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: {x1+3x2+10x3=18−2x1+7x2+32x3=29−x1+3x2+14x3=12x1+x2+2x3=8⇒[1310−2732−1314112][x1 x2x3]=[18 29128]⇒augmented matrix [131018−273229−1314121128]2R1+R2→R2,R1+R3→R3,−R1+R4→R4→[13101801352650624300−2−8−10]R3/6→R3,−R4/2→R4→[131018013526501450145]−3R3+R1→R1,−13R3+R2→R2,−R3+R4→R4→[10−23000001450000]R2↔R3→[10−23014500000000]⇒{x1−2x3=3x2+4x3=5⇒(x1,x2,x3,x4)∈{(2k+3,−4k+5,k)∣k∈R}解答: (a)|11312413a|=0⇒a=5⇒[11321243135b]R2−R1→R2,R3−R1→R3→[11320111022b−2]infinitely many sollutions ⇒12=1b−2⇒b=4⇒{a=5b=4(b) trivially,{a=5b≠4
解答: Let T:E→F and T=[a1a2a3b1b2b3c1c2c3]⇒T(vi)=ui,i=1,2,3⇒{[a1a2a3b1b2b3c1c2c3][111]=[110][a1a2a3b1b2b3c1c2c3][232]=[120][a1a2a3b1b2b3c1c2c3][154]=[121]⇒{[111232154][a1a2a3]=[111][111232154][b1b2b3]=[122][111232154][c1c2c3]=[001][111232154]−1=[2313−13−21073−4313]⇒{[a1a2a3]=[2313−13−21073−4313][111]=[23−143][b1b2b3]=[2313−13−21073−4313][122]=[23013][c1c2c3]=[2313−13−21073−4313][001]=[−13013]⇒T=[23−14323013−13013]x=3v1+2v2−v3=(3,3,3)+(4,6,4)−(1,5,4)=(6,4,3)=au1+bu2+cu3⇒{a+b+c=6a+2b+2c=4c=3⇒(a,b,c)=(8,−5,3)
解答: lim
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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