}1 & 0 & 0 & 0& 0 \\a_1 &1 & 0 & 0& 0 \\a_2 & a_3 & 1 & 0& 0 \\a_4 & a_5 &a_6 &1 & 0 \\ a_7 & a_8 & a_9 & a_{10}& 1 \end{matrix} \right] \left[ \begin{matrix}b_1 & b_2 & b_3 & b_4 \\0 &b_5 & b_6 & b_7 \\0 & 0 & b_8 & b_9 \\0 & 0 &0 &b_{10} \\ 0 & 0 & 0 & 0 \end{matrix} \right]\\ = \left[ \begin{matrix}b_1 & b_2 & b_3 & b_4 \\a_1b_1 &a_1b_2+ b_5 & a_1b_3+ b_6 & a_1b_4+ b_7 \\a_2 b_1 & a_2b_2+ a_3b_5 & a_2b_3+ a_3b_6+b_8 &a_2b_4 +a_3b_7+b_9\\ a_4b_1 & a_4b_2+ a_5b_5 &a_4b_3+ a_5b_6+b_8 &a_4b_4+ a_5b_7+ a_6b_9+b_{10} \\ a_7b_1 & a_7b_2+ a_8b_5 & a_7b_3+ a_8b_6+ a_9b_8 & a_7b_4+ a_8b_7+ a_9b_9+ a_{10}b_{10} \end{matrix} \right] \\ \Rightarrow \cases{b_1=2\\ b_2=-4\\ b_3=-2\\ b_4=3} \Rightarrow \cases{a_1b_1=6\\ a_1b_2+ b_5=-9\\ a_1b_3+b_6=-5\\ a_1b_4+b_7=8} \Rightarrow \cases{a_1=3\\ b_5=3\\ b_6=1\\ b_7=-1} \Rightarrow \cdots\\ B= \bbox[red, 2pt]{\left[ \begin{matrix}1 & 0 & 0 & 0 & 0 \\3 & 1 & 0 & 0 & 0 \\1 & -1 & 1 & 0 & 0 \\2 & 2 & -1 & 1 & 0 \\-3 & -3 & 2 & 0 & 1 \end{matrix} \right] \left[\begin{matrix}2 & -4 & -2 & 3 \\0 & 3 & 1 & -1 \\0 & 0 & 0 & 5 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{matrix} \right]}$$
解答: $$C=\left[\begin{matrix}1 & -1\\0 & 1\\1 & 0\end{matrix}\right] \Rightarrow W=CC^T =\left[ \begin{matrix}2 & -1 & 1\\-1 & 1 & 0\\1 & 0 & 1\end{matrix} \right]\\ \det(W-\lambda I)=0 \Rightarrow \text{eigenvalues: }\lambda =3,1,0 \Rightarrow \text{ eigenvectors:} \begin{bmatrix} 2\\-1\\ 1 \end{bmatrix}, \begin{bmatrix}0 \\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} -1\\ -1\\ 1 \end{bmatrix} \\ \xrightarrow{normalization} u_1= \begin{bmatrix} \sqrt 6/3\\-\sqrt 6/6\\ \sqrt 6/6 \end{bmatrix}, u_2= \begin{bmatrix}0 \\ \sqrt 2/2\\ \sqrt 2/2 \end{bmatrix}, u_3=\begin{bmatrix} -\sqrt 3/3\\ -\sqrt 3/3\\ \sqrt 3/3 \end{bmatrix}\\ \Rightarrow \text{square roots of the nonzero eigenvalues }\cases{\sigma_1=\sqrt 3\\ \sigma_2 = 1} \Rightarrow \sum =\begin{bmatrix}\sqrt 3 & 0 \\0 & 1\\ 0& 0 \end{bmatrix} \\ \Rightarrow U=[u_1 \mid u_2\mid u_3] =\begin{bmatrix}{\sqrt 6 \over 3} & 0 & -{\sqrt 3\over 3} \\-{\sqrt 6\over 6} & {\sqrt 2\over 2} &-{\sqrt 3\over 3} \\ {\sqrt 6\over 6} & {\sqrt 2\over 2} & {\sqrt 3\over 3}\end{bmatrix}\\ \Rightarrow \cases{v_1= {1\over \sigma_1} C^T \cdot u_1= \begin{bmatrix}\sqrt 2/2 \\-\sqrt 2/2 \end{bmatrix}\\ v_2= {1\over \sigma_2} C^T\cdot u_2= \begin{bmatrix}\sqrt 2/2 \\\sqrt 2/2 \end{bmatrix}} \Rightarrow V= \begin{bmatrix}\sqrt 2/2 & \sqrt 2/2 \\-\sqrt 2/2 & \sqrt 2/2\end{bmatrix} \\ \Rightarrow C=U \sum V^T = \bbox[red, 2pt]{\begin{bmatrix} {\sqrt 6 \over 3} & 0 & -{\sqrt 3\over 3} \\-{\sqrt 6\over 6} & {\sqrt 2\over 2} &-{\sqrt 3\over 3} \\ {\sqrt 6\over 6} & {\sqrt 2\over 2} & {\sqrt 3\over 3}\end{bmatrix} \begin{bmatrix}\sqrt 3 & 0 \\0 & 1\\ 0& 0 \end{bmatrix} \begin{bmatrix}\sqrt 2/2 & \sqrt 2/2 \\-\sqrt 2/2 & \sqrt 2/2\end{bmatrix}^T}$$
解答: $$x^2-6xy+y^2 =[x,y]\begin{bmatrix}1 & -3 \\-3 & 1 \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} =[x,y] \left[ \begin{matrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix} \right] \begin{bmatrix}-2 & 0 \\0 & 4 \end{bmatrix} \left[ \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix} \right] \begin{bmatrix}x \\y \end{bmatrix} \\ =[{1\over \sqrt 2}(x+y), {1\over \sqrt 2}(-x+y)] \begin{bmatrix}-2 & 0 \\0 & 4 \end{bmatrix} \begin{bmatrix}{1\over \sqrt 2}(x+y)\\{1\over \sqrt 2}(-x+y)\end{bmatrix} =[x',y']\begin{bmatrix}-2 & 0 \\0 & 4 \end{bmatrix} \begin{bmatrix}x' \\y' \end{bmatrix} \\ =-2x'^2+4y'^2 =5 為一雙曲線\\ 雙曲線主軸旋轉角度\theta \Rightarrow \cot 2\theta ={1-1\over 6} =0 \Rightarrow \theta =45^\circ ,因此圖形如下.$$
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