國立臺北科技大學115學年度碩士班招生考試
系所組別:自動化科技研究所 科目:工程數學
解答:$$\text{Given }\cases{\lambda_1=1, \lambda_2=4, \lambda_3=9\\ v_1=[1/\sqrt 2, 1/\sqrt 2,0]^T\\v_2=[1/\sqrt 2, -1/\sqrt 2,0]^T \\ v_3=[0,0,1]^T} \Rightarrow \cases{D= \begin{bmatrix}\lambda_1& 0& 0\\ 0& \lambda_2 & 0\\ 0&0& \lambda_3\end{bmatrix} \\Q=[v_1\; v_2\;v_3]} \\ \Rightarrow A= QDQ^T = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 9 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5/2 & -3/2 & 0 \\ -3/2 & 5/2 & 0 \\ 0 & 0 & 9 \end{bmatrix} \\ \Rightarrow \cases{a= b=5/2\\ c=9 \\d =e=0 \\ f=-3/2} \Rightarrow \text{The true statements are }\bbox[red, 2pt]{(AC)}.$$
解答:$$(A) \times:Q \text{ is orthogonal projection matrix} \Rightarrow Q^2=Q \text{ and } rank(Q)=tr(Q)\\\qquad Q^2=(I-uu^T)(1-uu^T) =I-2uu^T+ uu^Tuu^T =I-uu^T \Rightarrow u^Tu=1 \\ \qquad \Rightarrow rank(Q)=tr(Q)=tr(I-uu^T) =tr(I)-tr(uu^T)=tr(I)-tr(u^Tu)=4-1=3 \\ \qquad \Rightarrow \text{eigenvalues: }1,1,1,0 \Rightarrow \det(Q)=1\times1\times 1\times 0=0 \Rightarrow \det(Q)+ rank(Q)=0+3=3\ne 4 \\(B)\bigcirc: \begin{vmatrix} 1+\alpha_1& \alpha_2& \alpha_3 & \cdots& \alpha_n\\ \alpha_1& 1+\alpha_2& \alpha_3 & \cdots& \alpha_n \\\alpha_1& \alpha_2& 1+\alpha_3& \cdots& \alpha_n \\ \vdots& \vdots & \cdots& \cdots& \vdots\\ \alpha_1& \alpha_2& \alpha_3& \cdots& 1+\alpha_n\end{vmatrix} \xrightarrow{R_1-R_n\to R_1, R_2-R_n\to R_2, \dots, R_{n-1}-R_n\to R_{n-1}} \\ \begin{vmatrix} 1 & 0& 0 & \cdots& -1\\ 0& 1 & 0 & \cdots& -1 \\ 0& 0& 1 & \cdots& -1 \\ \vdots& \vdots & \cdots& \ddots & \vdots \\ \alpha_1& \alpha_2& \alpha_3& \cdots& 1+\alpha_n\end{vmatrix} \xrightarrow{R_n-\alpha_1R_1\to R_n} \begin{vmatrix} 1 & 0& 0 & \cdots& -1\\ 0& 1 & 0 & \cdots& -1 \\ 0& 0& 1 & \cdots& -1 \\ \vdots& \vdots & \cdots& \ddots & \vdots \\ 0& \alpha_2& \alpha_3& \cdots& 1+\alpha_1+ \alpha_n\end{vmatrix} \\ = \begin{vmatrix} 1 & 0 & \cdots& -1 \\ 0& 1 & \cdots& -1 \\ \vdots & \cdots& \ddots & \vdots \\ \alpha_2& \alpha_3& \cdots& 1+\alpha_1+ \alpha_n\end{vmatrix} \xrightarrow{R_{n-1}-\alpha_2 R_1\to R_{n-1}}\begin{vmatrix} 1 & 0 & \cdots& -1 \\ 0& 1 & \cdots& -1 \\ \vdots & \cdots& \ddots & \vdots \\ 0& \alpha_3& \cdots& 1+\alpha_1+ \alpha_2+ \alpha_n\end{vmatrix} \\= \begin{vmatrix} 1 &0& \cdots& -1 \\ \vdots & \cdots& \ddots & \vdots \\ \alpha_3& \alpha_4& \cdots& 1+\alpha_1+ \alpha_2+ \alpha_n\end{vmatrix} = \cdots = \begin{vmatrix} 1& -1\\ \alpha_{n-1} & 1+\alpha_1+\alpha_2+ \cdots+ \alpha_{n-2}+\alpha_n \end{vmatrix} \\=1+ \sum_{k=1}^n a_k \\(C)\bigcirc: Ax=c \Rightarrow \det(A+cd^T) =\det(A+Axd^T) = \det(A(I+xd^T)) =\det(A)\cdot \det(I+xd^T) \\\quad =\det(A)\cdot (1+d^Tx) \quad \text{by Sylvester's determinant theorem} \\(D)\times: A=A^H \Rightarrow \text{all eigenvalues of }A \text{ are real numbers} \Rightarrow \det(A) \text{ must be real}\\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(BC)}$$
解答:$$y'+2x^3y=0 \Rightarrow {1\over y}dy = -2x^3\,dx \Rightarrow \int{1\over y}\,dy= \int -2x^3\,dx \Rightarrow \ln|y|=-{1\over 2}x^4+c_1 \\ \Rightarrow y=c_2e^{-x^4/2} \Rightarrow y(0)=c_2=1 \Rightarrow \bbox[red, 2pt]{y=e^{-x^4/2}}$$
解答:$$Y(s) ={s^3\over s^4+4} ={s^3\over (s^2-2s+2)(s^2+2s+2)} ={1\over 2} \left( {s-1\over s^2-2s+2}+{s+1\over s^2+2s+2} \right) \\ ={1\over 2} \left( {s-1\over (s-1)^2+ 1}+{s+1\over (s+1)^2+ 1} \right) \Rightarrow y(t)=L^{-1}\{Y(s)\} ={1\over 2} \left( e^t\cos t+e^{-t}\cos t \right) \\= \left( e^t+e^{-t}\over 2 \right)\cos t =\cosh t\cos t \Rightarrow \bbox[red, 2pt]{y(t)=\cosh (t)\cos (t)}$$
解答:$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 6 \\ 0 & 0 & -7 \end{bmatrix} \Rightarrow \text{eigenvalues: } \lambda_1 = 1, \lambda_2 = 5, \lambda_3 = -7\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 0 & 2 & 3 \\ 0 & 4 & 6 \\ 0 & 0 & -8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow y=z=0 \\\qquad \Rightarrow v= x \begin{bmatrix}1\\0 \\0 \end{bmatrix} \Rightarrow \text{choosing }v_1= \begin{bmatrix}1\\0\\0 \end{bmatrix} \\ \lambda_2=5 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -4 & 2 & 3 \\ 0 & 0 & 6 \\ 0 & 0 & -12 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{y=2x\\ z=0} \\ \qquad \Rightarrow v =x \begin{bmatrix}1\\ 2\\ 0 \end{bmatrix} \Rightarrow \text{ choosing }v_2= \begin{bmatrix}1\\2\\0 \end{bmatrix} \\ \lambda_3=-7 \Rightarrow (A-\lambda_3 I) v=0 \Rightarrow \begin{bmatrix} 8 & 2 & 3 \\ 0 & 12 & 6 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{y=2x\\ z=-2y} \\ \qquad \Rightarrow v= y \begin{bmatrix}1/2\\ 1\\-2 \end{bmatrix} \Rightarrow \text{ choosing }v_3= \begin{bmatrix}1\\2\\-4 \end{bmatrix} \\ \bbox[red, 2pt]{\text{eigenvalues: 1,5, -7 and eigenvectors: }\begin{bmatrix}1\\0\\0 \end{bmatrix}, \begin{bmatrix}1\\2\\0 \end{bmatrix}, \begin{bmatrix}1\\2\\-4 \end{bmatrix}}$$
解答:$$A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I) = -\lambda(\lambda-2)(\lambda+1)=0 \Rightarrow \lambda=0,2, -1 \\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow v_1= \begin{bmatrix}1\\-1\\0 \end{bmatrix} \\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow v_2= \begin{bmatrix}1\\1\\2 \end{bmatrix} \\ \lambda_3=-1 \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow v_3= \begin{bmatrix}1\\1\\ -1 \end{bmatrix} \\ \Rightarrow Q= [v_1\; v_2\; v_3] = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 1 \\ 0 & 2 & -1 \end{bmatrix}, D = \begin{bmatrix}\lambda_1& 0& 0\\0& \lambda_2& 0\\ 0&0& \lambda_3 \end{bmatrix}= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{bmatrix} \\ \Rightarrow A=QDQ^{-1} \Rightarrow A^{100}=QD^{100}Q^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 1 \\ 0 & 2 & -1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 2^{100} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1/2& -1/2&0 \\ 1/6& 1/6& 1/3\\ 1/3& 1/3& -1/3 \end{bmatrix} \\= \begin{bmatrix} 0 & 2^{100} & 1 \\ 0 & 2^{100} & 1 \\ 0 & 2 \cdot 2^{100} & -1 \end{bmatrix} \begin{bmatrix}1/2& -1/2&0 \\ 1/6& 1/6& 1/3\\ 1/3& 1/3& -1/3 \end{bmatrix} = \bbox[red, 2pt]{\frac{1}{3} \begin{bmatrix} 2^{99} + 1 & 2^{99} + 1 & 2^{100} - 1 \\ 2^{99} + 1 & 2^{99} + 1 & 2^{100} - 1 \\ 2^{100} - 1 & 2^{100} - 1 & 2^{101} + 1 \end{bmatrix} }$$
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解題僅供參考，碩士班歷年試題及詳解
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