113年台大碩士班-基礎數學詳解

 國立臺灣大學 113 學年度碩士班招生考試

科目: 基礎數學

解答:$$A= \begin{bmatrix}2&-1\\-1&2 \end{bmatrix} \Rightarrow \det(A-\lambda I)= (\lambda-1)(\lambda-3)=0 \Rightarrow \lambda=1,3 \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)=0 \Rightarrow \begin{bmatrix} 1 & -1 \\-1 & 1\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=x_2 \Rightarrow v=x_2 \begin{pmatrix}1\\ 1 \end{pmatrix}, \text{ choose }v_1= \begin{pmatrix}1\\ 1 \end{pmatrix} \\ \lambda_2=3 \Rightarrow (A-\lambda_2 I)=0 \Rightarrow \begin{bmatrix} -1 & -1 \\-1 & -1\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1+x_2 =0\Rightarrow v=x_2 \begin{pmatrix}-1 \\ 1 \end{pmatrix}, \text{ choose }v_2= \begin{pmatrix}-1\\ 1 \end{pmatrix} \\ \bbox[red, 2pt]{A的特徵值為1,3，特徵向量為\begin{pmatrix}1\\ 1 \end{pmatrix}, \begin{pmatrix} -1\\ 1 \end{pmatrix}} \Rightarrow A^2(v) =A(\lambda v) =\lambda A(v)=\lambda^2 v \\ \Rightarrow A^2的特徵值為\lambda^2, 特徵向量與A相同\Rightarrow \bbox[red, 2pt]{A^2的特徵值為1,9，特徵向量為\begin{pmatrix}1\\ 1 \end{pmatrix}, \begin{pmatrix} -1\\ 1 \end{pmatrix}}\\ A(v)=\lambda v \Rightarrow A^{-1} A(v)= A^{-1}(\lambda v) \Rightarrow v= \lambda A^{-1}(v) \Rightarrow A^{-1}(v)={1\over \lambda}v \\ \Rightarrow A^{-1}的特徵值為{1\over \lambda}, 特徵向量與A相同 \Rightarrow \bbox[red, 2pt]{A^{-1}的特徵值為1,{1\over 3}，特徵向量為\begin{pmatrix}1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1\\ 1 \end{pmatrix}} \\ (A-4I)v= Av-4v=\lambda v-4v= (\lambda-4)v   \Rightarrow (A-4I)的特徵值為(\lambda-4), 特徵向量與A相同 \\\Rightarrow \bbox[red, 2pt]{A-4I的特徵值為-3,-1，特徵向量為\begin{pmatrix}1\\ 1 \end{pmatrix}, \begin{pmatrix} -1\\ 1 \end{pmatrix}}$$

解答:$$\det(A-\lambda I) = \begin{vmatrix}4-\lambda & -1 & -1 & -1 \\-1 & 4-\lambda & -1 & -1 \\-1 & -1 & 4-\lambda & -1 \\-1 & -1 & -1 & 4-\lambda \end{vmatrix} \xrightarrow{R_1-R_4\to R_1, R_2-R_4\to R_2, R_3-R_4\to R_1} \\\begin{vmatrix}5-\lambda & 0 & 0 & -5+\lambda \\0 & 5-\lambda & 0 & -5+\lambda \\0 & 0 & 5-\lambda & -5+\lambda \\-1 & -1 & -1 & 4-\lambda \end{vmatrix}  =(5-\lambda) \begin{vmatrix} 5-\lambda &0& -5+ \lambda \\ 0& 5-\lambda& -5+\lambda\\ -1&-1&4-\lambda \end{vmatrix} + \begin{vmatrix} 0&0& -5+\lambda\\ 5-\lambda&0& -5+ \lambda\\ 0& 5-\lambda&-5+\lambda\end{vmatrix} \\=(5-\lambda)   (\lambda-5)^2(2-\lambda) +(\lambda-5)^3 =(\lambda-5)^3(\lambda-1) \Rightarrow \lambda=1,5\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}3 & -1 & -1 & -1 \\-1 & 3 & -1 & -1 \\-1 & -1 & 3 & -1 \\ -1 & -1 & -1 & 3\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_4 \\ x_2=x_4\\ x_3=x_4} \\\qquad \Rightarrow v=x_4 \begin{pmatrix} 1\\1 \\ 1\\1\end{pmatrix}, \text{ choose }v_1= \begin{pmatrix} 1\\1 \\ 1\\1\end{pmatrix} \\ \lambda_2= 5\Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-1 & -1 & -1 & -1 \\-1 & -1 & -1 & -1 \\-1 & -1 & -1 & -1 \\ -1 & -1 & -1 & -1\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} =0 \Rightarrow x_1+x_2+ x_3+x_4=0 \\\qquad \Rightarrow v=x_2 \begin{pmatrix} -1\\1 \\ 0\\0\end{pmatrix} +x_3 \begin{pmatrix} -1\\0 \\ 1\\ 0\end{pmatrix} +x_4 \begin{pmatrix} -1\\0 \\ 0\\ 1\end{pmatrix}, \text{ choose }v_2= \begin{pmatrix} -1\\1 \\ 0\\0\end{pmatrix} ,v_3=\begin{pmatrix} -1\\0 \\ 1\\ 0\end{pmatrix}, v_4= \begin{pmatrix} -1\\0 \\ 0\\ 1\end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{特徵值:1,5，特徵向量:\begin{pmatrix} 1\\1 \\ 1\\1\end{pmatrix},  \begin{pmatrix} -1\\1 \\ 0\\0\end{pmatrix} , \begin{pmatrix} -1\\0 \\ 1\\ 0\end{pmatrix},   \begin{pmatrix} -1\\0 \\ 0\\ 1\end{pmatrix}}$$

解答:$$[A\mid I] = \left[ \begin{array}{rrrr|rrrr} 4 & -1 & -1 & -1 & 1 & 0 & 0 & 0\\-1 & 4 & -1 & -1 & 0 & 1 & 0 & 0\\-1 & -1 & 4 & -1 & 0 & 0 & 1 & 0\\-1 & -1 & -1 & 4 & 0 & 0 & 0 & 1 \end{array} \right] \xrightarrow{R_1+3R_2 \to R_1} \left[ \begin{array}{rrrr|rrrr} 1 & 11 & -4 & -4 & 1 & 3 & 0 & 0\\-1 & 4 & -1 & -1 & 0 & 1 & 0 & 0\\-1 & -1 & 4 & -1 & 0 & 0 & 1 & 0\\-1 & -1 & -1 & 4 & 0 & 0 & 0 & 1 \end{array} \right] \\\xrightarrow{R_1+R_2 \to R_2, R_1+R_3 \to R_3, R_1+R_4 \to R_4}  \left[ \begin{array}{rrrr|rrrr} 1 & 11 & -4 & -4 & 1 & 3 & 0 & 0\\0 & 15 & -5 & -5 & 1 & 4 & 0 & 0\\0 & 10 & 0 & -5 & 1 & 3 & 1 & 0\\0 & 10 & -5 & 0 & 1 & 3 & 0 & 1 \end{array} \right] \xrightarrow{R_2/15 \to R_2}  \\\left[ \begin{array}{rrrr|rrrr} 1 & 11 & -4 & -4 & 1 & 3 & 0 & 0\\0 & 1 & - \frac{1}{3} & - \frac{1}{3} & \frac{1}{15} & \frac{4}{15} & 0 & 0\\0 & 10 & 0 & -5 & 1 & 3 & 1 & 0\\0 & 10 & -5 & 0 & 1 & 3 & 0 & 1 \end{array} \right] \xrightarrow{R_1-11R_2 \to R_1, R_3-10R_2\to R_3, R_4-10R_2\to R_4} \\ \left[ \begin{array}{rrrr|rrrr} 1 & 0 & - \frac{1}{3} & - \frac{1}{3} & \frac{4}{15} & \frac{1}{15} & 0 & 0\\0 & 1 &- \frac{1}{3} & - \frac{1}{3} & \frac{1}{15} & \frac{4}{15} & 0 & 0\\0 & 0 & \frac{10}{3} & - \frac{5}{3} & \frac{1}{3} & \frac{1}{3} & 1 & 0\\0 & 0 & - \frac{5}{3} & \frac{10}{3} & \frac{1}{3} & \frac{1}{3} &0 & 1\ \end{array} \right] \xrightarrow{(3/10)R_3 \to R_3}  \left[ \begin{array}{rrrr|rrrr} 1 & 0 & - \frac{1}{3} & - \frac{1}{3} & \frac{4}{15} & \frac{1}{15} & 0 & 0\\0 & 1 &- \frac{1}{3} & - \frac{1}{3} & \frac{1}{15} & \frac{4}{15} & 0 & 0\\0 & 0 & 1 & - \frac{1}{2} & \frac{1}{10} & \frac{1}{10} & \frac{3}{10} & 0\\0 & 0 & - \frac{5}{3} & \frac{10}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 1 \end{array} \right] \\ \xrightarrow{R_1+ R_3/3 \to R_1, R_2+R_3/3 \to R_2, R_4+(5/3)R_3 \to R_4}  \left[ \begin{array}{rrrr|rrrr}1 & 0 & 0 & - \frac{1}{2} & \frac{3}{10} & \frac{1}{10} & \frac{1}{10} & 0\\0 & 1 &  0 & - \frac{1}{2} & \frac{1}{10} & \frac{3}{10} & \frac{1}{10} & 0\\0 & 0 & 1 & - \frac{1}{2} & \frac{1}{10} & \frac{1}{10} & \frac{3}{10} & 0\\0 & 0 & 0 & \frac{5}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1\end{array} \right] \xrightarrow{(2/5)R_4\to R_4} \\ \left[ \begin{array}{rrrr|rrrr}1 & 0 & 0 & - \frac{1}{2} & \frac{3}{10} & \frac{1}{10} & \frac{1}{10} & 0\\0 & 1 & 0 & - \frac{1}{2} & \frac{1}{10} & \frac{3}{10} & \frac{1}{10} & 0\\0 & 0 & 1 & - \frac{1}{2} & \frac{1}{10} & \frac{1}{10} & \frac{3}{10} & 0\\0 & 0 & 0 & 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{2}{5}\end{array} \right] \xrightarrow {R_1 +(1/2)R_4\to R_1, R_2+(1/2)R_4\to R_2,R_3+ (1/2)R_4\to R_3 } \\ \left[ \begin{array}{rrrr|rrrr}1 & 0 & 0 & 0 & \frac{2}{5} & \frac{1}{5} & \frac{1}{5} & {1\over 5}\\0 & 1 & 0 & 0 & \frac{1}{5} & \frac{2}{5} & \frac{1}{5} & {1\over 5}\\0 & 0 & 1 & 0 & \frac{1}{5} & \frac{1}{5} & \frac{2}{5} & {1\over 5}\\0 & 0 & 0 & 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{2}{5}\end{array} \right] \Rightarrow A^{-1} = \bbox[red, 2pt]{\left[\begin{matrix}\frac{2}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}\\\frac{1}{5} & \frac{2}{5} & \frac{1}{5} & \frac{1}{5}\\\frac{1}{5} & \frac{1}{5} & \frac{2}{5} & \frac{1}{5}\\\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{2}{5}
\end{matrix}\right]}$$

解答:$$\lim_{x\to 0}{e^x-1\over x} =\lim_{x\to 0}{e^x\over 1} = \bbox[red, 2pt]1$$

解答:$$L= (\cos(x))^{1/x^2} \Rightarrow \ln L={\ln \cos(x) \over x^2} \Rightarrow \lim_{x\to 0} \ln L= \lim_{x\to 0} {\ln \cos(x) \over x^2} =\lim_{x\to 0}{-\tan(x) \over 2x} \\=\lim_{x\to 0} {-\sec^2 x\over 2} =-{1\over 2} \Rightarrow \lim_{x\to 0}L = \bbox[red, 2pt]{e^{-1/2}}$$

解答:$$\cases{u=\ln(x) \\ dv=x^2\,dx} \Rightarrow \cases{du=dx/x \\ v=x^3/3} \Rightarrow \int x^2\ln (x)\,dx ={1\over 3}x^3\ln(x) - \int{1\over 3}x^2\,dx \\= \bbox[red, 2pt]{{1\over 3}x^3 \ln(x)-{1\over 9}x^3+C}$$

解答:$$\cases{u=e^x \\ dv=\cos(x)\,dx} \Rightarrow  \cases{du=e^x\,dx\\ v= \sin x} \Rightarrow I=\int e^x \cos(x)\,dx =e^x\sin x-\int e^x \sin x\,dx \\ \cases{u=e^x \\dv = \sin x\,dx } \Rightarrow \cases{du=e^x\,dx \\ v=-\cos x} \Rightarrow \int e^x \sin x\,dx =-e^x\cos x+\int e^x \cos x\,dx \\ \Rightarrow I=e^x\sin x- \left( -e^x\cos x+\int e^x \cos x\,dx \right) =e^x\sin x+ e^x\cos -I \Rightarrow 2I=e^x(\sin x+\cos x) \\ \Rightarrow I= \bbox[red, 2pt]{{1\over 2}e^x(\sin x+\cos x)+C}$$

解答:$$u =\sin x \Rightarrow du =\cos x\,dx \Rightarrow \int \sin^5(x) \cos(x)\,dx =\int u^5\,du ={1\over 6}u^6+C = \bbox[red, 2pt]{{1\over 6}\sin^6(x)+C}$$

解答:

$$R= \int_1^2 (x^2-(2-x))\,dx ={11\over 6}, S= \int_2^3 x^2\,dx ={19\over 3} \\ \Rightarrow R+S= \bbox[red, 2pt]{49\over 6}$$

解答:$$u=x-1 \Rightarrow \int_1^5 2\sqrt{x-1}\,dx =\int_0^4 2\sqrt u\,du = \left. \left[ {4\over 3}u^{3/2} \right] \right|_0^4 = \bbox[red, 2pt]{32\over 3}$$

解答:$$I=\int (\sin^2 x\cos x-3\sin x \cos^2 x)\,dx =\int \sin^2 x\cos x\,dx + \int -3\sin x\cos^2x\,dx \\ \cases{u=\sin x\\ v=\cos x}\Rightarrow I=\int u^2\,du +\int 3v^2\,dv = \bbox[red, 2pt]{{1\over 3}u^3+v^3+C={1\over 3}\sin^3x +\cos^3x +C}$$

解答:$$y={2u\over 1-4u} ={2(5x^2+1)^4 \over 1-4(5x^2+1)^4} \Rightarrow y'={8(5x^2+1)^3\cdot10x \over 1-4(5x^2+1)^4} -{2(5x^2+1)^4 \cdot -16(5x^2+1)^3\cdot 10x\over (1-4(5x^2+1)^4)^2} \\ \Rightarrow y'(0)= \bbox[red, 2pt]0$$

解答:$$ rref \left( \begin{bmatrix}1& 1&1\\ 1& 2& 2\\ 1& -1&1 \\ 0& 1& 1 \end{bmatrix} \right) = \begin{bmatrix}1&0& 0\\ 0&1& 0\\ 0& 0 & 1\\ 0&0&0 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{正確}$$

解答:$$A= \begin{bmatrix}2&-5\\ 1& -2 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2+1=0 \Rightarrow \lambda =\pm i\\ \lambda_1= i \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}2-i& -5\\ 1& -2-i \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=(2+i)x_2 \\ \qquad \Rightarrow v= x_2 \begin{pmatrix}2+i\\ 1 \end{pmatrix}, \text{ choose }v_1= \begin{pmatrix}2+i\\ 1 \end{pmatrix} \\ \lambda_2=-i\Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}2+i& -5\\ 1& -2+i \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=(2-i)x_2 \\ \qquad \Rightarrow v= x_2 \begin{pmatrix}2-i\\ 1 \end{pmatrix}, \text{ choose }v_2= \begin{pmatrix}2-i\\ 1 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{特徵值:i, -i, 特徵空間:E_A(i)= span \left(  \begin{bmatrix}2+i\\ 1 \end{bmatrix}\right), E_A(-i) =span\left( \begin{bmatrix}2-i\\ 1 \end{bmatrix} \right)}$$

解答:$$\cases{P(A\to A)= P(A\to B)= P(A\to C) = P(A\to D)=1/4 \\ P(B\to A) =P(B\to B) =1/2\\ P(C\to A) =P(C\to C) =P(C\to D) =1/3\\ P(D\to A) = P(D\to C) = P(D\to D)=1/3} \\ \Rightarrow 轉移矩陣T= \begin{bmatrix}1/4& 1/2 & 1/3 &1/3\\ 1/4& 1/2& 0& 0\\ 1/4& 0&1/3 & 1/3\\ 1/4& 0& 1/3& 1/3 \end{bmatrix} \\= \begin{bmatrix}0 & \frac{4}{3} & \frac{-\sqrt{73}-11}{6} & \frac{\sqrt{73}-11}{6} \\0 & \frac{2}{3} & \frac{\sqrt{73}-1}{6} & \frac{-\sqrt{73}-1}{6} \\-1 & 1 & 1 & 1 \\1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & \frac{-\sqrt{73}+5}{24} & 0 \\0 & 0 & 0 & \frac{\sqrt{73}+5}{24}\end{bmatrix} \begin{bmatrix}0 & 0 & \frac{-1}{2} & \frac{1}{2} \\\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{-5\sqrt{73}-73}{584} & \frac{19\sqrt{73}-73}{584} & \frac{-3\sqrt{73}+73}{584} & \frac{-3\sqrt{73}+73}{584} \\\frac{5\sqrt{73}-73}{584} & \frac{-19\sqrt{73}-73}{584} & \frac{3\sqrt{73}+73}{584} & \frac{3\sqrt{73}+73}{584} \end{bmatrix} \\ \Rightarrow T^\infty =\begin{bmatrix}0 & \frac{4}{3} & \frac{-\sqrt{73}-11}{6} & \frac{\sqrt{73}-11}{6} \\0 & \frac{2}{3} & \frac{\sqrt{73}-1}{6} & \frac{-\sqrt{73}-1}{6} \\-1 & 1 & 1 & 1 \\1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{bmatrix} \begin{bmatrix}0 & 0 & \frac{-1}{2} & \frac{1}{2} \\\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{-5\sqrt{73}-73}{584} & \frac{19\sqrt{73}-73}{584} & \frac{-3\sqrt{73}+73}{584} & \frac{-3\sqrt{73}+73}{584} \\\frac{5\sqrt{73}-73}{584} & \frac{-19\sqrt{73}-73}{584} & \frac{3\sqrt{73}+73}{584} & \frac{3\sqrt{73}+73}{584} \end{bmatrix} \\= \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \end{bmatrix} \Rightarrow \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} 6000\\ 0\\0\\0\end{bmatrix} = \begin{bmatrix} 2000 \\1000 \\1500 \\1500\end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{A鄉鎮2000隻\\ B鄉鎮1000隻\\ C鄉鎮1500隻\\ D鄉鎮1500隻}} \\\bbox[cyan, 2pt]{另解}:穩定狀態: T(v)=v \Rightarrow (T-I)(v)=0 \Rightarrow  \begin{bmatrix}-3/4& 1/2 & 1/3 &1/3\\ 1/4& -1/2& 0& 0\\ 1/4& 0&-2/3 & 1/3\\ 1/4& 0& 1/3& -2/3 \end{bmatrix} \begin{bmatrix}x\\ y\\ z\\w \end{bmatrix} =0 \\ 且x+y+z+w =1 \Rightarrow \cases{y=x/2\\ z=3x/4\\ w=3x/4} \Rightarrow \cases{x=1/3\\ y =1/6\\ z=1/4\\ w=1/4} \Rightarrow \cases{A鄉鎮病原數=6000\cdot {1\over 3} =2000 \\B鄉鎮病原數=6000\cdot {1\over 6} =1000 \\C鄉鎮病原數=6000\cdot {1\over 4} =1500 \\D鄉鎮病原數=6000\cdot {1\over 4} =1500 }$$

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