111年暨南大學電機系-工程數學詳解

 國立暨南國際大學111學年度碩士班入學考試試題

科目:工程數學(線性代數+微分方程)

解答: $$\textbf{(a)}\; k=13 \Rightarrow \det(A)=k-11=2 \ne 0 \Rightarrow \bbox[red,2pt]{A \text{ is nonsigular}} \\\textbf{(b)}\; k=13 \Rightarrow \det(A-\lambda I)=-\lambda^3+15\lambda^2-10\lambda+2=0 \Rightarrow \lambda =\cdots \\\textbf{(c)}\; \\\textbf{(d)}\; \det(A)=k-12=-9 \Rightarrow \bbox[red, 2pt]{k=3} \\\textbf{(e)}\; A=\begin{bmatrix}1 & 0 & 3 \\2 & 1 & 5 \\ 4& 1& 3\end{bmatrix} \Rightarrow A^T= \bbox[red, 2pt]{\begin{bmatrix}1 & 2 & 4 \\0 & 1 & 1 \\ 3& 5& 3\end{bmatrix}} \\\textbf{(f)}\; \det(A)=k-11=0 \Rightarrow k=\bbox[red, 2pt]{11} \\\textbf{(g)} \;A\mathbf x=B \Rightarrow \cases{x_1+3x_3=5 \cdots(1)\\ 2x_1+2x+ 3x_3=8 \cdots(2) \\ 4x_1 +x_2+11x_3=m \cdots(3)} \Rightarrow \cases{x_1=5-3x_3\\ x_2=3x_3-2} \\ \Rightarrow 18+2x_3=m, 取\bbox[red, 2pt]{m＝0}  \Rightarrow \cases{x_1=32 \\ x_2=-29 \\x_3=-9} \\ \mathbf{(h)}\; \bbox[red, 2pt]{ x=\begin{bmatrix}32\\ -29\\ -9 \end{bmatrix}}$$

解答: $$\textbf{(a)}\; \cos x+0\cdot x-\cos x=0 \Rightarrow (1,0,-1) \ne (0,0,0)  \Rightarrow \bbox[red, 2pt]{\text{linear dependent}} \\\textbf{(b)}\; f(x)= a(x^2-2x+5)+b(x^2-4x+10) \Rightarrow f'(x)=a(2x-2)+b(2x-4)\\\quad \cases{f(0)=0\\ f'(0)=0} \Rightarrow \cases{5a+10b=0\\ -2a-4b=0} \Rightarrow a=-2b \Rightarrow  \bbox[red, 2pt]{\text{linear dependent}}  \\\textbf{(c)}\; \cos(2x)=\cos^2x -\sin^2 x \Rightarrow (-1) \sin^2x +\cos^2x -\cos(2x)=0 \Rightarrow (-1,1,-1)\ne (0,0,0) \\\quad \Rightarrow \bbox[red, 2pt]{\text{linear dependent}}\\ \textbf{(d)}\; f(x)= a(x^2-2x+5)+ b(x^2-5x+10) +cx^2 \Rightarrow f'(x)=a(2x-2)+ b(2x-5)+2cx \\\quad \Rightarrow f''(x) = 2a+2b+2c \Rightarrow \cases{f(0)=0\\ f'(b)=0\\ f''(0)=0} \Rightarrow \cases{5a+10b=0 \\-2a-5b=0 \\ a+b+c=0} \Rightarrow \cases{a=0\\ b=0\\ c=0}\Rightarrow \bbox[red, 2pt]{\text{linear independent}} \\ \textbf{(e)}\; \sin^2x+\cos^2 x=1 \Rightarrow 0\cdot \sin x+\sin^2 x+\cos^2x -1=0 \\\quad \Rightarrow (0,1,1,-1) \ne (0,0,0,0) \Rightarrow  \bbox[red, 2pt]{\text{linear dependent}}  \\ \textbf{(e)}\; \sum_{k=1}^\infty a_ke^{kx} =0 \Rightarrow a_1+  \sum_{k=2}^\infty a_ke^{(k-1)x} =0 \Rightarrow \lim_{x\to -\infty} \left( a_1+  \sum_{k=2}^\infty a_ke^{(k-1)x}\right) =0 \Rightarrow a_1=0 \\ \quad \Rightarrow \sum_{k=2}^\infty a_ke^{kx} =0 \Rightarrow a_2+ \sum_{k=3}^\infty a_ke^{(k-2)x} =0 \Rightarrow \lim_{x\to -\infty} \left( a_2+  \sum_{k=3}^\infty a_ke^{(k-2)x}\right) =0 \Rightarrow a_2=0\\ \quad \Rightarrow \cdots \Rightarrow  a_k=0, k=1,2,\dots \Rightarrow  \bbox[red, 2pt]{\text{linear independent}}  $$

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